Assignment – I
MTH611A – 2020-21
1. Let R and A be rings and φ : R −→ A be a ring homomorphism. Define φe : R[x] −→ A[x]
Pn
Pn
i
i
e
by φe
i=0 ai x =
i=0 φ(ai )x . Show that φ is a ring homomorphism.
2. (a) Let I be an ideal of R. Put J = {f (x) ∈ R[x] | all coefficients of f (x) are in I}.
Show that J is an ideal of R[x], where R[x] is the polynomial ring in one variable x
over R. We denote this ideal as I[x].
(b) Show that if I is prime then so is I[x].
(c) Let A = R/I. Show that R[x]/I[x] ∼
= A[x] as rings.
3. Let a, b ∈ R such that a is a unit and b is nilpotent. Show that a + b is a unit.
Pn
i
4. Let f =
i=0 ai x ∈ R[x]. Then f is said to be primitive if (a0 , . . . , an ) = R . Let
f, g ∈ R[x]. Show that f and g are primitive iff f g is primitive.
Pn
i
5. Let f =
i=0 ai x ∈ R[x]. Show that f is a unit iff a0 is a unit and a1 , . . . , an are
nilpotents.
6. Show that nil(R[x]) = J(R[x]).
7. An element e ∈ R is called an idempotent if e2 = e. If in R every ideal which is not
contained in the nil(R) contains a nonzero idempotent then show that J(R) = nil(R).
8. A ring R has the property that for every element x ∈ R there exists some n > 1 such that
xn = x. Show that every prime ideal is maximal in R.
9. Show that the following statements are equivalent:
(a) R has exactly one prime ideal.
(b) Every element of R is either a unit or a nilpotent.
(c) R/nil(R) is a field.
10. Let e1 , . . . , en ∈ R be idempotents and let x ∈ R. Show that (a) There exists some
idempotent e ∈ R such that (e1 , . . . , en ) = (e). (b) (e1 , . . . , en , x) = (y) for some y ∈ R.
11. Show that if I is an ideal of R which is free as an R-module then I is principal.
12. Let I be an ideal of R. If I is a direct summand of R then show that I is generated by an
idempotent.
13. Let I be an ideal of R generated by an idempotent. Show that I is a direct summand of
R. Hence I is projective as an R-module. Show that if R contains an idempotent e such
that e 6= 0, 1 then I = (e) is a projective R-module which is not free.
14. Let I be the ideal generated by 6 in Z. Then e = 3 + I is an idempotent of R = Z/I and
e 6= 0, 1.
15. Let R = Z[x]. Show that the ideal (n, x) is not principal, where n ≥ 2 is an integer.
16. Let P be a projective module and let M be an R-module. Let f : M −→ P be an onto
R-linear map. Show that there exists a submodule N of M such that M = N + ker(f ),
N ∩ ker(f ) = 0 and P ∼
= N.
17. Let I be an ideal of R such that R/I is projective. Show that I is a direct summand of R.
18. Show that Q is not projective over Z.
19. An R-module M is called torsion-free over R if every nonzero divisor of R is also a nonzero
divisor of M (element a ∈ R is said to be a nonzero divisor of M if ax 6= 0 for every nonzero
element x of M ). Show that if M is flat then M is also torsion-free.
20. Let I 6= R be an ideal of R and let F = {P ⊆ R|P is a prime ideal of R and I ⊆ P }.
Show that F contains minimal elements.
21. Let A be an R-algebra and let M and N be an R-modules. Let f : M −→ N be an R-linear
map. Show that
(a) fe: A ⊗R M −→ A ⊗R N is A-linear.
(b) if M is generated by S = {xλ | λ ∈ Λ} as an R-module then A ⊗R M is generated by
S 0 = {1 ⊗ xλ | λ ∈ Λ} as an A-module.
(c) if M is finitely generated as an R-module then A ⊗R M is finitely generated as an
A-module.
22. Let A be an R-algebra and let M be a free R-module. Show that A ⊗R M is a free
A-module.
23. Let A be an R-algebra and let M be a projective R-module. Show that A ⊗R M is a
projective A-module.
2
Assignment – II
MTH611A – 2020-21
1. Let I and J be ideals of R and let M be an R-module. Show that J (M/IM ) = (JM +
IM )/IM .
2. (a) Let M and N be R-modules and let I = ann(M ) and J = ann(N ). If I + J = then
show that M ⊗ N = 0.
(b) Let I and J be ideals of R such that I + J = R. Show that (R/I) ⊗ (R/J) = 0.
(c) Let m, n be relatively prime integers. Let I = (m) and J = (n). Show that (Z/I) ⊗
(Z/J) = 0.
3. Let A be an R-algebra and let M and N be A-modules. Let f : M −→ N be an A-linear
map. Treat M and N as R-modules by restriction of scalars. Show that (a) f is R-linear.
(b) every A-submodule of M is also an R-submodule of M .
4. Let A be an R-algebra, M be an R-module and L, K be A-modules. Show that M ⊗R
(L ⊗A K) is naturally isomorphic to (M ⊗R L) ⊗A K as an A-module.
5. Let A be an R-algebra. Let M be a flat R-module. Show that A ⊗R M is flat as an
A-module.
6. Let {Mλ | λ ∈ Λ} be a family of R-modules and N be an R-module. Show that (⊕λ∈Λ Mλ )⊗
N is isomorphic to ⊕λ∈Λ (Mλ ⊗ N ).
7. Let {Mλ | λ ∈ Λ} be a family of R-modules. Show that ⊕λ∈Λ Mλ is flat iff each Mλ is flat.
(atiyah book page-31 chapt-2 question 4)
8. Let A be an R-algebra, M and N be A-modules. Show that there exists a natural R-linear
map φ : M ⊗R N −→ M ⊗A N . Furthermore, φ is onto. Also show that M is isomorphic
to a direct summand of A ⊗R M .
9. Let R be a local ring and let M, N be finitely generated R-modules. If M ⊗R N = 0 then
show that either M = 0 or N = 0. (atiyah book page-31 chapt-2 question 2)
10. Let A be an R-algebra which is flat as an R-module. Let N be a flat A-module. Show
that N is a flat R-module.
11. Let M and N be flat R-modules. Show that so is M ⊗ N .
12. Let f : M −→ N be an R-linear map. Let I be an ideal of R. Show that
(a) f induces an R-linear map f¯: M/IM −→ N/IN .
(b) if I is contained in J(R), f is onto and N is finitely generated then f is also onto.
13. Let m, n ∈ N and let f : Rm −→ Rn be R-linear. Show that
(a) if f is onto then n ≤ m.
(b) if f is an isomorphism then n = m.
(c) if f is one then m ≤ n.
f
g
14. Let M −→ N −→ P −→ 0 be an exact sequence of R-modules. Show that if M and P are
fg (finitely generated) then so is N . Also if N is fg then so is P .
15. Let f : R −→ A be a ring homomorphism and let S be a multiplicative set in R. Show
that T = f (S) is a multiplicative set in A. Let M be an A-module. Show that S −1 M and
T −1 M are isomorphic as S −1 R-modules.
16. Let S be a multiplicative set in R and let M be a fg R-module. Then S −1 M = 0 iff there
exists some s ∈ S such that sM = 0.
17. Let M be an R-module and
n x let S be
o a multiplicative set in R. Let T be a set of generators
0
of M . Show that T =
| x ∈ S is a set of generators of S −1 M as an S −1 R-module.
1
Hence if M is fg as an R-module then S −1 M as fg as an S −1 R-module.
18. Let I be an ideal and let S = 1 + I = {1 + a | a ∈ I}. Show that S −1 I ⊆ J(S −1 R). In
particular, if M is fg and M = IM then there exist a ∈ I such that (1 + a)M = 0.
19. Let S be a multiplicative set in R. Let M be an R-module. Show that every S −1 Rsubmodule of S −1 M is of the form S −1 N for some R-submodule N of M .
20. Let S be a multiplicative set in R. Let M be an R-module.
(a) Let N1 , . . . , Nk be submodules of M . Then S −1 ∩ki=1 Ni = ∩ki=1 S −1 Ni .
∼ ⊕λ∈Λ S −1 Mλ .
(b) Let {Mλ | λ ∈ Λ} be a family of R-modules. Then S −1 (⊕λ∈Λ Mλ ) =
√ √
(c) Let I be an ideal of R. Then S −1
I = S −1 I.
(d) S −1 (nil(R)) = nil(S −1 R).
(e) If M is fg then S −1 (annR (M )) = annS −1 R (M ).
2
Assignment – 3
MTH611A – 2020-21
1. Let I and J be ideal of R and let S be a multiplicative set in R. If J is fg then show that
(S −1 I : S −1 J) = S −1 (I : J).
2. Let S be a multiplicative set in R such that 0 6∈ S. Then there is a prime ideal P of R
such that P ∩ S = ∅.
3. Let F = {S | S is a multiplicative set in R and 0 6∈ S}. Show that F has maximal elements. Further show that S is maximal in F iff S = R \ P for some minimal prime P of
R.
4. Let S be a multiplicative set. Then S is called saturated if the following condition is
satisfied: For x, y ∈ R, xy ∈ S iff x, y ∈ S. Show that S is saturated iff R \ S is a union
of prime ideals.
5. Let S be a multiplicative set. Show that there exists a smallest saturated multiplicative
set S containing S. Show that R \ S equals the union of all prime ideal which do not
intersect with S. (This S is called the saturation of S.)
6. Show that S −1 R and S
−1
R are isomorphic as rings, where S is saturation of S.
7. Show that the set D of all zero-divisors of R is a union of prime ideals. Show that every
minimal prime of R is contained in D.
8. Let I1 , . . . , In be ideals of R and let S be a multiplicative set of R. Show that S −1 (I1 . . . In ) =
S −1 I1 . . . S −1 In .
9. Show that Q ⊗Z Z/nZ = 0 for all n ∈ N.
10. Show
that
Q
is
Q
Q ⊗Z
Z/nZ
=
6
0.
n∈N
flat
as
a
Z-module.
Also,
show
that
11. Let R be an integral domain and let S be a multiplicative set in R such that 0 6∈ S. Show
that
(a) The natural map : R −→ S −1 R is one-one.
(b) S −1 R is an integral domain.
12. Let R be an integral domain with the quotient field K. Let S be a multiplicative set in R
such that 0 6∈ S. Show that there exists a one-one ring homomorphism from S −1 R to K.
Further show that the quotient field of S −1 R is isomorphic to K.
n
13. Let R be an integral domain and let a ∈ R \ 0 be a non-unit such that ∩∞
n=1 (a ) = 0 (for
example, 2 in Z). Put In = (an ). Let S be a multiplicative set in R such that a ∈ S and
−1
−1
−1 (∩∞ I ) = 0.
0 6∈ S. Show that ∩∞
n=1 S In = S R but S
n=1 n
f
g
14. Let 0 −→ K −→ M −→ L −→ 0 be an exact sequence of R-modules. Let g1 : L −→ M be
an R-linear map such that g og1 = IL . Show that there exists an R-linear map f1 : M −→ K
such that f1 of = IK . Furthermore, if M is fg then so are K and L.
15. Let R be a local ring and P be a fg projective R-module. Show that P is free.
16. Let I be an ideal of R and let M be an R-module. Let X = {m ⊆ R |
m is a maximal ideal of R and I ⊆ m}. If Mm = 0 for all m ∈ X then show that M =
IM .
2
Assignment – 4
MTH612A – 2020-21
1. Let R be a ring and let S ⊆ R. Let F = {A ⊆ R | A is a subring of R and S ⊆ A}. Show
that F has a unique minimal element. (This element is called the subring generated by S
and denoted by RS .)
2. Let R be a ring and let A be a subring of R. Let S ⊆ R. Then the subring generated by
A ∪ S is also called the subring generated by S over A and it is denoted by A[S]. Show
that A[S ∪ T ] = (A[S])[T ] for all S, T ⊆ R. If S = {b1 , . . . , bn } then find the structure of
A[S]. (We denote A[S] by A[b1 , . . . , bn ] in this case.)
3. Let A ⊆ B be a ring extension. Let b1 , . . . , bn ∈ B be integral over A. Show that
A[b1 , . . . , bn ] is finitely generated as an A-module.
4. Let A be a ring and let B, C and D be A-algebras. Let f : B −→ C be a ring homomorphism which is also A-linear. If C is integral over B then show that so is C ⊗A D over
B ⊗A D.
5. Let A be a ring and let I = nil(A). Show that the integral closure of A in A[x] is A + I[x].
6. Let f ∈ A[x] be monic. Show that A[x] is integral over A[f ].
7. Let A ⊆ B be integral. Let a ∈ A be a unit in B. Show that a is also a unit in A. Hence,
show that J(A) = J(B) ∩ A.
Q
8. Let M1 , . . . , Mn be finitely generated R-modules. Show that ni=1 Mi is a finitely generated
R-module.
9. An A algebra B is called an integral A-algebra if B is integral over A. Let B1 , . . . , Bn be
Q
integral A-algebras. Show that ni=1 Bi is also an integral A-algebra.
10. Let A ⊆ B be an extension of rings such that B \ A is closed under multiplication. Show
that A is integrally closed in B.
11. Let f ∈ A[x] be a monic polynomial of degree ≥ 1 and let B = A[x]/(f ). Show that the
natural map from A to B is a one-one ring homomorphism and that B is integral over A.
12. Let A be a subring of an integral domain B. Let C be the integral closure of A in B. Let
f, g ∈ B[x] be monic polynomials such that f g ∈ C[x]. Show that f, g ∈ C[x].
13. Prove the previous results without assuming that B is an integral domain.
14. Let A ⊆ B be an extension of rings and let C be the integral closure of A in B. Show that
the integral closure of A[x] in B[x] is C[x].
15. Show that Z is integrally closed in Q.
16. Let (M, +) be an abelian group (and therefore, a Z-module). Let R be a ring and let
φ : R −→ HomZ (M, M ) be a group homomorphism such that φ(1) = IM and φ(ab) =
φ(a)oφ(b) for all a, b ∈ R. Show that M becomes an R-module via φ.
17. Let (R, +) be an abelian group (and therefore, a Z-module). Let φ : R −→ HomZ (R, R)
be a group homomorphism. Put φa = φ(a) for all a ∈ R. The map φ satisfies the following
conditions
(1) There exists an e ∈ R such that φe = IR
(2) φa oφb = φφa (b) for all a, b ∈ R.
(3) φa (b) = φb (a) for all a, b ∈ R.
Show that R becomes a commutative ring with unity via φ.
2
Assignment – 4
MTH612A – 2020-21
1. Let R be a ring and let S ⊆ R. Let F = {A ⊆ R | A is a subring of R and S ⊆ A}. Show
that F has a unique minimal element. (This element is called the subring generated by S
and denoted by RS .)
2. Let R be a ring and let A be a subring of R. Let S ⊆ R. Then the subring generated by
A ∪ S is also called the subring generated by S over A and it is denoted by A[S]. Show
that A[S ∪ T ] = (A[S])[T ] for all S, T ⊆ R. If S = {b1 , . . . , bn } then find the structure of
A[S]. (We denote A[S] by A[b1 , . . . , bn ] in this case.)
3. Let A ⊆ B be a ring extension. Let b1 , . . . , bn ∈ B be integral over A. Show that
A[b1 , . . . , bn ] is finitely generated as an A-module.
4. Let A be a ring and let B, C and D be A-algebras. Let f : B −→ C be a ring homomorphism which is also A-linear. If C is integral over B then show that so is C ⊗A D over
B ⊗A D.
5. Let A be a ring and let I = nil(A). Show that the integral closure of A in A[x] is A + I[x].
6. Let f ∈ A[x] be monic. Show that A[x] is integral over A[f ].
7. Let A ⊆ B be integral. Let a ∈ A be a unit in B. Show that a is also a unit in A. Hence,
show that J(A) = J(B) ∩ A.
Q
8. Let M1 , . . . , Mn be finitely generated R-modules. Show that ni=1 Mi is a finitely generated
R-module.
9. An A algebra B is called an integral A-algebra if B is integral over A. Let B1 , . . . , Bn be
Q
integral A-algebras. Show that ni=1 Bi is also an integral A-algebra.
10. Let A ⊆ B be an extension of rings such that B \ A is closed under multiplication. Show
that A is integrally closed in B.
11. Let f ∈ A[x] be a monic polynomial of degree ≥ 1 and let B = A[x]/(f ). Show that the
natural map from A to B is a one-one ring homomorphism and that B is integral over A.
12. Let A be a subring of an integral domain B. Let C be the integral closure of A in B. Let
f, g ∈ B[x] be monic polynomials such that f g ∈ C[x]. Show that f, g ∈ C[x].
13. Prove the previous results without assuming that B is an integral domain.
14. Let A ⊆ B be an extension of rings and let C be the integral closure of A in B. Show that
the integral closure of A[x] in B[x] is C[x].
2
Lecture – 01
1. Some elementary facts
In this course we shall learn about commutative rings with unity and
some related objects.
We start with recalling some definitions and elementary results.
Let (R, +, ·) be a ring. Then
(1) The symbol + is called addition and · is called multiplication.
(2) (R, +) is an abelian group.
(3) · is associative and distributes over +.
(4) Identity for + is denoted is denoted by 0.
(5) The inverse of a ∈ R with respect to + is denoted by −a.
(6) For a, b ∈ R, we write ab in stead of a · b.
(7) In general, R may not have identity for multiplication.
(8) If R has identity for multiplication, then it is unique. It is called
unity (or multiplicative identity) and denoted by 1.
(9) If R has identity for multiplication, then R is called a ring with
unity.
(10) If · is commutative then R is called a commutative ring.
(11) If R is a ring with unity such that 1 = 0 then R is singleton.
(12) For a ∈ R and n ∈ N, we denote a . . . a(n times) by an. If R has
unity then (in this course) a0 is defined to be 1 for all a ∈ R, whether
a = 0 or not.
Please brush up the follwing definitions: ideals, maximal ideals, zero
divisors, quotient rings, product of rings, integral domains, fields, etc.
Let R be a ring.
We prove a very elementary lemma now.
Lemma 1 (a) a.0 = 0 for all a ∈ R.
(b) (−a)b = −(ab) = a(−b) for all a, b ∈ R.
(c) 1 6= 0 in if R has at least two elements.
Proof.
(a) a.0 = a.(0 + 0) = a.0 + a.0 =⇒ a.0 = 0.
(b) (−a)b + ab = (−a + a)b = 0.b = 0 by (a).
Hence the additive inverse of ab is (−a)b.
Notationally, this means −(ab) = (−a)b.
Similarly, a(−b) = −(ab).
(c) Suppose, if possible, that 1 = 0.
This means the additive identity of R is same as the multiplicative
identity of R.
We shall now show that R is singleton.
Let a ∈ R. Then a = a.1 = a.0 = 0. Hence R contains only one
element, namely, 0. A contradiction as R has at least 2 elements. In view of above lemma, the set {0} is an ideal of R.
Furthermore, R is always an ideal of itself.
An ideal is called proper if it is not equal to R.
In particular, {0} is a proper ideal of R.
Recall that the intersection of any nonempty collection of ideals of R
is again an ideal of R.
There are many examples of commutative rings with unity like Z, R,
C. However, there are many rings which are not commutative and
many do not have unity.
(1) The ring of all n × n matrices over any field is noncommutative
for n ≥ 2. However, they do have unity.
(2) The set of all even integers is a commutative ring under usual
addition and multiplication of integers. However, it has no unity.
(3) The set of all n × n upper triangular matrices with all diagonal
entries 0 over any field is noncommutative and has no unity for n ≥ 3
for usual addition and multiplication of matrices.
Henceforth, we shall assume that all rings considered are commutative with unity having at least 2 elements. So that 1 6= 0.
By a ring we shall always mean a commutative ring with unity.
2. Ideal generated by a set
Comments. If we take infinitely many elements of R then their sum,
in general, can not be defined because it can not be computed. However, one may ask a question that in analysis we consider summation
of series, say
P∞
n=1 an , where an ∈ R. One should be clear here that
this notation is not for the sum of an’s but it represents the limit of
a particular sequence, namely, the limit of { n
i=1 ai }, if it converges.
This means, this sum makes sense with respect to the topology of R.
P
Not all series in R converge and therefore, even this sum makes sense
only for a certain class of series. Even here, the sum of uncountably
many elements is not defined. Since a ring is considered without any
reference to a topology on it, the sum of infinitely many elements of
the ring does not make sense. However, if we have infinitely elements,
say, {aλ | λ ∈ Λ} of R, such that all but finitely many of aλ are zero,
P
then the sum λ∈Λ aλ is assigned the meaning in the following way:
first throw away all zero terms in the summation, then whatever is
left is a finite sum, which can be computed. The sum
P
λ∈Λ aλ is
taken as the sum of these finitely many nonzero elements. If all aλ’s
P
are 0 then the sum λ∈Λ aλ is defined to be 0.
Definition.
Consider a subset {aλ | λ ∈ Λ} of R, where Λ is an
indexing set. By a linear combination of aλ’s we mean an element of
P
the type λ∈Λ bλaλ, where bλ’s are elements of R and they are zero
for all but for finitely values of λ. This means, all but finitely many
terms of
P
λ∈Λ bλ aλ are zero and therefore, it is well defined. We call
bλ as the coefficient of aλ in this linear combination.
Let S be a subset of R, may even be empty.
Let
F = {I ⊆ R | I is an ideal of R, S ⊆ I}.
Note that R ∈ F , and hence F is nonempty.
Hence, the intersection of all members of F is an ideal of R containing
S.
In this course we shall denote this ideal by I(S).
Note that I(S) is also an element of F and is contained in every
member of F .
In other words, if an ideal of R contains S then it also contains the
ideal I(S).
In this sense I(S) is the smallest ideal of R containing S.
The ideal I(S) is called the ideal generated by S.
If S is a finite set, say, {a, . . . , an}, we use the notation (a1, . . . , an) for
I(S).
We now look at some special cases of S.
(1) Consider the case when S = ∅. Then F contains all the ideals of
R. Hence, the ideal {0} as well. Therefore, I(S) = {0} in this case.
(2) Consider the case when S itself is an ideal. Then S ∈ F and hence
I(S) ⊆ S ⊆ I(S). That is, I(S) = S. In particular the ideal generated
by the set {0} is {0} and so the notation (0) also represents the ideal
{0}.
(3) Consider when S ⊆ T . As S ⊆ T ⊆ I(T ) we have I(S) ⊆ I(T ).
(4) We now determine the structure of I(S) when S is finite, that is,
S = {a1, . . . , an}. Put
J = {b1a1 + · · · + bnan | b1, . . . , bn ∈ R are arbitrary}.
Note that ai = 0a1 + · · · + 0ai−1 + 1ai + 0ai+1 + · · · + 0an.
=⇒ ai ∈ J for all i = 1, . . . , n.
=⇒ S ⊆ J.
We claim that J is an ideal of R. Let x, y ∈ J and c ∈ R.
Then there exist b1, . . . , bn, c1, . . . , cn ∈ R such that x = b1a1 +· · ·+bnan
and y = c1a1 + · · · + cnan.
=⇒ x − y = (b1 − c1)a1 + · · · + (bn − cn)an ∈ J and
cx = cb1a1 + · · · + cbnan ∈ J.
Since x, y, c are arbitrary, J is an ideal of R.
=⇒ I(S) ⊆ J as S ⊆ J.
Also a1, . . . , an ∈ S ⊆ I(S)
=⇒ x = b1a1 + · · · + bnan ∈ I(S) (as I(S) is an ideal).
This means J ⊆ I(S). Hence J = I(S).
(5) Consider now the case when S is infinite. Let
X = {T ⊆ S | T is finite}.
We shall show that I(S) = ∪T ∈X I(T ). Put J = ∪T ∈X I(T ).
Let x ∈ S. Then {x} ∈ X.
=⇒ x ∈ I({x}) ⊆ J.
=⇒ S ⊆ J.
We now show that J is an ideal of R.
Let x, y ∈ J and c ∈ R.
=⇒ There exist finite subsets T1 and T2 of S such that x ∈ I(T1) and
y ∈ I(T2).
=⇒ x, y ∈ I(T1 ∪ T2).
As T1 ∪ T2 is a finite subset of S, we have x − y ∈ I(T1 ∪ T2) ⊆ J and
cx ∈ I(T1) ⊆ J.
Hence J is an ideal.
As S ⊆ J, we have I(S) ⊆ J.
Conversely, let T ∈ X.
Then T ⊆ S and hence I(T ) ⊆ I(S).
=⇒ J = ∪T ∈X I(T ) ⊆ I(S).
=⇒ I(S) = J = ∪T ∈X I(T ).
Therefore, if x ∈ I(S) then there exist finitely many elements a1, . . . , an
of S such that x is a linear combination of a1, . . . , an. Hence x is a linear
combination of elements of S. Conversely, every linear combination
of elements of S is in I(S). Hence I(S) is precisely the set of all linear
combination of elements of S.
3. Sum of ideals
Let Λ be an indexing set.
Suppose that for every λ ∈ Λ we are given an ideal Iλ.
Thus we have a family {Iλ | λ ∈ Λ} of ideals of R.
Then their sum is defined to be the set
{
X
aλ | aλ ∈ Iλ, aλ = 0 for all but finitely many values of λ}.
λ∈Λ
This set is denoted by
P
λ∈Λ Iλ .
P
λ∈Λ Iλ is an ideal of R.
P
Let x, y ∈ λ∈Λ Iλ.
(1)
P
P
=⇒ aλ, bλ ∈ Iλ such that x = λ∈Λ aλ, y = λ∈Λ bλ and aλ, bλ’s are
zero for all but finitely many values of λ.
Note that
(a) aλ − bλ ∈ Iλ for all λ ∈ Λ.
(b) Let X = {λ ∈ Λ | aλ 6= 0} and Y = {λ ∈ Λ | bλ 6= 0}. Then X and
Y are finite, and therefore, so is X ∪ Y .
(c) For λ ∈ Λ \ (X ∪ Y ), aλ = 0 = bλ, that is, aλ − bλ = 0.
=⇒ aλ − bλ = 0 for all but finitely values of λ.
=⇒ x − y =
P
P
(a
−
b
)
∈
λ
λ∈Λ λ
λ∈Λ Iλ .
P
P
Let α ∈ R. Then αx = λ∈Λ αaλ ∈ λ∈Λ Iλ, as αaλ ∈ Iλ, and αaλ = 0
for all λ ∈ Λ \ X.
P
As x, y and α are arbitrary, λ∈Λ Iλ is an ideal of R.
P
(2) Iµ ⊆ λ∈Λ Iλ for all µ ∈ Λ.
Let a ∈ Iµ. Define bλ = 0 if λ 6= µ and bµ = a.
P
P
=⇒ a = λ∈Λ bλ ∈ λ∈Λ Iλ.
=⇒ Iµ ⊆
P
λ∈Λ Iλ .
(3) J + I = I + J for all ideals I and J of R.
It is obvious as addition is commutative in R.
(4) Sum of ideals is associative.
Let I, J, K be ideals of R.
I + (J + K).
Let x ∈ (I + J) + K.
We need to show that (I + J) + K =
=⇒ There exist y ∈ I + J and c ∈ K such that x = y + c.
=⇒ There exist a ∈ I and b ∈ J such that y = a + b (y ∈ I + J).
=⇒ x = (a + b) + c = a + (b + c) ∈ I + (J + K).
=⇒ (I + J) + K ⊆ I + (J + K).
Similarly, I + (J + K) ⊆ (I + J) + K.
Hence, (I + J) + K = I + (J + K).
(5) I(S) + I(T ) = I(S ∪ T ) for all subsets S and T of R.
Note that as S ⊆ S ∪ T ⊆ I(S ∪ T ), we have I(S) ⊆ I(S ∪ T ).
Similarly, I(T ) ⊆ I(S ∪ T ).
Let x ∈ I(S) + I(T ).
=⇒ There exist y ∈ I(S) and z ∈ I(T ) such that x = y + z.
=⇒ y, z ∈ I(S ∪ T ).
=⇒ x = y + z ∈ I(S ∪ T ) (as I(S ∪ T ) is an ideal).
=⇒ I(S) + I(T ) ⊆ I(S ∪ T ).
Conversely, S ⊆ I(S) ⊆ I(S) + I(T ) and T ⊆ I(T ) ⊆ I(S) + I(T ).
=⇒ SU T ⊆ I(S) + I(T ).
=⇒ I(SU T ) ⊆ I(S) + I(T ) (as I(S) + I(T ) is an ideal).
Hence they are equal.
Lecture – 02
4. Product of ideals
Let I1, . . . , In be ideals of R. Then the ideal generated by the set
{a1 . . . an | a1 ∈ I1, . . . , an ∈ In} is called the product of ideals I1, . . . , In
and is denoted by I1 . . . In. Note that I1 . . . In is, by definition, an ideal
and it need not be equal to the set {a1 . . . an | a1 ∈ I1, . . . , an ∈ In}.
We now look at the structure of the ideal I1 . . . In.
Lemma 2 Let I1, . . . , In be ideals of R and let S = {a1 . . . an | a1 ∈
I, . . . , an ∈ In}. Then I1 . . . In = {x1 + · · · + xk | xi ∈ S, k ≥ 1}.
Proof. Let a ∈ R and let x ∈ S.
=⇒ There exist aj ∈ Ij such that x = a1 . . . an.
=⇒ ax = (aa1)a2 . . . an ∈ S as aa1 ∈ I1.
=⇒ S is closed under multiplication by elements of R.
Put X = {x1 + · · · + xk | xi ∈ S, k ≥ 1}.
Note that X is precisely the set of all elements of R which are finite
sums of elements of S.
Note also that S ⊆ X and X ⊆ I1 . . . In.
Claim. X is an ideal of R.
Let x, y ∈ X and a ∈ R.
=⇒ There exist some l, m ≥ 1, x1, . . . , xl , y1, . . . , ym ∈ S such that
x = x1 + · · · + xl and y = y1 + · · · + ym.
=⇒ x + y is a sum of finitely many elements of S.
=⇒ x + y ∈ X.
Also, ax = ax1 + · · · + axn ∈ X as axi ∈ S for all i = 1, . . . , l.
Similarly, −x = (−1)x ∈ X.
=⇒ X is an ideal of R.
As X contains S, I1 . . . In = I(S) ⊆ X.
=⇒ I1 . . . In = X.
We now look at some properties of the product of ideals.
(1) Let σ be a permutation of 1, . . . , n. Then I1 . . . In= Iσ(1) . . . Iσ(n).
It is obvious as R is commutative and therefore, the sets {a1 . . . an |
a1 ∈ I1, . . . , an ∈ In} and {x1 . . . xn | x1 ∈ Iσ(1), . . . , xn ∈ Iσ(n)} are identical.
(2) Let S and T be subsets of R. Let X = {st | s ∈ S, t ∈ T }.
Then I(S)I(T ) = I(X).
Put Y = {ab | a ∈ I(S), b ∈ I(T )}.
=⇒ I(S)I(T ) = I(Y ) and X ⊆ Y as S ⊆ I(S) and T ⊆ I(T ).
=⇒ I(X) ⊆ I(Y ) = I(S)I(T ).
We shall show that Y ⊆ I(X).
Let y ∈ Y .
=⇒ There exist a ∈ I(S) and b ∈ I(T ) such that y = ab.
=⇒ There exist a1, . . . , an ∈ S and x1, . . . , xn ∈ R such that a =
Pn
i=1 xi ai .
Similarly, there exist b1, . . . , bm ∈ T and y1, . . . , ym ∈ R such that
Pm
b = i=1 yibi.
=⇒ y =
Pn Pm
i=1 j=1 xi yj ai bj .
=⇒ y ∈ I(X) (as aibj ∈ X for all i, j).
=⇒ Y ⊆ I(X).
=⇒ I(S)I(T ) = I(Y ) ⊆ I(X).
=⇒ I(S)I(T ) = I(X).
(3) Product of ideals is associative. Let I, J, K be ideals of R.
We need to show that (IJ)K = I(JK).
Put X = {ab | a ∈ I, b ∈ J}.
=⇒ IJ is generated by X.
As K is generated by K itself, (IJ)K is generated by Y = {(ab)c | a ∈
I, b ∈ J, c ∈ K}.
Similarly, I(JK) is also generated by Y .
=⇒ (IJ)K = I(JK).
Lemma 3 Let I1, . . . , In be ideals of R. Then I1 . . . In ⊆ I1 ∩ . . . ∩ In.
Proof. Let S = {a1 . . . an | a1 ∈ I1, . . . , an ∈ In}.
Then I1 . . . In = I(S).
We shall show that S ⊆ Ij for all j = 1, . . . , n.
Let x ∈ S.
=⇒ x = a1 . . . an for some a1 ∈ I1, . . . , an ∈ In.
Note that x = a1 . . . aj . . . an ∈ Ij for all j = 1, . . . , n.
=⇒ S ⊆ Ij for all j = 1, . . . , n.
=⇒ I1 . . . In = I(S) ⊆ Ij for all j = 1, . . . , n.
=⇒ I1 . . . In ⊆ I1 ∩ . . . ∩ In.
5. Maximal ideals
Definition. Let R be a ring and I be an ideal of R such that I 6= R.
Then I is said to be maximal if it is not properly contained any ideal
of R other than R. More precisely, I is maximal if there is no ideal J
of R such that I ( J ( R.
We shall show that R admits maximal ideals.
We prove a simple lemma now.
Lemma 4 Let I be an ideal of R. Then I = R iff I contains a unit.
Proof.
Suppose first that I = R. Then 1 ∈ I, which is a unit.
Conversely, suppose that I contains a unit a.
=⇒ There exists b ∈ R such that ab = 1.
=⇒ 1 ∈ I.
Let x ∈ R.
=⇒ x = x1 ∈ I.
=⇒ R ⊆ I ⊆ R, that is, I = R.
We prove another lemma which shall be used in many results.
Lemma 5 Let X be a nonempty collection of ideals of R. Then X is
partially ordered by inclusion. Let T be a chain in X. Let J = ∪I∈T I.
Then J is an ideal of R.
Proof.
Let a, b ∈ J and c ∈ R.
=⇒ There exist I1, I2 ∈ T such that a ∈ I1 and b ∈ I2.
Since T is a chain, either I1 ⊆ I2 or I2 ⊆ I1. Assume that I1 ⊆ I2.
=⇒ a, b ∈ I2.
=⇒ ca, a − b ∈ I2 ⊆ J.
=⇒ J is an ideal.
We shall now show that R has maximal ideals.
Theorem. Let I be a proper ideal of R. Then there exists a maximal
ideal m of R such that I ⊆ m.
Proof.
We shall use Zorn’s Lemma to prove this result.
Let
F = {J ⊆ R | I ⊆ J, J is a proper ideal of R}.
As I ∈ F , F is nonempty.
Order F by inclusion.
Let T be a chain in F .
Put J0 = ∪J∈T J.
Clearly, I ⊆ J0 and J0 is an ideal of R.
Claim. J0 6= R.
Suppose, if possible, that J0 = R.
=⇒ 1 ∈ J0.
=⇒ There exists some J ∈ T such that 1 ∈ J.
=⇒ J = R, a contradiction.
=⇒ J0 =
6 R.
=⇒ J0 ∈ F .
=⇒ J0 upper bound of T in F .
=⇒ Every chain in F has an upper bound in F .
=⇒ By Zorn’s Lemma, F has maximal elements.
Let m be a maximal element of F .
=⇒ m is a proper ideal of R and I ⊆ m.
We show that m is a maximal ideal of R.
To show this let J be an ideal of R such that m ( J ( R.
=⇒ I ⊆ m ⊆ J 6= R.
=⇒ J ∈ F .
A contradiction, as J contains m properly and m is a maximal element
of F .
=⇒ m is a maximal ideal of R which contains I.
Since R does have proper ideals (for example {0}), R also has maximal
ideals.
The following result is useful in constructing fields.
Lemma 6 Let I be a proper ideal of R. Then I is maximal iff R/I
is a field.
Proof.
Suppose that I is maximal.
To show that R/I is a field.
We need to show that every nonzero element of R/I is a unit.
Let x ∈ R/I, x 6= 0.
=⇒ There exists a ∈ R such that x = a + I.
=⇒ a 6∈ I as x 6= 0.
Put J = I + (a).
Then J is an ideal of R and it contains I.
Since a = 0 + 1a ∈ J and a 6∈ I, J contains I properly.
=⇒ J = R.
=⇒ 1 ∈ J.
=⇒ There exist b ∈ I and c ∈ R such that 1 = b + ca.
=⇒ 1 + I = ab + I = (a + I)(b + I) = x(b + I).
=⇒ x is a unit of R/I.
Since x is arbitrary, every nonzero element of R/I is a unit.
=⇒ R/I is a field.
Conversely, let R/I be a field.
Let J be an ideal of R containing I properly.
We need to show that J = R.
Let a ∈ J \ I.
Consider x = a + I. Then x is a nonzero element of R/I.
=⇒ x is a unit of R/I.
=⇒ There exists some y ∈ R/I such that xy = 1 + I, the unity of R/I.
Let b ∈ R such that y = b + I.
=⇒ (a + I)(b + I) = 1 + I.
=⇒ 1 − ab ∈ I ⊆ J.
=⇒ 1 = ab + 1 − ab ∈ J (as a ∈ J).
=⇒ J = R.
=⇒ I is maximal.
Comment. Let a ∈ R be a nonunit. Then the ideal (a) 6= R. Hence
there exists a maximal ideal m such that (a) ⊆ m. That is a ∈ m. In
other words, every nonunit of R is an element of some maximal ideal
of R. Conversely, if I is a maximal ideal and b ∈ I then b is a nonunit
as I is a proper ideal. Hence the set of all nonunits of R is equal to
the union of all maximal ideals of R.
6. Prime ideals
An ideal I is said to be a prime ideal of R if I is proper and the
following condition is satisfied: if a, b ∈ R such that ab ∈ I then either
a ∈ I or b ∈ I.
The notion of prime ideals is a generalisation of prime elements of Z.
We shall see some examples after proving the following lemma.
Lemma 7 Let I be a proper ideal of R. Then I is prime iff R/I is
an integral domain.
Proof.
First suppose that I is a prime ideal.
Let x, y ∈ R/I be nonzero elements of R/I.
We need to show that xy 6= 0.
Suppose, if possible, that xy = 0.
Let a, b ∈ R such that x = a + I and y = b + I.
=⇒ I = 0 = xy = (a + I)(b + I) = ab + I.
=⇒ ab ∈ I.
=⇒ Either a ∈ I or b ∈ I (as I is a prime ideal).
=⇒ Either x = 0 or y = 0, a contradiction.
=⇒ R/I is an integral domain.
Conversely, suppose that R/I is an integral domain.
To show that I is a prime ideal.
Let a, b ∈ R such that ab ∈ I.
=⇒ (a + I)(b + I) = ab + I = I = 0.
=⇒ Either a + I = 0 or b + I = 0 (as R/I is an integral domain).
=⇒ Either a ∈ I or b ∈ I.
=⇒ I is prime.
Comments. Therefore, all maximal ideals are prime. Thus every
commutative ring with unity having at least two elements admits
prime ideals.
The ideal {0} is prime in Z.
It easily follows that if n ∈ Z then (n) is a prime ideal iff n is a prime
number.
In every integral domain, the ideal {0} is a prime ideal (why?).
Prime ideals play a very important role in determing the structure of
the ring itself. We shall see that later.
We prove a small lemma which shall be useful later.
Lemma 8 Let I be prime ideal of R.
(a) Let a1, . . . , an ∈ R such that a1 . . . an ∈ I. Then there exists a
j ∈ {1, . . . , n} such that aj ∈ I.
(b) Let a ∈ R such that an ∈ I for some n ∈ N. Then a ∈ I.
Proof. (a) Clearly, (a1 + I) . . . (an + I) = a1 . . . an + I = I = 0 in R/I.
=⇒ There exists some j ∈ {1, . . . , n}
such that aj + I = 0 = I in R/I (as R/I is an integral domain).
=⇒ aj ∈ I.
(b) a . . . a(n times) = an ∈ I. Hence by (a), a ∈ I.
Lecture – 03
7. The colon ideals
Let I be an ideal of R and a ∈ R. Then aI denotes the set {ax | x ∈ I}.
Let I and J be ideals of R. Consider
{a ∈ R | aJ ⊆ I}.
We denote this set by (I : J) and call it the colon ideal of I and J (or
ideal quotient of I and J).
Claim. (I : J) is an ideal of R.
Let a, b ∈ (I : J) and c ∈ R.
Let x ∈ J.
=⇒ (a − b)x = ax − bx ∈ I (as ax, bx ∈ I).
Also cax = a(cx) ∈ I as cx ∈ J.
=⇒ (a − b)J ⊆ I and (ca)J ⊆ I.
=⇒ a − b, ca ∈ (I : J).
=⇒ (I : J) is an ideal as a, b, c are arbitrary.
We shall, now look at some properties of colon ideals.
Lemma 9 Let I, J, K, I1, . . . , In, J1, . . . , Jm be ideals of R.
(1) I ⊆ (I : J).
(2) (I : J)J ⊆ I.
(3) ((I : J) : K) = (I : JK) = ((I : K) : J).
n
(4) (∩n
r=1 Ir : J) = ∩r=1 (Ir : J).
(5) (I :
Proof.
Pm
m
r=1 Jr ) = ∩r=1 (I : Jr ).
(1) Let a ∈ I.
=⇒ aJ ⊆ I.
=⇒ a ∈ (I : J).
=⇒ I ⊆ (I : J).
(2) Let a ∈ (I : J) and b ∈ J.
Then ab ∈ I for all a ∈ (I : J) and b ∈ J.
That is, {ab | a ∈ (I : J), b ∈ J} ⊆ I.
=⇒ The ideal generated by {ab | a ∈ (I : J), b ∈ J} is contained in I.
=⇒ (I : J)J ⊆ I.
(3) Let a ∈ ((I : J) : K), b ∈ J and c ∈ K.
=⇒ ac ∈ (I : J).
=⇒ acb ∈ I for all b ∈ J and c ∈ K.
Let x ∈ JK. Then there exist b1, . . . , bk ∈ J and c1 . . . , ck ∈ K such
Pk
that x = i=1 bici.
=⇒ ax =
Pk
i=1 abi ci ∈ I.
=⇒ a(JK) ⊆ I, that is, a ∈ (I : JK).
=⇒ ((I : J) : K) ⊆ (I : JK), as a is arbitrary.
Conversely, let a ∈ (I : JK).
Let b ∈ J and c ∈ K.
Then bc ∈ JK.
=⇒ abc ∈ I for all b ∈ J.
=⇒ ac ∈ (I : J) for all c ∈ K.
=⇒ a ∈ ((I : J) : K) for all a ∈ (I : JK).
=⇒ (I : JK) ⊆ ((I : J) : K).
So we have ((I : J) : K) = (I : JK).
Furthermore, ((I : J) : K) = (I : JK) = (I : KJ) = ((I : K) : J).
(4) Let a ∈ (∩n
r=1 Ir : J).
=⇒ aJ ⊆ ∩n
r=1 Ir or aJ ⊆ Ir for all r = 1, . . . , n.
=⇒ a ∈ (Ir : J) for all r = 1, . . . , n.
=⇒ a ∈ ∩n
r=1 (Ir : J).
n
=⇒ (∩n
r=1 Ir : J) ⊆ ∩r=1 (Ir : J).
Conversely, let a ∈ ∩n
r=1 (Ir : J).
Then a ∈ (Ir : J) for all r = 1, . . . , n.
=⇒ aJ ⊆ Ir for all r = 1, . . . , n.
=⇒ aJ ⊆ ∩n
r=1 Ir .
=⇒ a ∈ (∩n
r=1 Ir : J).
n I : J).
=⇒ ∩n
(I
:
J)
⊆
(∩
r
r=1
r=1 r
n
=⇒ ∩n
r=1 (Ir : J) = (∩r=1 Ir : J).
Pm
(5) Let a ∈ (I : r=1 Jr ) and b ∈ Jr .
P
=⇒ b ∈ m
r=1 Jr and hence ab ∈ I.
=⇒ aJr ⊆ I or a ∈ (I : Jr ) for all r = 1, . . . , m.
=⇒ a ∈ ∩m
r=1 (I : Jr ) for all a ∈ (I :
Pm
r=1 Jr ).
Pm
=⇒ (I : r=1 Jr ) ⊆ ∩m
r=1 (I : Jr ).
Conversely, let a ∈ ∩m
r=1 (I : Jr ).
=⇒ a ∈ (I : Jr ) for all r = 1, . . . , m.
=⇒ aJr ⊆ I for all r = 1, . . . , m.
Let
=⇒
=⇒
=⇒
=⇒
Pm
x ∈ r=1 Jr .
P
x= m
r=1 xr , where xr ∈ Jr for all r = 1, . . . , m.
Pm
Pm
ax = r=1 axr ∈ I for all x ∈ r=1 Jr .
P
m (I : J ).
a ∈ (I : m
J
)
for
all
a
∈
∩
r
r
r=1
r=1
Pm
∩m
(I
:
J
)
⊆
(I
:
r
r=1 Jr ). Hence, they are equal.
r=1
8. Isaac’s theorem
We call an ideal I finitely generated if there exists a finite set S such
that I = I(S). In other words, I = (a1, . . . , an) for some a1, . . . , an ∈ I.
We prove an interesting theorem now, known as Isaac’s theorem.
Theorem. Let R be a ring in which all prime ideals are finitely generated. Then all ideals of R are finitely generated.
Proof.
We use a contrapositive argument here. Let
F = {I ⊆ R | I is an ideal of R, I is not finitely generated}.
We want to show that F is empty.
Suppose, if possible, that F is nonempty.
Order F by inclusion.
We shall use Zorn’s Lemma to get to a contradiction.
Let T be a chain in F and let J = ∪I∈T I.
=⇒ J is ideal of R.
Claim. J is not finitely generated.
Suppose, if possible, that J is finitely generated, say, J = (a1, . . . , an).
As ar ∈ J = ∪I∈T I, there exists some Ir ∈ T such that ar ∈ Ir for all
r = 1, . . . , n.
=⇒ {a1, . . . , ar } ⊆ ∪n
r=1 Ir .
Claim. ∪n
r=1 Ir = Ik for some k ∈ {1, . . . , n}.
We prove this by induction on n.
If n = 1 then there is nothing to prove.
Assume now that n ≥ 2 and that the result is true for n − 1.
=⇒ ∪n−1
r=1 Ir = Il for some l ∈ {1, . . . , n − 1}.
=⇒ ∪n
r=1 Ir = Il ∪ In .
Since T is chain, either Il ⊆ In or In ⊆ Il .
n I =I ∪I =I.
=⇒ ∪n
I
=
I
∪
I
=
I
or
∪
r
n
n
n
l
l
l
r=1
r=1 r
So k is either l or n.
=⇒ {a1, . . . , an} ⊆ ∪n
r=1 Ir = Ik .
=⇒ J = (a1, . . . , an) ⊆ Ik ⊆ J.
=⇒ Ik = J = (a1, . . . , an), that is, Ik is finitely generated, a contradiction.
=⇒ J is not finitely generated and so J ∈ F .
=⇒ Every chain in F has an upper bound in F .
=⇒ By Zorn’s Lemma, F has a maximal element, say, P .
Claim. P is a prime ideal.
As R is finitely generated by {1}, P 6= R.
Suppose, if possible, that P is not a prime ideal of R.
=⇒ There exist a, b ∈ R such that ab ∈ P but a 6∈ P and b 6∈ P .
Put I = (a).
=⇒ P ⊆ P + I and P 6= P + I as a ∈ P + I.
=⇒ P + I 6∈ F as P is a maximal element of F .
=⇒ P + I is finitely generated, that is, P + I = (x1, . . . , xn).
As xi ∈ P + I, there exist, pi ∈ P and αi ∈ R such that xi = pi + αia
for all 1 = 1, . . . , n.
Now consider the ideal (P : I).
We know that P ⊆ (P : I).
We show that P 6= (P : I).
Let x ∈ I.
=⇒ x = ta for some t ∈ R.
=⇒ bx = tab ∈ P , for all x ∈ I.
=⇒ bI ⊆ P , that is, b ∈ (P : I).
As P ⊆ (P : I) and b ∈ (P : I) \ P , P 6= (P : I).
=⇒ (P : I) 6∈ F .
=⇒ (P : I) is finitely generated, say, (P : I) = (y1, . . . , ym).
Put S = {p1, . . . , pn, ay1, . . . , aym}.
Claim. P = I(S).
Note that yj ∈ (P : I).
=⇒ ayj ∈ P for all j = 1, . . . , m (as a ∈ I).
=⇒ S = {p1, . . . , pn, ay1, . . . , aym} ⊆ P .
=⇒ I(S) ⊆ P .
Conversely, let y ∈ P .
=⇒ y ∈ P + I.
=⇒ There exist β1, . . . , βn ∈ R such that
Pn
Pn
Pn
Pn
β
x
=
β
(p
+
α
a)
=
β
p
+
(
i
i=1 i i
i=1 i i
i=1 i i
i=1 βi αi )a.
Pn
Put t = ( i=1 βiαi).
P
=⇒ ta = y − n
i=1 βi pi ∈ P .
y=
Let z ∈ I. Then z = ua for some u ∈ R.
=⇒ tz = tua ∈ P for all z ∈ I.
=⇒ t ∈ (P : I) = (y1, . . . , ym).
=⇒ There exist γj ∈ R such that t =
P
=⇒ y = n
i=1 βi pi + ta
Pm
j=1 γj yj .
Pn
Pm
= i=1 βipi + a j=1 γj yj
P
Pm
= n
β
p
+
i=1 i i
j=1 γj ayj
=⇒ y ∈ I(S).
As y ∈ P is arbitrary, P ⊆ I(S).
=⇒ P = I(S), that is, P is finitely generated.
A contradiction again as P ∈ F .
Therefore, P must be a prime ideal.
=⇒ P must again be finitely generated.
=⇒ P 6∈ F , which is absurd.
=⇒ F is empty and all ideals of R are finitely generated.
9. Cohen’s theorem
An ideal I is said to be principal if it is generated by a singleton set.
More precisely, I = (a) for some a ∈ I. The ring R is said to be
principal ideal ring if all ideals of R are principal.
The following theorem is known as Cohen’s theorem.
Theorem.
Let R be a ring in which all prime ideals are principal.
Then all ideals of R are principal.
Proof.
We use a contrapositive argument here. Let
F = {I ⊆ R | I is an ideal of R, I is not principal}.
We want to show that F is empty.
Suppose, if possible, that F is nonempty.
Order F by inclusion.
We shall use Zorn’s Lemma to get to a contradiction.
Let T be a chain in F and let J = ∪I∈T I.
Then J is an ideal of R.
Claim. J is not principal.
Suppose, if possible, that J is principal, say, J = (a).
=⇒ There exists I ∈ T such that a ∈ I (as a ∈ J).
=⇒ J = (a) ⊆ I ⊆ J, that is, I = J = (a).
=⇒ I is principal, a contradiction.
=⇒ J is not principal and so J ∈ F .
=⇒ Every chain in F has an upper bound in F .
=⇒ By Zorn’s Lemma, F has a maximal element, say, P .
Claim. P is a prime ideal.
As R is generated by {1}, R is principal. Hence P 6= R.
Suppose, if possible that P is not prime.
=⇒ There exists a, b ∈ R such that ab ∈ P but a 6∈ P and b 6∈ P .
Put I = (a).
=⇒ P ⊆ P + I and P 6= P + I as a ∈ P + I.
=⇒ P + I 6∈ F as P is a maximal element of F .
=⇒ P + I is principal, that is, P + I = (x).
=⇒ There exist p ∈ P and α ∈ R such that x = p + αa.
Put J = (x).
Now consider the ideal (P : J).
Let t ∈ J.
=⇒ t = sx for some s ∈ R.
=⇒ bt = sbx = sbp + αsab ∈ P , for all t ∈ J.
=⇒ bJ ⊆ P , that is, b ∈ (P : J).
As P ⊆ (P : J) and b ∈ (P : J) \ P , we have P 6= (P : J).
=⇒ (P : J) 6∈ F .
=⇒ (P : J) is principal, say, (P : J) = (y).
Claim. P = (xy).
Note that y ∈ (P : J) and x ∈ J. Thus xy ∈ P .
=⇒ (xy) ⊆ P .
Conversely, let z ∈ P .
=⇒ z ∈ P + I = (x).
=⇒ There exist β ∈ R such that z = βx.
Let u ∈ J.
=⇒ u = vx for some v ∈ R.
=⇒ βu = βvx = vβx = vz ∈ P for all u ∈ J.
=⇒ β ∈ (P : J) = (y).
=⇒ There exist γ ∈ R such that β = γy.
=⇒ z = βx = γyx ∈ (xy).
As z ∈ P is arbitrary, P ⊆ (xy).
=⇒ P = (xy), that is P is principal.
A contradiction again as P ∈ F .
=⇒ P must be a prime ideal.
But then P must again be principal and hence P 6∈ F , which is absurd.
=⇒ F is empty and all ideals of R are principal.
Lecture – 04
10. Radical of an ideal
Let I be an ideal. If I 6= R then I is contained in prime ideals. We
define radical of I to be the intersection of all prime ideals containing
I in this case. If I = R then radical of I is defined to be R itself. The
radical of I is denoted by rad(I). Sometimes the radical of I is also
called the prime radical of I.
For an ideal I we consider the set {a ∈ R | an ∈ I for some n ∈ N}.
√
This set is denoted by I and is called root ideal of I.
Lemma 10 Let I, J be ideals of R. Then
√
√
(a) I ⊆ I and I is an ideal of R.
√
√
(b) I ⊆ J then I ⊆ J.
q √
√
(c) ( I) = I.
√
Proof. (a) Clearly, I ⊆ I.
√
Let a, b ∈ I and let c ∈ R.
=⇒ There exist m, n ∈ N such that an, bm ∈ I.
Pn+m−1 n+m−1 i n+m−1−i
n+m−1
Note that (a + b)
= i=0
ab
.
i
Consider a term n+m−1
aibn+m−1−i of this sum:
i
n+m−1 i n+m−i
i
(1) if i ≥ n then a ∈ I and therefore,
ab
∈I
i
(2) if i < n then m + n − 1 − i ≥ m and therefore, bn+m−i ∈ I, which
n+m−1 i n+m−i
means
ab
∈ I.
i
Pn+m−1 n+m−1 i n+m−i
=⇒ All terms of i=0
ab
are in I.
i
=⇒ (a + b)n+m−1 ∈ I.
=⇒ a + b ∈
√
I.
Also (−a)n = ±an ∈ I and hence −a ∈
√
I.
Furthermore, (ca)n = cnan ∈ I and hence ca ∈
As a, b ∈
√
I and c ∈ R are arbitrary,
√
(b) Let a ∈ I.
√
√
I.
I is an ideal of R.
=⇒ There exists some n ∈ N such that an ∈ I ⊆ J.
√
=⇒ a ∈ J.
√
√
=⇒ I ⊆ J.
(c) As I ⊆
√
√
I, we have I ⊆
q√
I.
Conversely, let a ∈
q√
I.
=⇒ There exists some n ∈ N such that an ∈
√
I.
=⇒ There exists some m ∈ N such that (an)m ∈ I.
=⇒ amn ∈ I.
√
=⇒ a ∈ I.
q√
q√
√
√
=⇒
I ⊆ I or I =
I.
Now we prove a theorem.
Theorem. Let I be an ideal of R. Then rad(I) =
√
I.
Proof. First assume that I = R.
√
√
=⇒ R = I ⊆ I and hence, I = R = rad(I).
So we are done in this case.
Now assume that I 6= R.
√
Let a ∈ I.
Then there exist some n ∈ N such that an ∈ I.
We want to show that a ∈ rad(I).
Let P be a prime ideal containing I.
=⇒ an ∈ P .
=⇒ a ∈ P (by an earlier result).
=⇒ a is in the intersection of all prime ideals containing I.
=⇒ a ∈ rad(I) for all a ∈
√
=⇒ I ⊆ rad(I).
√
I.
√
We now want to show that rad(I) ⊆ I.
√
Suppose, if possible, that rad(I) 6⊆ I.
=⇒ There exists some a ∈ rad(I) such that a 6∈
√
I.
This means no positive power of a belongs to I.
Put S = {a, a2, a3, . . .}.
Then S ∩ I = ∅.
Define
F = {J ⊆ R | J is an ideal of R, I ⊆ J and J ∩ S = ∅}.
Clearly, I ∈ F and therefore, F 6= ∅.
Order F by inclusion.
We shall use Zorn’s Lemma to arrive at a contradiction.
Let T be a chain in F and let J0 = ∪J∈T J.
=⇒ J0 is an ideal of R and J0 ∩ S = (∪J∈T J) ∩ S = ∪J∈T (J ∩ S) =
∪J∈T ∅ = ∅.
=⇒ J0 ∈ F (as I ⊆ J0).
=⇒ Every chain in F has an upper bound in F .
=⇒ By Zorn’s Lemma F has a maximal element, say, P .
Then P is an ideal of R, I ⊆ P and P ∩ S = ∅.
Claim. P is a prime ideal of R.
Assume the contrary.
Then there exist x, y ∈ R such that xy ∈ P but x 6∈ P and y 6∈ P .
Consider the ideal P + (x).
Then x ∈ P + (x) \ P and P ⊆ P + (x).
As P is maximal in F , P + (x) 6∈ F .
Note that P + (x) is an ideal of R, I ⊆ P ⊆ P + (x).
=⇒ S ∩ (P + (x)) can not be empty.
=⇒ There exists some n ∈ N such that an ∈ P + (x).
=⇒ There exist some α ∈ P and β ∈ R such that an = α + βx.
Similarly, there exist some m ∈ N, γ ∈ P and δ ∈ R such that
am = γ + δy.
=⇒ an+m = anam = (α + βx)(γ + δy) = αγ + αδy + βγx + βδxy ∈ P .
=⇒ S ∩ P 6= ∅, a contradiction.
=⇒ P is a prime ideal.
Since a ∈ rad(I), I ⊆ P and P is prime we must have a ∈ P , again a
contradiction.
√
=⇒ rad(I) ⊆ I.
So we have rad(I) =
√
I.
11. Prime avoidance theorem
Comments. We recall a result from group theory: Let G be a group
and H, H1, H2, . . . , Hn be subgroups of G such that H ⊆ H1 ∪ H2 ∪
. . . ∪ Hn. If n = 2 then H ⊆ H1 or H ⊆ H2. However, if n ≥ 3 then
there is no guarantee that H will be contained in one of the Hi’s. For
example take G = {1, a, b, c}, the Klien’s 4-group, H1 = (a), H2 = (b)
and H3 = (c). Then G ⊆ H1 ∪ H2 ∪ H3 but G is not contained in any
of Hi’s. In ring theory we have a similar result if we replace G by a
commutative ring with unity and subgroups by prime ideals with no
restriction on the value of n. This is known as the prime avoidance
theorem. We discuss this result below.
Let I be an ideal and P1, . . . , Pn be prime ideals such that I ⊆ P1 ∪
. . . ∪ Pn. Then it is true that I is contained in one of the Pi’s. Equivalently, if I avoids being contained in all of the Pi’s then it avoids being
contained in their union. This result is known as the prime avoidance
theorem.
We prove the following more general variant of prime avoidance theorem.
Theorem. Let I, I1, I2 be ideals and P1, . . . , Pn be prime ideals such
that I 6⊆ I1, I 6⊆ I2 and I 6⊆ Pj for all j = 1, . . . , n, where n ≥ 0. Then
I 6⊆ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pn.
Proof. We use induction on n.
First suppose that n = 0.
As I 6⊆ I1, there exists some a ∈ I such that a 6∈ I1.
If a 6∈ I2 then we are done.
Suppose that a ∈ I2.
As I 6⊆ I2, there exists some b ∈ I such that b 6∈ I2.
If b 6∈ I1 then again we are done.
Suppose that b ∈ I1.
Then a − b 6∈ I1 and a − b 6∈ I2 but a − b ∈ I.
=⇒ I 6⊆ I1 ∪ I2.
Assume now that n ≥ 1. Since all prime ideals are also ideals, we may
treat some of them purely as ideals if the need arises.
By induction, I 6⊆ I2 ∪ P1 ∪ . . . ∪ Pn (here we treat P1 purely as an ideal
and P2, . . . , Pn as prime ideals).
=⇒ There exists a1 ∈ I such that a1 6∈ I2 ∪ P1 ∪ . . . ∪ Pn.
Similarly, there exists a2 ∈ I such that a2 6∈ I1 ∪ P1 ∪ . . . ∪ Pn.
Let j ∈ {1, . . . , n}. Again by induction, I 6⊆ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pj−1 ∪
Pj+1 ∪ . . . ∪ Pn.
=⇒ There exists bj ∈ I such that bj 6∈ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pj−1 ∪ Pj+1 ∪
. . . ∪ Pn.
We do this for all j ∈ {1, . . . , n}.
So we get elements b1, . . . , bn ∈ I such that
bj 6∈ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pj−1 ∪ Pj+1 ∪ . . . ∪ Pn for all j = 1, . . . , n.
Therefore,
(1) if a1 6∈ I1 then we are done. So assume that a1 ∈ I1.
(2) if a2 6∈ I2 then we are done. So assume that a2 ∈ I2.
(3) if for some j ∈ {1, . . . , n}, bj 6∈ Pj then we are done. So assume
that bj ∈ Pj for every j ∈ {1, . . . , n}.
Consider the element a = b1 + a1a2b2 . . . bn.
Note that a ∈ I.
Claim. a 6∈ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pn.
Assume, if possible, that a ∈ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pn. Then we see that
(1) if a ∈ I1, then b1 = a − a1a2b2 . . . bn ∈ I1 as a1 ∈ I1. But b1 6∈ I1 so
a 6∈ I1.
(2) if a ∈ I2, then b1 = a − a1a2b2 . . . bn ∈ I2 as a2 ∈ I2. But b1 6∈ I2 so
a 6∈ I2.
(3) if a ∈ P1, then a1a2b2 . . . bn = a − b1 ∈ P1 as b1 ∈ P1. But none of
a1, a2, b2, . . . , bn is in P1 and hence a1a2b2 . . . bn can not be in P1 (as
P1 is a prime ideal). So a 6∈ P1.
(4) if a ∈ Pj , for some j = 2, . . . , n, then b1 = a − a1a2b2 . . . bn ∈ Pj as
bj ∈ Pj . But b1 6∈ Pj for j = 2, . . . , n. So a 6∈ Pj for any j = 2, . . . , n.
Therefore, a 6∈ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pn.
But as a ∈ I, I 6⊆ I1 ∪ I2 ∪ P1 ∪ . . . ∪ Pn.
12. Jacobson radical and the nil radical
The Jacobson radical of a ring R, denoted by J(R), is defined to be
the intersection of all maximal ideals of R whereas the nil radical of
R is defined to be the intersection of all prime ideals of R. The nil
radical of R is denoted by nil(R). As every maximal ideal is also a
prime ideal, nil(R) ⊆ J(R). As evey prime ideal contains {0} ideal,
nil(R) =
q
{0} = {a ∈ R | an = 0 for some n ∈ N}.
Elements of nil(R) are called nilpotent elements of R.
In the following lemma we give a similar description of J(R).
Lemma 11 J(R) = {a ∈ R | 1 + ba is a unit of R for all b ∈ R}.
Proof. Put X = {a ∈ R | 1 + ba is a unit of R for all b ∈ R}.
Let a ∈ J(R). Suppose, if possible, that a 6∈ X.
=⇒ There exists some b ∈ R such that 1 + ba is not a unit of R.
=⇒ There exists a maximal ideal m of R such that 1 + ba ∈ m.
However, as m is maximal, J(R) ⊆ m.
=⇒ a ∈ m as a ∈ J(R).
=⇒ 1 = 1 + ba − ba ∈ m or m = R, a contradiction.
=⇒ a ∈ X, that is, J(R) ⊆ X.
Conversely, let a ∈ X. Suppose, if possible, that a 6∈ J(R).
=⇒ There exists a maximal ideal m of R such that a 6∈ m.
=⇒ m + (a) = R (as m is properly contained in m + (a)).
=⇒ There exists some x ∈ m and b ∈ R such that x + ba = 1.
=⇒ 1 + (−b)a = x ∈ m.
But 1 + (−b)a is a unit of R so m = R, a contradiction.
=⇒ a ∈ J(R), that is, X ⊆ J(R).
So J(R) = X = {a ∈ R | 1 + ba is a unit of R for all b ∈ R}.
Lecture – 05
13. Ring homomorphism
Comments. (1) In this course, a ring homomorphism shall always
be assumed to take unity of the domain of definition to the unity of
codomain. More precisely, if R and R0 are rings then φ : R −→ R0 is a
called ring homomorphism if φ(x + y) = φ(x) + φ(y), φ(xy) = φ(x)φ(y)
and φ(1) = 1 for all x, y ∈ R.
(2) Let R be a ring. Then a nonempty subset A of R is called a
subring of R if
(a) (A, +) is a subgroup of (R, +).
(b) A is closed under multiplication of R, that is xy ∈ A for all x, y ∈ A.
(c) 1 ∈ A.
(d) A is a commutative ring with unity in its own right with respect to
the addition and multiplication of R. (This condition actually follows
from (a),(b) and (c).)
Lemma 12 Let R and R0 be rings and φ : R −→ R0 be a ring homomorphism. Then
(1) φ(R) is a subring of R0.
(2) if J is an ideal of R0 then φ−1(J) is an ideal of R.
(3) if P is a prime ideal of R0 then φ−1(P ) is a prime ideal of R.
Proof. (1) Clearly, (φ(R), +) is a subgroup of (R0, +) and it is closed
under multiplication of R0. Furthermore, 1 = φ(1) ∈ φ(R). It is obvious that φ(R) is a subring of R0.
(2) Clearly, φ−1(J) is a subgroup of (R, +).
Let a ∈ R and x ∈ φ−1(J).
=⇒ φ(x) ∈ J.
=⇒ φ(ax) = φ(a)φ(x) ∈ J.
=⇒ ax ∈ φ−1(J) for all a ∈ R and x ∈ φ−1(J).
=⇒ φ−1(J) is an ideal of R.
(3) By (2), φ−1(P ) is an ideal of R.
As P 6= R0, φ(1) = 1 6∈ P .
=⇒ 1 6∈ φ−1(P ), that is, φ−1(P ) 6= R0.
Let a, b ∈ R such that ab ∈ φ−1(P ).
=⇒ φ(a)φ(b) ∈ P .
=⇒ Either φ(a) ∈ P or φ(b) ∈ P (as P is prime in R0).
=⇒ Either a ∈ φ−1(P ) or b ∈ φ−1(P ).
=⇒ φ−1(P ) is a prime ideal of R.
Comments. Let R and R0 be rings and φ : R −→ R0 be a ring homomorphism.
(1) If I is a maximal ideal of R0 then φ−1(I) need not be a maximal
ideal of R. For example {0} is maximal ideal in Q but its inverse image in Z under the inclusion map : Z −→ Q is not a maximal ideal of Z.
(2) The inverse image of a maximal ideal is necessarily a prime ideal.
(3) If I is an ideal of R then φ(I) need not be an ideal of R0. For
example I = 2Z is an ideal in Z but its image in Q under the inclusion
map : Z −→ Q is not an ideal of Q. The reason is that it is not closed
under multiplication by elements of Q.
(4) If I is a prime ideal of R then φ(I) need not be a prime ideal of
R0. See, the example in (3) above.
(5) If φ is onto and I is an ideal of R then φ(I) is an ideal of R0.
We now prove some results.
Lemma 13 Let R and R0 be rings and φ : R −→ R0 be an onto ring
homomorphism. Let Y be the set of all ideals of R0 and X = {I ⊆
R | I is an ideal of R containing ker(φ)}. Then the map θ : Y −→ X
given by θ(J) = φ−1(J) for all J ∈ Y is a bijection.
Proof. Let J ∈ Y . Then φ−1(J) is an ideal of R.
Furthermore, ker(φ) ⊆ φ−1(J) as 0 ∈ J.
Therefore, φ−1(J) ∈ X.
=⇒ θ is well defined.
(1) θ is one-one. Let J1, J2 ∈ Y such that θ(J1) = θ(J2).
=⇒ φ−1(J1) = φ−1(J2).
Let x ∈ J1.
As φ is onto, there exists some y ∈ R such that φ(y) = x.
=⇒ y ∈ φ−1(J1) = φ−1(J2).
=⇒ x = φ(y) ∈ J2 for all x ∈ J1.
=⇒ J1 ⊆ J2.
Similarly, J2 ⊆ J1, that is, J1 = J2.
Therefore, θ is one-one.
(2) θ is onto. Let I ∈ X and let J = φ(I).
Then J is an ideal of R0 as φ is onto.
We show that I = φ−1(J) = φ−1(φ(I)).
Let x ∈ φ−1(J).
=⇒ φ(x) ∈ J = φ(I).
=⇒ There exists some y ∈ I such that φ(x) = φ(y).
=⇒ x − y ∈ ker(φ) ⊆ I.
=⇒ x = x − y + y ∈ I for all x ∈ φ−1(J).
=⇒ φ−1(J) ⊆ I
Conversely, let x ∈ I.
=⇒ φ(x) ∈ φ(I) = J.
=⇒ x ∈ φ−1(J) for all x ∈ I.
=⇒ I ⊆ φ−1(J), that is, I = φ−1(J) = θ(J).
=⇒ I ∈ Im(θ) for all I ∈ X.
=⇒ θ is onto.
Therefore, θ is a bijection.
Lemma 14
Let R and R0 be rings and φ : R −→ R0 be an onto
ring homomorphism. Let Y be the set of all prime ideals of R0 and
X = {P ⊆ R | P is a prime ideal of R containing ker(φ)}. Then the
map θ : Y −→ X given by θ(Q) = φ−1(Q) for all Q ∈ Y is a bijection.
Proof. Let Q ∈ Y . Then by Lemma 12 and Lemma 13, φ−1(Q) is a
prime ideal of R containing ker(φ).
=⇒ θ is well defined.
(1) θ is one-one. Note that θ here is same as θ in Lemma 13 with
different domain and codomain. Hence θ is one-one here as well.
(2) θ is onto. Let P ∈ X and let Q = φ(P ).
Then Q is an ideal of R0 as φ is onto.
Furthermore, as in Lemma 13, φ−1(Q) = P .
Claim. Q is a prime ideal of R0.
Suppose, if possible, that Q = R0.
=⇒ 1 ∈ Q.
=⇒ φ(1) = 1 ∈ Q
=⇒ 1 ∈ φ−1(Q) = P , that is, P = R.
This is a contradiction. Therefore, Q 6= R0.
Let x, y ∈ R0 such that xy ∈ Q.
=⇒ There exist a, b ∈ R such that x = φ(a) and y = φ(b) as φ is onto.
=⇒ φ(ab) = φ(a)φ(b) = xy ∈ Q.
=⇒ ab ∈ φ−1(Q) = P .
=⇒ a ∈ P or b ∈ P .
=⇒ x = φ(a) ∈ Q or y = φ(b) ∈ Q.
=⇒ Q is a prime ideal of R0, that is, Q ∈ Y .
As θ(Q) = φ−1(Q) = P , P ∈ Im(θ) for all P ∈ X.
=⇒ θ is onto.
Therefore, θ is a bijection.
Lemma 15 Let R and R0 be rings and φ : R −→ R0 be an onto ring
homomorphism. Let Y be the set of all maximal ideals of R0 and
X = {P ⊆ R | P is a maxima ideal of R containing ker(φ)}. Then
the map θ : Y −→ X given by θ(Q) = φ−1(Q) for all Q ∈ Y is a
bijection.
Proof.
Let Q ∈ Y . Then by Lemma 14, φ−1(Q) is a prime ideal of
R containing ker(φ) (as every maximal ideal is a prime ideal).
We need to show that φ−1(Q) is a maximal ideal of R.
Note that φ−1(Q) 6= R as φ−1(Q) is a prime ideal.
Let I be an ideal of R such that φ−1(Q) ⊆ I and φ−1(Q) 6= I.
We shall show that I = R.
As is Lemma 13, Q = φ(φ−1(Q)) 6= φ(I) and clearly, Q = φ(φ−1(Q)) ⊆
φ(I).
=⇒ φ(I) = R0 as Q is a maximal ideal of R0.
=⇒ φ(1) = 1 ∈ φ(I) and hence 1 ∈ φ−1(φ(I)) = I (by Lemma 13).
=⇒ I = R.
Therefore, φ−1(Q) is a maximal ideal.
=⇒ θ is well defined.
(1) θ is one-one. Note that θ here is same as θ in Lemma 13 with
different domain and codomain. Hence θ is one-one here as well.
(2) θ is onto. Let P ∈ X and let Q = φ(P ).
Then by Lemma 14, Q is a prime ideal of R0 and φ−1(Q) = P as φ is
onto. In particular, Q 6= R0.
Claim. Q is a maximal ideal of R0.
We already know that Q 6= R0.
Let J be an ideal of R0 such that Q ⊆ J and Q 6= J.
=⇒ P = φ−1(Q) ⊆ φ−1(J) and by Lemma 13, P = φ−1(Q) 6= φ−1(J).
As P is maximal, φ−1(J) = R.
=⇒ 1 ∈ φ−1(J) and therefore, 1 = φ(1) ∈ J.
=⇒ J = R0.
=⇒ Q is a maximal ideal of R0, that is, Q ∈ Y .
As θ(Q) = φ−1(Q) = P , P ∈ Im(θ) for all P ∈ X.
=⇒ θ is onto.
Therefore, θ is a bijection.
14. Chinese remainder theorem
Comments. In number theory we have the famous Chinese remainder theorem which states the following: Let m1, . . . , mr be pairwise
coprime integers and a1, . . . , ar ∈ Z. Then there exists a ∈ Z such that
a ≡ ai (mod mi) for all i = 1, . . . , r.
We shall prove the algebraic version of this result.
We shall need some definitions and elementary results.
Ideal I and J are called comaximal if I + J = R. For example two
distinct maximal ideals are always comaximal. This is algebraic equivalent of the notion of coprime integers in Z.
We prove two lemmas now.
Lemma 16 Let I1, . . . , In be ideals of R. Let P be a prime ideal of
R such that I1 . . . In ⊆ P . Then there exists some j ∈ {1, . . . , n} such
that Ij ⊆ P .
Proof. Suppose, if possible, that none of the Ij ’s is contained in P .
Let j ∈ {1, . . . , n}.
=⇒ There exists some aj ∈ Ij such that aj 6∈ P .
But a1 . . . an ∈ I1 . . . In ⊆ P .
=⇒ There exists some k ∈ {1, . . . , n} such that ak ∈ P , a contradiction.
=⇒ There exists some j ∈ {1, . . . , n} such that Ij ⊆ P .
Lemma 17 Let I1, . . . , In be pairwise comaximal ideals. Then
(a) ∩n−1
j=1 Ij and In are comaximal.
(b) ∩n
j=1 Ij = I1 . . . In .
(a) Suppose, if possible, that ∩n−1
j=1 Ij and In are not
comaximal.
Proof.
=⇒ ∩n−1
j=1 Ij + In 6= R.
=⇒ There exists a maximal ideal m such that ∩n−1
j=1 Ij + In ⊆ m.
Clearly, ∩n−1
j=1 Ij ⊆ m and In ⊆ m.
=⇒ I1 . . . In−1 ⊆ ∩n−1
j=1 Ij ⊆ m
=⇒ There exists some j ∈ {1, . . . , n − 1} such that Ij ⊆ m.
=⇒ R = Ij + In ⊆ m.
=⇒ m = R, a contradiction.
=⇒ ∩n−1
j=1 Ij and In are comaximal.
(b) We use induction of n.
If n = 1 then there is nothing to prove.
So let n = 2.
=⇒ I1 + I2 = R.
=⇒ There exist x ∈ I1 and y ∈ I2 such that x + y = 1.
Let a ∈ I1 ∩ I2.
=⇒ a = xa + ay.
Note that as x ∈ I1 and a ∈ I2, xa ∈ I1I2.
Similarly, ay ∈ I1I2.
=⇒ a ∈ I1I2, that is, I1 ∩ I2 ⊆ I1I2.
As I1I2 ⊆ I1 ∩ I2, we are done in this case.
Now let n ≥ 3.
n−1
Then by induction, I1 . . . In−1 = ∩j=1
Ij .
n−1
n−1
n−1
=⇒ ∩n
I
=
(∩
I
)
∩
I
=
(∩
I
)I
as
(∩
n
n
j
j
j
j=1
j=1
j=1
j=1 Ij ) and In are comaximal.
=⇒ ∩n
j=1 Ij = (I1 . . . In−1 )In = I1 . . . In .
Theorem. [Chinese remainder theorem] Let I1, . . . , In be pairwise comaximal ideals of R. Then the natural ring homomorphism φ : R −→
Qn
n
j=1 R/Ij is onto with kernel ∩j=1 Ij .
Proof. The map φ is given by φ(a) = (a + I1, . . . , a + In).
Let k ∈ {1, . . . , n}.
Put Pk = ∩n
j=1,j6=k Ij .
As Pk and Ik are comaximal, Pk + Ik = R.
=⇒ There exist ak ∈ Pk and bk ∈ Ik such that ak + bk = 1.
We construct ak and bk for all k = 1, . . . , n.
Note that ak ∈ Ij for all j = 1, . . . , k − 1, k + 1, . . . , n and bk ∈ Ik .
We shall now prove that φ is onto.
Qn
Let x ∈ j=1 R/Ij .
Then there exist x1, . . . , xn ∈ R such that x = (x1 + I1, . . . , xn + In).
Put a =
Pn
k=1 ak xk .
Let l ∈ {1, . . . , n}.
P
Pn
=⇒ a + Il = ( n
a
x
)
+
I
=
l
k=1 k k
k=1 (ak xk + Il ).
Consider the term ak xk + Il of this summation. We note that
(1) if k 6= l, then ak ∈ Il and hence ak xk + Il = Il
(2) if k = l then al xl + Il = (1 − bl )xl + Il = (xl − bl xl ) + Il =
xl + Il − bl xl + Il = xl + Il as bl ∈ Il .
=⇒ a + Il =
Pn
k=1 (ak xk + Il ) = xl + Il for all l = 1, . . . , n.
=⇒ x = (x1 + I1, . . . , xn + In) = (a + I1, . . . , a + In) = φ(a).
=⇒ x is in the image of φ.
AS x is arbitrary, φ is onto.
We now show that ker(φ) = ∩n
j=1 Ij .
Let a ∈ ker(φ).
=⇒ (a + I1, . . . , a + In) = φ(a) = (I1, . . . , In).
=⇒ a + Ij = Ij or aj ∈ Ij for all j = 1, . . . , n.
n
=⇒ a ∈ ∩n
j=1 Ij or ker(φ) ⊆ ∩j=1 Ij .
Conversely, let a ∈ ∩n
j=1 Ij .
=⇒ φ(a) = (a + I1, . . . , a + In) = (I1, . . . , In).
=⇒ a ∈ ker(φ).
n I = ker(φ).
=⇒ ∩n
I
⊆
ker(φ)
or
∩
j
j=1
j=1 j
Lecture – 06
15.
Modules
In linear algebra we have vector-spaces over a field. We have similar
objects, called R-modules over an arbitrary commutative ring R with
unity. These R-modules have a similar definition. However, they do
not exhibit all properties of vector-spaces as in an arbitrary commutative ring with unity nonzero elements are not necessarily invertible.
Let M be an abelian group. We denote the binary operation of M by
+ and call it addition. The identity of + shall be denoted by 0 and
the additive inverse of x ∈ M shall be denoted by −x. For x, y ∈ M ,
by x − y we mean x + (−y).
Definition. Let (M, +) be an abelian group and let R be a ring (commutative with unity). Then M is said to be an R-module or a module
over R if there exists a map . : R × M −→ M (we denote the image of
(a, x) under . by a.x) such that the following conditions are satisfied:
(1) 1.x = x for all x ∈ M .
(2) a.(x + y) = a.x + a.y for all a ∈ R and x, y ∈ M .
(3) (a + b).x = a.x + b.x for all a, b ∈ R and x ∈ M .
(4) a.(b.x) = (ab).x for all a, b ∈ R and x ∈ M .
Comments. (1) Clearly, the above is same as the scalar multiplication defined in case of vector spaces.
(2) There are two binary operation in above definition which are denoted by +, namely, addition of R and addition of M . However, it
causes no ambiguity if we observe the elements between whom this
operation is performed. For example in (3), a and b are elements of
R and therefore the + between a and b is of R. Similarly in (2) the
plus between x and y is addition of M . Furthermore, in the right hand
side of (2) and (3), + is the addition of M . After a while, with more
familiarity, any doubt about this situation will vanish.
We look at some examples:
(1) R is a module over itself for a.x = ax, the multiplication in R.
(2) Let n ∈ N. Let M = Rn = R × . . . × R (n times). Then Rn is an
abelian group under componentwise addition. Define a.(x1, . . . , xn) =
(ax1, . . . , axn). Then Rn becomes an R-module.
(3) Let (M, +) be an abelian group. Let n ∈ Z and x ∈ M . Define
n.x = nx. Then M becomes a Z-module. In particular, Q, R, C are
Z-modules.
(4) Let M be the trivial group {0}. Then M is abelian. Let R be any
ring. Define a.0 = 0 for all a ∈ R. Then M becomes an R-module.
This M is called the trivial R-module and is denoted by 0.
(5) Let I be an ideal of R. Then I is an R-module for a.x = ax, the
multiplication in R.
(6) Vector-spaces over a field are also modules over the field.
(7) Let R and A be rings and let φ : R −→ A be a ring homomorphism. For a ∈ R and x ∈ A define a.x = φ(a)x. Then A becomes an
R-module. When A is treated as an R-module in this fashion, A is
also called an R-algebra.
Comments.
(1) An abelian group may be a module over many
rings.
(2) There may be many different R-module structures on a given
abelian group M over a given ring R.
We prove a lemma now.
Lemma 18 Let M be an R-module. Then
(a) a.0 = 0 for all a ∈ R.
(b) 0.x = 0 for all x ∈ M .
(c) (−a).x = −(a.x) = a.(−x) for all a ∈ R and all x ∈ M .
Proof. (1) a.0 = a.(0 + 0) = a.0 + a.0 =⇒ a.0 = 0.
(2) 0.x = (0 + 0).x = 0.x + 0.x =⇒ 0.x = 0.
(3) a.x + (−a).x = (a + (−a)).x = 0.x = 0.
=⇒ The additive inverse of a.x is (−a).x.
Notationally, this means −(a.x) = (−a).x.
Similarly, −(a.x) = a.(−x).
Comments. (1) Let M be an R-module. Let a ∈ R and x ∈ M . Then
instead of a.x we write ax.
(2) What we have defined above is a left R-module, because the
elements of ring appear in the left. In a similar fashion we can define
right R-modules. Then the map . will be from M × R to M and the
image of (x, a) shall be denoted by xa. However, everything we prove
about left R-modules shall be analogously true for right R-modules
as well and vice versa. Hence we shall consider only left R-modules
and by an R-module we shall always mean a left R-module.
(3) Let a ∈ R and x ∈ M . Then ax = 0 6=⇒ either a = 0 or x = 0.
16. Submodules and quotient modules
Let M be an R-module and let N be a subgroup of M . If ax ∈ N for
all a ∈ R and x ∈ N , then the map R × N −→ N give by (a, x) −→ ax,
induces an R-module structure on N . Therefore, N becomes an Rmodule in its own right.
Such an N is called an R-submodule or
simply a submodule of M .
Comments. (1) Note that for N to be submodule of M the R-module
structure on N must be induced by the R-module structure of M . If
N is considered with some other R-module structure then N will not
be called a submodule of M .
(2) Every ideal of R is a submodule of R.
(3) M is a submodule of M .
(4) The trivial subgroup is also a submodule of M by Lemma 18.
(5) The intersection of any family of submodule of M is again a
submodule of M .
We now prove a simple but a useful lemma.
Lemma 19 Let M be an R-module and let N ⊆ M be a nonempty
set. Then the following statements are equivalent:
(a) N is a submodule of M .
(b) ax + y ∈ N for all a ∈ R and x, y ∈ N .
Proof.
(a) =⇒ (b) Let a ∈ R and x, y ∈ N . As N is a submodule,
ax ∈ N . Therefore, ax + y ∈ N .
(b) =⇒ (a) We need to show that N is a subgroup of M and that
ax ∈ N for all a ∈ R and x ∈ N .
As N is nonempty, there exists some z ∈ N .
=⇒ (−1)z + z ∈ N (taking x = z = y and a = −1).
=⇒ 0 ∈ N .
Let b ∈ R and u, v ∈ N .
=⇒ −u = (−1)u + 0 ∈ N (taking x = u and a = −1),
u + v = 1u + v ∈ N (taking x = u, y = v and a = 1) and
bu = bu + 0 ∈ N (taking x = u, y = 0 and a = b).
As b ∈ R and u, v ∈ N are arbitrary, N is a submodule of M .
Definition. Let M be an R-module and let N be a submodule of M .
Treating M as group and N as a subgroup of M , we construct the
quotient group M/N . In M/N there is a natural R-module structure
induced by M , which we discuss below.
Let a ∈ R and X ∈ M/N .
Then there exists some x ∈ M (not unique) such that X = x + N .
Let y ∈ M such that X = y + N as well.
=⇒ x − y ∈ N .
=⇒ a(x − y) ∈ N .
But a(x − y) = a(x + (−y)) = ax + a(−y) = ax − ay.
=⇒ ax − ay ∈ N .
=⇒ ax + N = ay + N .
=⇒ ax + N does not depend on the choice of x.
We define a.X = ax + N .
Note that this is well defined by the above discussion.
Let X, Y ∈ M/N and let a, b ∈ R.
Let x, y ∈ M such that X = x + N and Y = y + N .
Then we have
(1) 1.X = (1x) + N = x + N = X
(2) a.(X + Y ) = a((x + N ) + (y + N ))
= a((x + y) + N )
= a(x + y) + N
= ax + ay + N
= (ax + N ) + (ay + N )
= a.X + a.Y .
(3) (a + b).X = (a + b)x + N
= ax + bx + N
= (ax + N ) + (bx + N )
= a.X + b.X.
(4) a.(b.X) = a.(bx + N ) = a(bx) + N = (ab)x + N = (ab).X.
As X, Y ∈ M/N and a, b ∈ R are arbitrary, M/N is an R-module.
17. Submodule generated by a set.
Comments. If we take infinitely elements of M then their sum (as
in the case of a ring), in general, can not be defined because it
can not be computed. However, if we have infinitely elements, say,
{xλ | λ ∈ Λ} of M , such that all but finitely many of xλ’s are zero, then
P
the sum λ∈Λ xλ is assigned the meaning in the following way: first
throw away all zero terms in the summation, then whatever is left
P
is a finite sum, which cab be computed. The sum λ∈Λ xλ is taken
as the sum of these finitely many nonzero elements. If all terms in
above summation are 0 then the sum is defined to be 0.
Consider a subset {xλ | λ ∈ Λ} of M , where Λ is an
indexing set. By a linear combination of xλ’s we mean an element of
Definition.
P
the type λ∈Λ aλxλ, where aλ’s are elements of R and they are zero
for all but for finitely values of λ. This means, all but finitely many
terms of
P
λ∈Λ aλ xλ are zero and therefore, it is well defined.
What we discuss now is similar to the ideal generated by a set. Let
S be a subset of an R-module M , may even be empty. Let
F = {N ⊆ M | N is a submodule of M , S ⊆ N }.
Then M ∈ F and hence F is nonempty.
The intersection of all members of F is again a submodule of M con-
taining S. We call this intersection as the submodule generated by S.
e
In this course we shall denote it by S.
Note that Se is also an element of F and is contained in every member
of F .
In other words, if any submodule of M contains S then it also contains
e
the submodule S.
In this sense Se is the smallest submodule of M containing S.
We now look at some special cases of S.
(1) Consider the case when S = ∅. Then F contains all the submodules of M . Hence, the submodule {0} as well. Therefore, Se = {0} in
this case.
(2) Consider the case when S itself is a submodule of M . Then S ∈ F
e That is, S
e = S.
and hence Se ⊆ S ⊆ S.
(3) Consider when S ⊆ T . As S ⊆ T ⊆ Te we have Se ⊆ Te .
We now determine the structure of Se when S is finite, that is, S =
{x1, . . . , xn}. Put
N = {a1x1 + · · · + anxn | a1, . . . , an ∈ R are arbitrary}.
Note that xi = 0a1 + · · · + 0xi−1 + 1xi + 0xi+1 + · · · + 0xn.
=⇒ xi ∈ N for all i = 1, . . . , n.
=⇒ S ⊆ N .
Claim. N is a submodule of M .
Let x, y ∈ N and c ∈ R.
Then there exist a1, . . . , an, b1, . . . , bn ∈ R such that x = a1x1+· · ·+anxn
and y = b1x1 + · · · + bnxn.
=⇒ cx + y = (ca1 + b1)x1 + · · · + (can + bn)xn ∈ N .
Since x, y, c are arbitrary, N is submodule of M .
=⇒ Se ⊆ N as S ⊆ N .
Also x1, . . . , xn ∈ S ⊆ Se
=⇒ x = a1x1 + · · · + anxn ∈ Se (as Se is a submodule).
e Hence N = S.
e
This means N ⊆ S.
Consider now the case when S is infinite. Let
X = {T ⊆ S | T is finite}.
We shall show that Se = ∪T ∈X Te . Put N = ∪T ∈X Te .
Let x ∈ S. Then {x} ∈ X.
g ⊆ N.
=⇒ x ∈ {x}
=⇒ S ⊆ N .
We now show that N is a submodule of M .
Let x, y ∈ N and c ∈ R.
=⇒ There exist finite subsets T1 and T2 of S such that x ∈ Tf1 and
y ∈ Tf2.
^
=⇒ x ∈ Tf1 ⊆ T^
1 ∪ T2 as T1 ⊆ T1 ∪ T2 , and similarly, y ∈ T1 ∪ T2 .
=⇒ cx + y ∈ T^
1 ∪ T2 ⊆ N as T1 ∪ T2 is a finite subset of S.
Hence N is a submodule of M .
As S ⊆ N , we have Se ⊆ N .
Conversely, let T ∈ X.
e
Then T ⊆ S and hence Te ⊆ S.
=⇒ N = ∪T ∈X Te ⊆ Se or Se = N = ∪T ∈X Te .
Therefore, if x ∈ Se then there exist a finite subset T of S such that
x ∈ Te . Let T = {xλ1 , . . . , xλn }. Then there exist a1, . . . , an ∈ R such
P
that x = a1xλ1 + · · · + anxλn . Then x = λ∈Λ bλxλ, where bλ = 0 for all
λ ∈ Λ \ {λ1, . . . , λn} and bλi = ai for all i = 1, . . . , n. Hence x is a linear
combination of elements of S. Conversely, every linear combination
e Hence S
e is precisely the set of all linear
of elements of S is in S.
combinations of elements of S.
Lecture – 07
18. Homomorphism
Let M and L be R-modules. A map φ : M −→ L is said to be a homomorphism of R-modules or an R-linear map (or simply a linear map)
if φ(x + y) = φ(x) + φ(y) and φ(ax) = aφ(x) for all x, y ∈ M and a ∈ R.
We look at some examples.
(1) The map IM : M −→ M given by IM (x) = x is R-linear. This map
is called the identity map of M .
(2) The φ : M −→ L given by φ(x) = 0 for all x ∈ M is R-linear. This
map is called the zero map and is denoted by 0.
(3) Let a ∈ R. The map φ : M −→ M given by φ(x) = ax is R-linear.
This map is called multiplication by a.
(4) Let N be a submodule of M . Then the inclusion map : N −→ M
and the natural map π : M −→ M/N are R-linear.
Comments. Let M and L be R-modules and φ : M −→ L be R-linear.
(1) ker(φ) is a submodule of M .
(2) Im(φ) is a submodule of L.
(3) Let N be a submodule of M . Then φ(N ) is a submodule of L.
(4) Let K be a submodule of L. Then φ−1(K) is a submodule of M .
We now prove some simple lemmas.
Lemma 20
Let M, L be R-modules and let φ : M −→ L be a (set
theoretic) map. Then the following statements are equivalent:
(a) φ is R-linear.
(b) φ(ax + y) = aφ(x) + φ(y) for all a ∈ R and x, y ∈ M .
Proof. (a) =⇒ (b) Let a ∈ R and x, y ∈ M .
=⇒ φ(ax + y) = φ(ax) + φ(y) = aφ(x) + φ(y).
(b) =⇒ (a) Let a ∈ R and x, y ∈ M .
=⇒ φ(x + y) = φ(1x + y) = 1φ(x) + φ(y) = φ(x) + φ(y) for x, y ∈ M .
In particular, φ is a group homomorphism of (M, +) to (L, +).
=⇒ φ(0) = 0.
Also, φ(ax) = φ(ax + 0) = aφ(x) + φ(0) = aφ(x) + 0 = aφ(x) for all
a ∈ R and x ∈ M .
Therefore, φ is R-linear.
Lemma 21 Let M, L be R-modules and φ, ψ : M −→ L be R-linear
maps. Let S be a set of generators of M . If φ(x) = ψ(x) for all x ∈ S
then φ = ψ.
Proof. Let N = {x ∈ M | φ(x) = ψ(x)}.
=⇒ S ⊆ N .
Claim. N is a submodule of M .
Clearly, 0 ∈ N as φ(0) = 0 = ψ(0).
Let a ∈ R and x, y ∈ N .
=⇒ φ(x) = ψ(x) and φ(y) = ψ(y).
=⇒ φ(ax + y) = aφ(x) + φ(y) = aψ(x) + ψ(y) = ψ(ax + y).
=⇒ ax + y ∈ N .
As a ∈ R and x, y ∈ N are arbitrary, N is a submodule of M .
=⇒ The submodule generated by S is contained in N , that is, M ⊆
N ⊆ M.
=⇒ N = M
=⇒ φ(x) = ψ(x) for all x ∈ M .
=⇒ φ = ψ.
Comments. (1) Composition of R-linear maps is again an R-linear
map.
More precisely, let M, L, K be R-module, φ : M −→ L and
ψ : L −→ K be R-linear maps. Then ψ oφ : M −→ K is R-linear.
We check this- Let x, y ∈ M and a ∈ R.
=⇒ (ψ oφ)(ax + y) = ψ(φ(ax + y))
= ψ(aφ(x) + φ(y))
= aψ(φ(x)) + ψ(φ(y))
= a(ψ oφ)(x) + (ψ oφ)(y).
As x, y ∈ M and a ∈ R are arbitrary, ψ oφ is R-linear.
(2) If either ψ or φ is the 0 map then so is ψ oφ, that is, ψ o0 = 0 and
0oφ = 0.
Let M, L be R-modules and let φ : M −→ L be an R-linear map.
Then φ is called an isomorphism if φ is one-one and onto. If φ is
an isomorphism then φ−1 is also R-linear. Note that φ−1 is a group
homomorphism and φ(aφ−1(x)) = aφ(φ−1(x)) = ax or equivalently,
φ−1(ax) = aφ−1(x).
19. Fundamental theorem of homomorphism
We have fundamental theorem of homomorphism in group theory and
ring theory. We also have a module theoretic version of this result.
More precisely we have the following variant:
Theorem. Let M and L be R-modules and let φ : M −→ L be an Rlinear map. Let N be a submodule of M such that N ⊆ ker(φ). Then
there exist a unique R-linear map ψ : M/N −→ L such that ψ oπ = φ,
where π : M −→ M/N is the natural map. Furthermore,
(a) Im(φ) = Im(ψ).
(b) ψ is one-one iff N = ker(φ).
(c) ψ is onto iff φ is onto.
Proof.
The proof of this result is nearly identical to that of group-
theoretic case.
Let X ∈ M/N .
=⇒ There exists x ∈ M (not necessarily unique) such that X = x + N .
Claim. φ(x) does not depend on the choice of x.
Let y ∈ M such that X = y + N .
=⇒ x − y ∈ N ⊆ ker(φ).
=⇒ φ(x − y) = 0 = φ(x) − φ(y) or φ(x) = φ(y).
Therefore, we are done.
Define ψ : M/N −→ L by ψ(X) = φ(x) if X = x + N , for all X ∈ M/N .
=⇒ ψ is well defined.
Claim. ψ is R-linear.
Let X, Y ∈ M/N and a ∈ R.
=⇒ There exist x, y ∈ M such that X = x + N and Y = y + N .
=⇒ aX + Y = a(x + N ) + (y + N ) = (ax + N ) + (y + N ) = (ax + y) + N .
=⇒ ψ(X) = φ(x), ψ(Y ) = φ(y) and ψ(aX + Y ) = φ(ax + y).
=⇒ ψ(aX + Y ) = φ(ax + y) = aφ(x) + φ(y) = aψ(X) + ψ(Y ).
As X, Y ∈ M/N and a ∈ R are arbitrary, ψ is R-linear.
Claim. ψ oπ = φ and ψ is unique with this property.
Let x ∈ M .
=⇒ (ψ oπ)(x) = ψ(π(x)) = ψ(x + N ) = φ(x) for all x ∈ M .
=⇒ ψ oπ = φ.
Uniqueness.
Let ψ 0 : M/N −→ L be an R-linear map such that
ψ 0oπ = φ.
We need to show that ψ 0 = ψ.
Let X ∈ M/N and let x ∈ M such that X = x + N .
=⇒ ψ 0(X) = ψ 0(x + N ) = ψ 0(π(x)) = (ψ 0oπ)(x)
= φ(x) = (ψ oπ)(x) = ψ(π(x)) = ψ(x + N ) = ψ(X).
As X ∈ M/N is arbitrary, ψ 0 = ψ.
We now prove the other parts.
(a) Let y ∈ Im(φ).
=⇒ The exists some x ∈ M such that y = φ(x)
=⇒ y = (ψ oπ)(x) = ψ(π(x)) = ψ(x + N ).
=⇒ y ∈ Im(ψ) for all y ∈ Im(φ).
=⇒ Im(φ) ⊆ Im(ψ).
Similarly, Im(ψ) ⊆ Im(φ), that is, Im(φ) = Im(ψ).
(b) First suppose that N = ker(φ).
We need to show that ψ is one-one.
It suffices to show that ker(ψ) = 0.
Let X ∈ ker(ψ) and let x ∈ M such that X = x + N .
=⇒ φ(x) = ψ(X) = 0.
=⇒ x ∈ ker(φ) = N .
=⇒ X = x + N = 0.
=⇒ ψ is one-one.
Conversely, let ψ is one-one.
We need to show that ker(φ) = N .
Let x ∈ ker(φ).
=⇒ ψ(x + N ) = φ(x) = 0.
=⇒ x + N ∈ ker(ψ).
=⇒ x + N = 0 = N as ψ is one-one.
=⇒ x ∈ N for all x ∈ ker(φ).
=⇒ ker(φ) ⊆ N .
As N ⊆ ker(φ), we have ker(φ) = N .
(c) By (a), Im(φ) = Im(ψ).
=⇒ Im(φ) = L iff Im(ψ) = L.
=⇒ φ is onto iff ψ is onto.
20. Annihilator
Let M be an R-module and let I = {a ∈ R | ax = 0 for all x ∈ M }.
Then 0 ∈ I.
Let α ∈ R and a, b ∈ I.
Then (a − b)x = ax − bx = 0 and (αa)x = α(ax) = α0 = 0 for all
x ∈ M.
Therefore, a − b ∈ I and αa ∈ I for all α ∈ R and a, b ∈ I.
=⇒ I is an ideal of R
The ideal I is called the annihilator of M and is denoted by annR (M )
(or ann(M ), if there is no ambiguity about the ring under consideration).
Let J ⊆ ann(M ) be an ideal of R.
We shall define an R/J-module structure on M as follows:
Define (a + J).x = ax for all a ∈ R and x ∈ M .
Note that ax corresponds to the R-module structure of M .
We need to show that above is well defined.
Let a + J = b + J.
=⇒ a − b ∈ J ⊆ ann(M ).
=⇒ (a − b)x = 0 or ax = bx for all x ∈ M .
Therefore, the above structure is well defined.
Claim. M is an R/J-module.
(1) (1 + J).x = 1x = x for all x ∈ M .
(2) (a + J).(x + y) = a(x + y)
= ax + ay
= (a + J).x + (a + J).y
for all a + J ∈ R/J and x, y ∈ M .
(3) ((a + J) + (b + J)).x = (a + b + J).x
= (a + b)x
= ax + bx
= (a + J).x + (b + J).x
for all a + J, b + J ∈ R/J and x ∈ M .
(4) ((a + J)(b + J)).x = (ab + J).x
= (ab)x = a(bx)
= (a + J).(bx) = (a + J).((b + J).x)
for all a + J, b + J ∈ R/J and x ∈ M .
Therefore, M is also an R/J-module.
Comments. (1) All R-submodules of M are also R/J-submodules of
M and vice versa.
Proof.
Let N be an R-submodule of M .
=⇒ N is a subgroup of M and ax ∈ N for all a ∈ R.
=⇒ N is a subgroup of M and (a + J).x = ax ∈ N for all a + J ∈ R/J.
=⇒ N is an R/J-submodule of M .
Similarly, if K is an R/J-submodule of M then it is also an Rsubmodule of M .
(2) If S is a set of generators of M considered as an R-module then
S is a also set of generators of M considered as an R/J-module and
vice versa.
Proof. Let S be a generators of M considered as an R-module. Let
x ∈ M . Then there exist x1, . . . , xn ∈ S and a1, . . . , an ∈ R such that
x = a1x1 + · · · + anxn.
=⇒ x = (a1 + J).x1 + · · · + (an + J).xn.
=⇒ S generates M as an R/J-module.
Similarly, if T generates M as an an R/J-module it also generates M
as an R-module.
(3) Let I be an ideal of R. Then I ⊆ annR (M/IM ) and hence M/IM
has an R/I-module structure.
Proof. Let a ∈ I and let X ∈ M/IM . Then there exists some x ∈ M
such that X = x + IM .
As ax ∈ IM , aX = a(x + IM ) = ax + IM = IM = 0 for all X ∈ M/IM .
=⇒ a ∈ annR (M/IM ) for all a ∈ I.
=⇒ I ⊆ annR (M/IM ).
Hence M/IM has an R/I-module structure.
(4) Let m be a maximal ideal of R. Then M/mM has an R/m-module
structure. In other words, M/mM is a vectorspace over the field R/m.
Lecture – 08
21. A special class of submodules
We discuss a certain class of submodules which can be constructed
in any module.
Let M be an R-module and let I be an ideal of R. Then the submodule
generated by the set {ax | a ∈ I, x ∈ M } is denoted by IM . Note that
IM is, by definition, a submodule of M and it need not be equal to
the set {ax | a ∈ I, x ∈ M }.
Furthermore, for a ∈ R, we denote the set {ax | x ∈ M } by aM .
We now look at the structure of the submodule IM .
Lemma 22 Let M be an R-module and let I be an ideal of R. Let
S = {ax | a ∈ I, x ∈ M }. Then IM = {x1 + · · · + xk | xi ∈ S, k ≥ 1}.
Proof. Let a ∈ R and let x ∈ S.
=⇒ There exist b ∈ I and y ∈ M such that x = by.
=⇒ ax = aby = (ab)y ∈ S as ab ∈ I.
=⇒ S is closed under multiplication by elements of R.
Put X = {x1 + · · · + xk | xi ∈ S, k ≥ 1}.
=⇒ S ⊆ X and X ⊆ IM .
Note that elements of X are precisely those which are sums of finitely
many elements of S.
Claim. X is a submodule of M .
Let x, y ∈ X and a ∈ R.
=⇒ There exist some l, m ≥ 1, x1, . . . , xl , y1, . . . , ym ∈ S such that
x = x1 + · · · + xl and y = y1 + · · · + ym.
=⇒ ax + y = ax1 + · · · + axl + y1 + · · · + ym is a sum of finitely many
elements of S.
=⇒ ax + y ∈ X.
=⇒ X is a submodule of M .
As X contains S, IM = Se ⊆ X.
=⇒ IM = X.
Comments. (1) Suppose that M is generated by the set S = {xλ |
λ ∈ Λ}, where Λ is an indexing set. Let I be an ideal of R.
Let x ∈ M and let a ∈ I.
P
Then x = λ∈Λ aλxλ, where aλ’s are zero for all but finitely many
values of λ ∈ Λ.
P
Then ax = λ∈Λ aaλxλ.
As aaλ ∈ I for all λ ∈ Λ, ax is a linear combination of xλ’s with coefficients coming from I.
Therefore, every element of IM is a linear combination of xλ’s with
coefficients coming from I, as it is finite sums of elements of the form
by, where b ∈ I and y ∈ M .
(2) Let a ∈ R and let I = (a). Then IM is generated by the set
S = {bx | b ∈ I, x ∈ M }.
Let b ∈ I, x ∈ M .
Then b = ca for some c ∈ R.
Therefore, bx = cax = a(cx).
This means every element of S is obtained by multiplying a with some
element of M .
As all elements of IM are sums of finitely many such elements, they
also have the same form. Hence IM = aM .
Lemma 23 Let M be an R-module and I, J be ideals of R. Then
I(JM ) = (IJ)M .
Proof. Let x ∈ I(JM ).
=⇒ There exist a1, . . . , an ∈ I and x1, . . . , xn ∈ JM such that x =
Pn
j=1 aj xj .
=⇒ For every j ∈ {1, . . . , n}, there exist bj1, . . . , bjmj ∈ J and
Pmj
yj1, . . . , yjmj ∈ M such that xj = k=1
bjk yjk .
Pm
Pn
Pmj
Pn
j
=⇒ x = j=1 aj
b
y
=
j=1 k=1 (aj bjk )yjk ∈ (IJ)M .
k=1 jk jk
As x ∈ I(JM ) is arbitrary, I(JM ) ⊆ (IJ)M .
Put S = {ax | a ∈ IJ, x ∈ M }.
As (IJ)M is generated by S, in order to show that (IJ)M ⊆ I(JM ) it
suffices to show that S ⊆ I(JM ).
Let x ∈ S.
=⇒ There exist b ∈ IJ and y ∈ M such that x = by
=⇒ There exist a1, . . . , an ∈ I and b1, . . . , bn ∈ J such that b =
Pn
i=1 ai bi .
P
Pn
n
=⇒ x = by =
a
b
y
=
i=1 i i
i=1 ai (bi y).
=⇒ For all i = 1, . . . , n, ai(biy) ∈ I(JM ) as biy ∈ JM and ai ∈ I.
=⇒ x ∈ I(JM ) for all x ∈ S.
=⇒ S ⊆ I(JM ).
=⇒ (IJ)M = Se ⊆ I(JM ).
=⇒ (IJ)M = I(JM ).
Definition. Let M be an R-module and let N be a submodule of M .
e If there exists a finite
We say that N is generated by a set S if N = S.
set S such that N = Se then we say that N is finitely generated.
Lemma 24 Let M be an R-module generated by a set S. Let L be
an R-module and let φ : M −→ L be an onto R-linear map. Then L is
generated by T = φ(S) = {φ(x) | x ∈ S}.
Proof. Let y ∈ L.
=⇒ There exists some x ∈ M such that φ(x) = y.
=⇒ There exist a1, . . . , an ∈ R and y1, . . . , yn ∈ S such that x =
Pn
i=1 ai yi .
Pn
=⇒ y = φ(x) = i=0 aiφ(yi) ∈ Te for all y ∈ L.
=⇒ L ⊆ Te ⊆ L, that is, Te = L.
In other words, L is generated by T .
Lemma 25 Let M and L be R-modules and φ : M −→ L be an onto
R-linear map.
Let I be an ideal of R such that M = IM .
Then
L = IL.
Proof. As M = IM , M is generated by S = {ax | a ∈ I, x ∈ M }.
As φ is onto, L is generated by T = {φ(y) | y ∈ S}.
We show that T ⊆ IL.
Let z ∈ T .
=⇒ There exist some a ∈ I and x ∈ M such that z = φ(ax) = aφ(x).
=⇒ z ∈ IL for all z ∈ T . Therefore, T ⊆ IL.
=⇒ L = Te ⊆ IL ⊆ L, that is, L = IL.
22. Nakayama Lemma and its applications
Lemma 26 [Nakayama Lemma] Let M be a finitely generated Rmodule such that M = IM , where I is ideal contained in the Jacobson
radical J(R) of R. Then M = 0.
Proof. We use a contrapositive argument.
Assume, if possible, that M 6= 0.
Since M is finitely generated, there are many finite subsets of M which
generate M . Among all such subsets, choose the one which has least
number of elements and denote it by S.
e S 6= ∅.
As M 6= 0 and M = S,
Let |S| = n ≥ 1.
Note that by choice of S, no subset of M containing less than n elements can generate M .
Let S = {x1, . . . , xn}.
Pn
As xn ∈ M = IM , there exist a1, . . . , an ∈ I such that xn = i=1 aixi.
P
=⇒ (1 − an)xn = n−1
j=1 aj xj .
As an ∈ I ⊆ J(R), 1 − an is a unit.
Therefore, there exists some c ∈ R such that c(1 − an) = 1.
=⇒ xn = 1xn = c(1 − an)xn =
Pn−1
j=1 caj xj . . . . . . . . . . . . . . . . . . . . . . . (1)
Put T = {x1, . . . , xn−1}. Then T ⊆ S.
e
=⇒ Te ⊆ S.
On the other hand, by (1), xn ∈ Te .
=⇒ S = T ∪ {xn} ⊆ Te .
=⇒ Se ⊆ Te .
=⇒ M = Se = Te .
This means, M is generated by the set T which has only n − 1 elements, a contradiction.
=⇒ M = 0.
Sum of submodules. Let Λ be an indexing set. Suppose that for
every λ ∈ Λ we are given a submodule Nλ of an R-module M . Thus
we have a family {Nλ | λ ∈ Λ} of submodules of M . Then their sum
is defined to be the set
{
X
xλ | xλ ∈ Nλ, xλ = 0 for all but finitely many values of λ}.
λ∈Λ
This set is denoted by
P
P
N
.
Then
λ∈Λ λ
λ∈Λ Nλ is a submodule of M
(the proof is identical to the proof given to prove that the sum of
ideals is an ideal).
Lemma 27 Let M be a finitely generated R-module and let N be
a submodule of M . Let I ⊆ J(A) be an ideal such that M = N + IM .
Then M = N
Proof.
phism.
Let L = M/N and let π : M −→ L be the natural homomor-
Then π is onto.
Let M be generated by a finite set S = {x1, . . . , xn}.
=⇒ L is generated by the set T = {π(x1), . . . , π(xn)}, by Lemma 24.
=⇒ L is finitely generated.
We now show that L = IL.
Let X ∈ L.
=⇒ There exists some x ∈ M such that X = x + N .
=⇒ There exist some y ∈ N and z ∈ IM such that x = y + z.
=⇒ There exist some z1, . . . , zm ∈ M and a1, . . . , am ∈ I such that
P
z= m
i=1 ai zi .
=⇒ X = x + N = (y +
Pm
Pm
a
z
)
+
N
=
(y
+
N
)
+
((
i=1 i i
i=1 ai zi ) + N ).
Pm
=⇒ X = i=1 ai(zi + N ) ∈ IL for all X ∈ L.
=⇒ L ⊆ IL ⊆ L, that is, L = IL.
By Nakayama Lemma, L = 0.
=⇒ M = N .
Definition. A ring R is said to be local if R has exactly one maximal
ideal.
Comments. Let R be a local ring and let m be its unique maximal
ideal. Then the Jacobson radical of R, J(R) = m.
Lemma 28
Let R be a local ring with unique maximal ideal m.
Let M be a finitely generated R-module.
Then M/mM is a finite
dimensional vector space over R/m. Let {x1 + mM, . . . , xn + mM } be
a basis of M/mM over R/m. Then M is generated by {x1, . . . , xn}.
Proof. Note that M/mM is finitely generated as an R-module.
=⇒ M/mM is finitely generated as an R/m-module.
=⇒ M/mM is finitely dimensional vector space over R/m.
Let N be the submodule generated by {x1, . . . , xn}.
Claim. M = N + mM .
Let x ∈ M .
=⇒ x + mM is a linear combination of {x1 + mM, . . . , xn + mM } over
R/m.
Pn
=⇒ There exist b1, . . . , bn ∈ R/m such that x + mM = 1=1 bi.(xi +
mM ).
Choose ai ∈ R such that bi = ai + m, i = 1, . . . , n.
P
=⇒ x + mM = n
1=1 (ai + m).(xi + mM )
Pn
= 1=1 ai(xi + mM )
P
= n
1=1 ai xi + mM .
Pn
1=1 ai xi ∈ mM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Pn
Pn
=⇒ x = 1=1 aixi + (x − 1=1 aixi) ∈ N + mM for all x ∈ M .
=⇒ x −
=⇒ M ⊆ N + mM ⊆ M , that is, M = N + mM .
As J(R) = m, by previous lemma, M = N .
Lecture – 09
Theorem. Let M a finitely generated R-module and I be an ideal
of R such that M = IM . Then there exists some a ∈ I such that
(1 − a)M = 0.
Proof. Let M be generated by a finite set S. Suppose that |S| = n.
We use induction on n.
If n = 0 then S is empty.
=⇒ M = 0. So we can take a = 0.
Therefore, we are done in this case.
Let n ≥ 1 and S = {x1, . . . , xn}.
Consider the case when n = 1.
As x1 ∈ IM , there exist a ∈ I such that x1 = ax1.
=⇒ (1 − a)x1 = 0.
Let x ∈ M . Then there exists some b ∈ R such that x = bx1.
=⇒ (1 − a)x = (1 − a)bx1 = b(1 − a)x1 = b0 = 0 for all x ∈ M .
=⇒ (1 − a)M = 0.
Assume now that n ≥ 2.
Let N be the submodule generated by {x1, . . . , xn−1}.
Put L = M/N . Let π : M −→ L be the natural map.
Then, by Lemma 24, L is generated by {π(x1), . . . , π(xn)}.
Note that π(x1) = · · · = π(xn−1) = 0 and zeros do not contribute to
a linear combination.
Therefore, L is also generated by {π(xn)}.
Also, by Lemma 25, L = IL.
By n = 1 case, there exists, some α ∈ I such that (1 − α)L = 0.
Let x ∈ M . Then (1 − α)(x + N ) = 0 = N , that is, ((1 − α)x) + N = N .
=⇒ (1 − α)x ∈ N for all x ∈ M .
=⇒ (1 − α)M ⊆ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Put I1 = (1 − α), the ideal generated by {1 − α}.
Then I1M = (1 − α)M ⊆ N .
=⇒ I1M = I1(IM ) = (I1I)M = (II1)M = I(I1M ) ⊆ IN . . . . . . . . . (2)
Let x ∈ N .
=⇒ (1 − α)x ∈ I1M as 1 − α ∈ I1 and x ∈ M .
=⇒ (1 − α)x ∈ IN , by (2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Also αx ∈ IN as α ∈ I and x ∈ N .
=⇒ x = (1 − α)x + αx ∈ IN for all x ∈ N .
=⇒ N ⊆ IN ⊆ N , that is, N = IN .
As N is finitely generated by n−1 elements, x1, . . . , xn−1, by induction,
there exists some β ∈ I such that (1 − β)N = 0.
Let x ∈ M . Then by (1), (1 − α)x ∈ N .
=⇒ (1 − β)(1 − α)x = 0.
Put a = α + β − αβ.
Then a ∈ I and (1 − a)x = 0 for all x ∈ M .
=⇒ (1 − a)M = 0.
Theorem. Let M be a finitely generated R-modules and let φ : M −→
M be an onto R-linear map. Then φ is also one-one.
Proof. Let A = R[t], the polynomial ring in one variable t over R.
We shall introduce an A-module structure on M via φ.
Let φ0 = IM and φk = φo · · · oφ (k times) for all k ≥ 1.
For f (t) = a0 + a1t + a2t2 + · · · + ar tr ∈ A and x ∈ M , define
f (t).x = a0x + a1φ(x) + a2φ2(x) + · · · + ar φr (x) . . . . . . . . . . . . . . . . . (1)
We want to show that this defines an A-module structure on M .
We now have
(1) 1.x = x for all x ∈ M (is obvious from (1)).
Pn
(2) Let f (t) = i=0 aiti and x, y ∈ M .
P
i (x + y)
=⇒ f (t).(x + y) = n
a
φ
i
i=0
P
i (x) + φi (y))
= n
a
(φ
i
i=0
P
i (x) + Pn
i (y)
= n
a
φ
a
φ
i
i
i=0
i=0
(as φi’s are R-linear)
= f (t).x + f (t).y for all f (t) ∈ A and x, y ∈ M .
(3) Let f (t) =
Pn
i , g(t) = Pm b ti ∈ A and x ∈ M .
a
t
i=0 i
i=0 i
Assume that n ≥ m.
Put bi = 0 for i = m + 1, . . . , n and write g(t) =
P
i
=⇒ (f (t) + g(t)).x = ( n
i=0 (ai + bi )t ).x
Pn
= i=0(ai + bi)φi(x)
Pn
i
i=0 bi t .
Pn
Pn
i
= i=0 aiφ (x) + i=0 biφi(x)
= f (t).x + g(t).x for all f (t), g(t) ∈ A and x ∈ M .
Pn
i , g(t) = Pm b ti ∈ A and x ∈ M .
a
t
i=0 i
i=0 i
Pn Pm
Then f (t)g(t) = i=0 j=0 aibj ti+j (we multiply term by term but do
(4) Let f (t) =
not simplify).
Pm
P
i+j ).x
a
b
t
=⇒ (f (t)g(t)).x = ( n
i
j
i=0 j=0
Pn Pm
i+j ).x
i=0 j=0 (ai bj t
Pn Pm
= i=0 j=0 aibj φi+j (x)
Pn Pm
= i=0 j=0 aibj φi(φj (x))
Pn Pm
= i=0 j=0 aiφi(bj φj (x))
=
(by (3) above)
(as φi+j = φioφj )
(as φi is R-linear)
Pn
i (Pm b φj (x))
a
φ
i=0 i
j=0 j
Pn
= i=0 aiφi(g(t).x)
=
(as φi is R-linear)
= f (t).(g(t).x) for all f (t), g(t) ∈ A and x ∈ M .
Therefore, M is an A-module.
Let a ∈ R. Treating a as an element of A, we see that a.x = ax for
all a ∈ R and x ∈ M .
Let M be generated by S = {x1, . . . , xn} as an R-module.
=⇒ Every element of M is a linear combination of x1, . . . , xn, with
coefficients in R.
=⇒ Every element of M is a linear combination of x1, . . . , xn, with
coefficients in A (as a.x = ax for all a ∈ R and x ∈ M ).
=⇒ M is finitely generated as an A-module as well.
Let I denote the ideal of A generated by t and let x ∈ M .
Then there exists some y ∈ M such that φ(y) = x (as φ is onto).
=⇒ t.y = x, that is, x ∈ IM .
As x ∈ M is arbitrary, M ⊆ IM ⊆ M and therefore, M = IM .
=⇒ There exists some g(t) ∈ I such that (1 − g(t))M = 0.
As g(t) ∈ I, there exists some h(t) ∈ A such that g(t) = th(t).
Let h(t) = b0 + b1t + b2t2 + · · · + bsts.
Let x ∈ ker(φ).
As (1 − g(t))x = 0, we have x = g(t).x.
=⇒ x = g(t).x = b0φ(x) + b1φ2(x) + · · · + bsφs+1(x).
As x ∈ ker(φ), φ(x) = 0 = φ2(x) = · · · = φs+1(x).
=⇒ x = 0, that is, ker(φ) = 0. Hence φ is one-one.
23. Free modules
Let M be an R-module and let S = {x1, . . . , xn} ⊆ M .
Then S is said to be linearly dependent (LD) over R if there exist
Pn
a1, . . . , an ∈ R such that i=1 aixi = 0 but at least one of the ai’s is
nonzero.
If S is not LD then it is called linearly independent (LI).
Empty set is defined to be LI.
An infinite set S is said to be LI if every finite subset of S is LI.
The module M is said to be free if there exists an LI subset S of M
such that M is generated by S. Such a set S is called a basis of M .
There are some rings and modules over them, which are not free.
The trivial R-module is free over R with basis as empty set.
The ring R is free as a module over itself with basis {1}.
The R-module Rn is free with basis {e1, . . . , en}, where ej is element
which has 1 at the jth place and 0 elsewhere.
Let M be a free R-module with a basis B. Then every x ∈ M has a
unique expression as a linear combination of elements of B.
A result from linear algebra is also valid here, namely,
Lemma 29 Let M be a free R-module with a basis B = {xλ | λ ∈ Λ}.
Let L be an R-module. Choose yλ ∈ L for all λ ∈ Λ. Then there exist
an R-linear map φ : M −→ L such that φ(xλ) = yλ for all λ ∈ Λ.
We now aim to prove the following result: Let S be a nonempty set.
Then there exists an R-module M containing S such that S is a basis
of M .
Let
L = {φ : S −→ R | φ(x) = 0 for all but finitely many x ∈ S}.
Consider θ : S −→ R given by θ(x) = 0 for all x ∈ S.
=⇒ θ ∈ L and therefore, L is nonempty.
For φ, ψ ∈ L define, θφψ : S −→ R by setting
θφψ (x) = φ(x) + ψ(x)
for all x ∈ M .
Let Xφ = {x ∈ S | φ(x) 6= 0} and Xψ = {x ∈ S | ψ(x) 6= 0}.
=⇒ φ(x) = 0 = ψ(x) for all x ∈ S \ (Xφ ∪ Xψ ).
=⇒ θφψ (x) = φ(x) + ψ(x) = 0 for all x ∈ S \ (Xφ ∪ Xψ ).
As Xφ and Xψ are finite, so is Xφ ∪ Xψ .
=⇒ θφψ (x) = 0 for all but finitely many values of x ∈ S.
=⇒ θφψ ∈ L for all φ, ψ ∈ L.
Define a binary operation ⊕ in L, by setting φ⊕ψ = θφψ for all φ, ψ ∈ L.
Note that (φ ⊕ ψ)(x) = θφψ (x) = φ(x) + ψ(x) for all x ∈ S.
Claim. L is an abelian group under ⊕.
(1) ⊕ is commutative. Let φ, ψ ∈ L.
=⇒ (φ ⊕ ψ)(x) = φ(x) + ψ(x) = ψ(x) + φ(x) = (ψ ⊕ φ)(x) for all x ∈ S.
=⇒ φ ⊕ ψ = ψ ⊕ φ for all φ, ψ ∈ L.
=⇒ ⊕ is commutative.
(2) ⊕ is associative. Let φ, ψ, η ∈ L.
=⇒ ((φ ⊕ ψ) + η)(x)
= (φ ⊕ ψ)(x) + η(x)
= (φ(x) + ψ(x)) + η(x)
= φ(x) + (ψ(x) + η(x))
= φ(x) + (ψ + η)(x)
= (φ + (ψ + η))(x) for all x ∈ S.
=⇒ (φ ⊕ ψ) + η = φ + (ψ + η) for all φ, ψ, η ∈ L.
=⇒ ⊕ is associative.
(3) Identity for ⊕. We have the map, θ ∈ L, defined above.
Let φ ∈ L.
=⇒ (φ ⊕ θ)(x) = φ(x) + θ(x) = φ(x) for all x ∈ S.
=⇒ φ ⊕ θ = φ = θ ⊕ φ for all φ ∈ L.
=⇒ Identity for ⊕ is θ.
(4) Inverse with respect to ⊕. Let φ ∈ L.
Define φ0 : S −→ R by setting φ0(x) = −φ(x) for all x ∈ S.
=⇒ φ0(x) = 0 iff φ(x) = 0.
=⇒ φ0(x) = 0 for all but finitely many values of x ∈ S.
=⇒ φ0 ∈ L.
Also (φ ⊕ φ0)(x) = φ(x) + φ0(x) = φ(x) − φ(x) = 0 = θ(x) for all x ∈ S.
=⇒ φ ⊕ φ0 = θ = φ0 ⊕ φ.
=⇒ φ0 is the inverse of φ with respect to ⊕.
Therefore, (L, ⊕) is an abelian group.
Henceforth, we shall use the standard notation in L, that is, write +
for ⊕, 0 for θ and −φ for φ0.
For a ∈ R and φ ∈ L, we define πaφ : S −→ R by setting
πaφ(x) = aφ(x)
for all x ∈ S.
=⇒ πaφ(x) = 0 whenever φ(x) = 0.
=⇒ πaφ(x) = 0 for all but finitely many values of x ∈ S.
=⇒ πaφ ∈ L.
Claim. L is an R-module.
For a ∈ R and φ ∈ L, define a.φ = πaφ.
Note that (a.φ)(x) = πaφ(x) = aφ(x) for all x ∈ S.
(1) Let φ ∈ L.
=⇒ (1.φ)(x) = 1φ(x) = φ(x) for all x ∈ S.
=⇒ 1.φ = φ
(2) Let a ∈ R and φ, ψ ∈ L.
=⇒ (a.(φ + ψ))(x) = a(φ + ψ)(x)
= a(φ(x) + ψ(x)).
= aφ(x) + aψ(x)
= (a.φ)(x) + (a.ψ)(x)
= (a.φ + a.ψ)(x) for all x ∈ S.
=⇒ a.(φ + ψ) = a.φ + a.ψ for all a ∈ R and φ, ψ ∈ L.
(3) Let a, b ∈ R and φ ∈ L.
=⇒ ((a + b).φ)(x) = (a + b)φ(x)
= aφ(x) + bφ(x))
= (a.φ)(x) + (b.φ)(x)
= (a.φ + b.φ)(x) for all x ∈ S.
=⇒ (a + b).φ = a.φ + b.φ for all a, b ∈ R and φ ∈ L.
(4) Let a, b ∈ R and φ ∈ L.
=⇒ ((ab).φ)(x) = (ab)φ(x) = a(bφ(x))
= a((b.φ)(x)) = (a.(b.φ))(x) for all x ∈ L.
=⇒ (ab).φ = a.(b.φ) for all a, b ∈ R and φ ∈ L.
Therefore, L is an R-module.
Henceforth, we shall be writing aφ instead of a.φ.
We shall now construct a basis of L.
For x ∈ S, define φx : S −→ R by setting

1 if y = x,
φx(y) =
0 if y =
6 x.
Clearly, φx ∈ L.
Note that if x 6= y then φx(x) = 1 6= 0 = φy (x).
=⇒ If x, y ∈ S, x 6= y then φx 6= φy .
Claim. B = {φx | x ∈ S} is LI.
=⇒ Let φx1 , . . . , φxn be distinct.
Then x1, . . . , xn ∈ S are also distinct.
Pn
Let a1, . . . , an ∈ R such that i=1 aiφxi = 0 (0 is same as the map θ
above).
Let j ∈ {1, . . . , n}.
=⇒ ( n
i=1 ai φxi )(xj ) = 0(xj ) = 0.
Pn
=⇒ i=1 aiφxi (xj ) = 0.
P
=⇒ aj = 0 for all j = 1, . . . , n.
=⇒ Every finite subset of B is LI.
=⇒ B is LI.
Claim. B generates L.
Let φ ∈ L.
If φ = 0 then obviously, it is linear combination of elements of B.
Assume that φ 6= 0 and let Xφ = {x ∈ S | φ(x) 6= 0}.
=⇒ Xφ is nonempty and finite, say, Xφ = {x1, . . . , xn}.
Pn
Claim. φ = i=1 φ(xi)φxi .
Let x ∈ S \ Xφ.
=⇒ φ(x) = 0 and ( n
i=1 φ(xi )φxi )(x) =
Let x ∈ Xφ.
P
Pn
i=1 φ(xi )φxi (x) = 0.
=⇒ x = xj for some j = 1, . . . , n.
n
Also, ( n
i=1 φ(xi )φxi )(xj ) = i=1 φ(xi )φxi (xj ) = φ(xj ).
Pn
=⇒ φ(x) = ( i=1 φ(xi)φxi )(x) for all x ∈ S.
P
P
Pn
=⇒ φ = i=1 φ(xi)φxi .
As φ ∈ L is arbitrary, B generates L.
=⇒ L is a free R-module with basis B.
Consider the map : S −→ B given by (x) = φx for all x ∈ S.
Then is a bijection.
=⇒ L is a free R-module with basis B and B is bijective to S.
Comments. We have only been able to construct a free R-module L
with basis B ∼ S. From here there a way to construct a free module
M which contains S and S is a basis of M . We will not discuss it for
now and take it as known for the time being.
Lemma 30
Let M be an R-module. Then there exists a free R-
module F and an onto R-linear map φ : F −→ M . That is, every R
module is a homomorphic image of a free R-module.
Proof.
Let S be a set of generators of M . For example S may be
taken as whole of M . Let F be a free R-module containing S such
that S is a basis of F . Let φ : F −→ M be the R-linear map given
by φ(x) = x for all x ∈ S. Then Im(φ) contains S and therefore, the
whole of M , as Im(φ) is a submodule of M . Therefore, φ is onto.
Note that there is no uniqueness about F .
Lecture – 10
24. Direct sum and direct product
Let {Mλ | λ ∈ Λ} be a family of R-modules.
Then their Cartesian product
M =
Y
Mλ
λ∈Λ
is called the direct product of Mλ’s.
Note that M is an abelian group under componentwise addition.
The R-module structure in M is given by a.(xλ) = (axλ) for all a ∈ R,
xλ ∈ Mλ and λ ∈ Λ.
Consider the subset
N = {(xλ) ∈ M | xλ =0 for all but finitely many values of λ}
of M .
Claim. N is a submodule of M .
Clearly, the element (0λ) ∈ N , where, for all λ ∈ Λ, 0λ = 0 is the
additive identity of Mλ.
Therefore, N is nonempty.
Let x = (xλ), y = (yλ) ∈ N and a ∈ R.
=⇒ xλ and yλ are 0 for all but finitely many values of λ.
Put X = {λ ∈ Λ | xλ 6= 0} and Y = {λ ∈ Λ | yλ 6= 0}.
Then X and Y are finite.
=⇒ X ∪ Y is finite.
Also, xλ = 0 = yλ for all λ ∈ Λ \ (X ∪ Y ).
=⇒ axλ + yλ = 0 for all λ ∈ Λ \ (X ∪ Y ).
=⇒ axλ + yλ = 0 for all but finitely many values of λ.
=⇒ ax + y = (axλ + yλ) ∈ N .
=⇒ N is a submodule of M .
The submodule N is called the direct sum of Mλ’s and is denoted by
⊕λ∈ΛMλ.
Fix some µ ∈ Λ and define pµ : M −→ Mµ by setting pµ((xλ)) = xµ for
all (xλ).
The map pµ is called the projection in µth component.
Claim. pµ is R-linear.
Let (xλ), (yλ) ∈ M and let a ∈ R.
=⇒ pµ(a(xλ) + (yλ)) = pµ((axλ) + (yλ))
= pµ((axλ + yλ))
= axµ + yµ
= apµ((xλ)) + pµ((yλ))
for all (xλ), (yλ) ∈ M and a ∈ R.
=⇒ pµ is R-linear.
We have the inclusion map : ⊕λ∈Λ Mλ −→
Q
λ∈Λ Mλ , which is R-
linear.
=⇒ pµo : ⊕λ∈Λ Mλ −→ Mµ is also R-linear.
This map is again called the projection from ⊕λ∈ΛMλ to µth component Mµ.
We also have inclusion µ : Mµ −→ ⊕λ∈ΛMλ given by µ(x) = (xλ),
where xλ = 0 if λ 6= µ and xµ = x.
It is clearly R-linear.
=⇒ oµ : Mµ −→
Q
λ∈Λ Mλ is R-linear.
Note that (1) pµo(oν ) : Mν −→ Mµ.
(2) pµo(oν ) = 0 if µ 6= ν and pµo(oµ) is identity of Mµ.
25. Hom
Let M, L be R-module. Consider the set
HomR (M, L) = {φ | φ : M −→ L is R-linear}.
This set is clearly nonempty as it contains the zero map, namely, 0.
Sometimes we may denote this set by Hom(M, L) if there is no ambiguity about the ring under consideration.
We define a binary operation ⊕ in HomR (M, L) as follows:
Let φ : M −→ L and ψ : M −→ L be R-linear maps.
Define θφψ : M −→ L by setting
θφψ (x) = φ(x) + ψ(x)
for all x ∈ M .
Let x, y ∈ M and a ∈ R. Then
θφψ (ax + y) = φ(ax + y) + ψ(ax + y)
= aφ(x) + φ(y) + zψ(x) + ψ(y)
= a(φ(x) + ψ(x)) + φ(y) + ψ(y)
= aθφψ (x) + θφψ (y).
As x, y ∈ M and a ∈ R are arbitrary, θφψ is R-linear.
=⇒ θφψ ∈ HomR (M, L).
Define φ ⊕ ψ = θφψ .
Claim. HomR (M, L) is an group under ⊕.
Note that (φ⊕ψ)(x) = θφψ (x) = φ(x)+ψ(x) for all φ, ψ ∈ HomR (M, L)
and all x ∈ M .
(1) ⊕ is commutative. Let φ, ψ ∈ HomR (M, L).
=⇒ (φ ⊕ ψ)(x) = φ(x) + ψ(x)
= ψ(x) + φ(x)
= (ψ ⊕ φ)(x) for all x ∈ M .
=⇒ φ ⊕ ψ = ψ ⊕ φ for all φ, ψ ∈ HomR (M, L).
=⇒ ⊕ is commutative.
(2) ⊕ is associative. Let φ, ψ, η ∈ HomR (M, L).
=⇒ ((φ ⊕ ψ) + η)(x) = (φ ⊕ ψ)(x) + η(x)
= (φ(x) + ψ(x)) + η(x)
= φ(x) + (ψ(x) + η(x))
= φ(x) + (ψ + η)(x)
= (φ + (ψ + η))(x) for all x ∈ M .
=⇒ (φ ⊕ ψ) + η = φ + (ψ + η) for all φ, ψ, η ∈ HomR (M, L).
=⇒ ⊕ is associative.
(3) Identity for ⊕. We have the zero map, 0 ∈ HomR (M, L).
Let φ ∈ HomR (M, L).
=⇒ (φ ⊕ 0)(x) = φ(x) + 0(x) = φ(x) for all x ∈ M .
=⇒ φ ⊕ 0 = φ = 0 ⊕ φ for all φ ∈ HomR (M, L).
=⇒ Identity for ⊕ is 0.
(4) Inverse with respect to ⊕. Let φ ∈ HomR (M, L).
Define φ0 : M −→ L by setting φ0(x) = −φ(x) for all x ∈ M .
=⇒ φ0(ax + y) = −φ(ax + y)
= −(aφ(x) + φ(y))
= a(−φ(x)) − φ(y)
= aφ0(x) + φ0(y) for all x, y ∈ M and all a ∈ R.
=⇒ φ0 is R-linear, that is, φ0 ∈ HomR (M, L).
Also (φ + φ0)(x) = φ(x) + φ0(x) = φ(x) − φ(x) = 0 = 0(x) for all x ∈ M .
=⇒ φ ⊕ φ0 = 0 = φ0 ⊕ φ.
=⇒ φ0 is the inverse of φ with respect to ⊕
Therefore, (HomR (M, L), ⊕) is an abelian group.
Henceforth, we shall use the standard notation in HomR (M, L). In
particular, we shall write + for ⊕.
We now define an R-module structure on HomR (M, L).
For a ∈ R and R-linear map φ : M −→ L, we define πaφ : M −→ L by
setting
πaφ(x) = aφ(x)
for all x ∈ M .
Let b ∈ R and let x, y ∈ M .
Then πaφ(bx + y) = aφ(bx + y)
= a(bφ(x) + φ(y))
= baφ(x) + aφ(y)
= bπaφ(x) + πaφ(y).
As x, y ∈ M and b ∈ R are arbitrary, πaφ is R-linear.
=⇒ πaφ ∈ HomR (M, L)
Claim. HomR (M, L) is an R-module.
For a ∈ R and φ ∈ HomR (M, L), define a.φ = πaφ.
=⇒ (a.φ)(x) = πaφ(x) = aφ(x) for all x ∈ M .
(1) Let φ ∈ HomR (M, L).
=⇒ (1.φ)(x) = 1φ(x) = φ(x) for all x ∈ M .
=⇒ 1.φ = φ
(2) Let a ∈ R and φ, ψ ∈ HomR (M, L).
=⇒ (a.(φ + ψ))(x) = a(φ + ψ)(x)
= a(φ(x) + ψ(x))
= aφ(x) + aψ(x)
= (a.φ)(x) + (a.ψ)(x)
= (a.φ + a.ψ)(x) for all x ∈ M .
=⇒ a.(φ + ψ) = a.φ + a.ψ for all a ∈ R and φ, ψ ∈ HomR (M, L).
(3) Let a, b ∈ R and φ ∈ HomR (M, L).
=⇒ ((a + b).φ)(x) = (a + b)φ(x)
= aφ(x) + bφ(x)
= (a.φ)(x) + (b.φ)(x)
= (a.φ + b.φ)(x) for all x ∈ M .
=⇒ (a + b).φ = a.φ + b.φ for all a, b ∈ R and φ ∈ HomR (M, L).
(4) Let a, b ∈ R and φ ∈ HomR (M, L).
=⇒ ((ab).φ)(x) = (ab)φ(x)
= a(bφ(x))
= a((b.φ)(x))
= (a.(b.φ))(x) for all x ∈ M .
=⇒ (ab).φ = a.(b.φ) for all a, b ∈ R and φ ∈ HomR (M, L).
Therefore, HomR (M, L) is an R-module.
We prove an elementary result about Hom.
∼ M.
Lemma 31 Let M be an R-module.Then HomR (R, M ) =
Proof. Here we are treating R as a module over itself.
Let φ ∈ HomR (R, M ).
Then φ is an R-linear map from R to M .
=⇒ φ(1) ∈ M .
Define θ : HomR (R, M ) −→ M by setting θ(φ) = φ(1) for all φ ∈
HomR (R, M ).
Claim. θ is R-linear.
Let a ∈ R and φ, ψ ∈ HomR (R, M ).
=⇒ θ(aφ + ψ) = (aφ + ψ)(1)
= (aφ)(1) + ψ(1)
= aφ(1) + ψ(1)
= aθ(φ) + θ(ψ) for all a ∈ R and φ, ψ ∈ HomR (R, M )
=⇒ θ is R-linear.
We now show that θ is one-one.
Let φ ∈ ker(θ).
=⇒ θ(φ) = 0 = φ(1).
=⇒ φ(a) = φ(a1) = aφ(1) = a0 = 0 for all a ∈ R as φ is R-linear.
=⇒ φ = 0.
=⇒ θ is one-one.
We now show that φ is onto.
Let x ∈ M .
Define φ : R −→ M by setting φ(a) = ax for all a ∈ R.
Let α, a, b ∈ R.
=⇒ φ(αa + b) = (αa + b)x = αax + bx = αφ(a) + φ(b) for all α, a, b ∈ R.
=⇒ φ is R-linear.
Also, θ(φ) = φ(1) = 1x = x.
=⇒ θ is onto as x ∈ M is arbitrary.
=⇒ θ is an isomorphism.
The following discussion shall be useful later: Let M, L, K, P be Rmodules and let f : M −→ L and g : L −→ K be R-linear maps.
Let φ ∈ HomR (P, M ).
=⇒ f oφ is an R-linear map from P to L.
=⇒ f oφ ∈ HomR (P, L) for all φ ∈ HomR (P, M ).
Define f ∗ : HomR (P, M ) −→ HomR (P, L) by setting f ∗(φ) = f oφ.
Claim. f ∗ is R-linear.
Let a ∈ R and φ, ψ ∈ HomR (P, M ).
=⇒ f ∗(aφ + ψ) = f o(aφ + ψ).
We need to show that f o(aφ + ψ) = af oφ + f oψ.
Let x ∈ P .
=⇒ (f o(aφ + ψ))(x) = f ((aφ + ψ)(x))
= f ((aφ)(x) + ψ(x))
= af (φ(x)) + f (ψ(x))
(as f is R-linear)
= a(f oφ)(x) + (f oψ)(x)
= (a(f oφ))(x) + (f oψ)(x)
= (a(f oφ) + (f oψ))(x) for all x ∈ P .
=⇒ f o(aφ + ψ) = a(f oφ) + (f oψ).
=⇒ f ∗(aφ + ψ) = af ∗(φ) + f ∗(ψ) for all a ∈ R and φ, ψ ∈ HomR (P, M ).
=⇒ f ∗ is R-linear.
We now prove some properties of this map.
(1) Let f : M −→ L and g : L −→ K be R-linear.
=⇒ g of : M −→ K is R-linear.
This means we have
f ∗ : HomR (P, M ) −→ HomR (P, L)
g ∗ : HomR (P, L) −→ HomR (P, K)
(g of )∗ : HomR (P, M ) −→ HomR (P, K) and
g ∗of ∗ : HomR (P, M −→ HomR (P, K).
Let φ ∈ HomR (P, M ).
=⇒ (g ∗of ∗)(φ) = g ∗(f ∗(φ)) = g ∗(f oφ) = g o(f oφ) = (g of )oφ = (g of )∗(φ).
As φ ∈ HomR (P, M ) is arbitrary, we have (g ∗of ∗) = (g of )∗.
(2) Let f be the map 0. Let φ ∈ HomR (P, M ). Then f ∗(φ) = f oφ =
0oφ = 0 = 0(φ) for all φ ∈ HomR (P, M ).
As φ ∈ HomR (P, M ) is arbitrary, we have f ∗ = 0, that is, 0∗ = 0.
(3) Let L = M and f = IM .
Let φ ∈ HomR (P, M ).
Then f ∗(φ) = f oφ = IM oφ = φ for all φ ∈ HomR (P, M ).
=⇒ f ∗ = IM ∗ is the identity map of HomR (P, M ).
(4) Assume that f : M −→ L is an isomorphism. Then g = f −1 : L −→
M is R-linear and g of = IM and f og = IL.
=⇒ g ∗of ∗ = (g of )∗ is the identity map of HomR (P, M ) and f ∗og ∗ is
the identity map of HomR (P, L).
=⇒ f ∗ is an isomorphism with inverse g ∗ = (f −1)∗.
Let φ ∈ HomR (L, P ). Then
Therefore, φof : M −→ P is also
We now consider a different case.
φ : L −→ P is an R-linear map.
an R-linear map, that is, φof ∈ HomR (M, P ). Thus we get a map
f∗ : HomR (L, P ) −→ HomR (M, P ) by setting f∗(φ) = φof . As in the
previous case, f∗ turns out to be R-linear and
(1) (g of )∗ = f∗og∗.
(2) 0∗ = 0.
(3) If f is the identity of M then f∗ is identity of HomR (M, M ).
(4) If f is an isomorphism then so is f∗ with inverse (f −1)∗.
Lecture – 11
26. Tensor product
Let M and N be R-modules.
Treat M × N purely as a set for time being.
Then there exist a free module F containing M × N and M × N is a
basis of F .
In F , consider the following subsets
S1 = {(x + x1, y) − (x, y) − (x1, y) | x, x1 ∈ M, y ∈ N },
S2 = {(x, y + y1) − (x, y) − (x, y1) | x ∈ M, y, y1 ∈ N },
S3 = {(ax, y) − a(x, y) | x ∈ M, y ∈ N, a ∈ R} and
S4 = {(x, ay) − a(x, y) | x ∈ M, y ∈ N, a ∈ R}.
Put S = S1 ∪ S2 ∪ S3 ∪ S4.
Let F0 be the submodule of F generated by S.
Before we proceed further we make some comments.
(1) Let x ∈ M and y ∈ N . Then (x, y) ∈ M × N is also an element of
F . As F is an R-module, −(x, y) is the additive inverse of (x, y) in F .
Therefore, (x + x1, y) − (x, y) − (x1, y) is a well defined element of F
for all x, x1 ∈ M, y ∈ N .
(2) Let a ∈ R, x ∈ M and y ∈ N .
Then ax ∈ M .
Therefore,
(x, y), (ax, y) ∈ M × N are also elements of F . The element a(x, y) is
determined by the R-module structure of F and −a(x, y) is its additive
inverse in F . Therefore, (ax, y) − a(x, y) is also a well defined element
of F for all x, ∈ M, y ∈ N and a ∈ R.
(3) Since M × N is a basis of F , every element of F is a linear combination of elements of M × N .
We denote by M ⊗R N the quotient module F/F0. Sometimes we may
drop R and write M ⊗ N is there is no ambiguity about the ring under
consideration. We call M ⊗R N the tensor product of M and N and
pronounce it as M tensored N .
We shall now determine the form of the elements in M ⊗R N .
For (x, y) ∈ M × N ⊆ F , we write x ⊗ y for the element (x, y) + F0 of
M ⊗R N = F/F0.
For all x, x1 ∈ M , y, y1 ∈ N and a ∈ R, we have
(1) As (x + x1, y) − (x, y) − (x1, y) ∈ S1 ⊆ S ⊆ F0, we have (x + x1, y) +
F0 = ((x, y) + (x1, y)) + F0 = ((x, y) + F0) + ((x1, y)) + F0).
=⇒ (x + x1) ⊗ y = x ⊗ y + x1 ⊗ y .
(2) Similarly, x ⊗ (y + y1) = x ⊗ y + x ⊗ y1
(3) As (ax, y) − a(x, y) ∈ S3 ⊆ S ⊆ F0, we have (ax, y) + F0 =
a(x, y) + F0 = a((x, y) + F0).
=⇒ ax ⊗ y = a(x ⊗ y).
(4) Similarly, x ⊗ ay = a(x ⊗ y). Hence ax ⊗ y = x ⊗ ay. Because of
this, we loosely say that elements of R can move around the symbol ⊗.
(5) Let X ∈ M ⊗R N .
=⇒ There exists some Y ∈ F such that X = Y + F0.
=⇒ There exist x1, . . . , xn ∈ M , y1, . . . , yn ∈ N and a1, . . . , an ∈ R such
Pn
that Y = i=1 ai(xi, yi).
P
=⇒ X = n
i=1 ai (xi , yi ) + F0
Pn
= i=1 ai((xi, yi) + F0)
P
= n
i=1 ai (xi ⊗ yi )
Pn
= i=1(aixi) ⊗ yi.
P
fi ⊗ yi, where x
fi = aixi for all i = 1, . . . , n.
= n
i=1 x
=⇒ X is a sum of finitely many elements of the form x ⊗ y.
(6) For all x ∈ M , x ⊗ 0 = x ⊗ (0 + 0) = x ⊗ 0 + x ⊗ 0. Thus x ⊗ 0 = 0.
(7) Similarly, 0 ⊗ y = 0 for all y ∈ N .
(8) x ⊗ y = 0 does not mean that either x = 0 or y = 0.
Comments. It is not obvious to construct R-linear maps from a tensor product. However, there is a way to do so, which we discuss below.
Let M, N, L be R-modules and let θ : M × N −→ L. Then θ is said to
be R-bilinear if the following conditions are satisfied:
(1) θ(x + x1, y) = θ(x, y) + θ(x1, y) for all x, x1 ∈ M and y ∈ N .
(2) θ(x, y + y1) = θ(x, y) + θ(x, y1) for all x ∈ M and y, y1 ∈ N .
(3) θ(ax, y) = aθ(x, y) = θ(x, ay) for all (x, y) ∈ M , y ∈ N and a ∈ R.
Universal property of tensor product. Let M, N, L be R-modules
and let θ : M × N −→ L be a R-bilinear map. Then θ induces R-linear
map φ : M ⊗ N −→ L such that φ(x ⊗ y) = θ(x, y) for all (x, y) ∈ M × N .
Let F , S1, S2, S3, S4, S and F0 be as before.
As M ×N is a basis of F , the map ψ : F −→ L given by ψ(x, y) = θ(x, y)
is a well-defined R-linear map.
Claim. S ⊆ ker(ψ).
We have
(1) ψ(x + x1, y) = θ(x + x1, y) = θ(x, y) + θ(x1, y) = ψ(x, y) + ψ(x1, y)
for all x, x1 ∈ M and y ∈ N .
=⇒ ψ((x + x1, y) − (x, y) − (x1, y)) = 0 for all x, x1 ∈ M and y ∈ N .
=⇒ (x + x1, y) − (x, y) − (x1, y) ∈ ker(ψ) for all x, x1 ∈ M and y ∈ N .
=⇒ S1 ⊆ ker(ψ).
Similarly, S2 ⊆ ker(ψ).
(2) ψ(ax, y) = θ(ax, y) = aθ(x, y) = aψ(x, y) for all x ∈ M , y ∈ N and
a ∈ R.
=⇒ ψ((ax, y) − a(x, y)) = 0 for all x ∈ M and y ∈ N .
=⇒ (ax, y) − a(x, y) ∈ ker(ψ) for all x ∈ M and y ∈ N .
=⇒ S3 ⊆ ker(ψ).
Similarly, S4 ⊆ ker(ψ).
=⇒ S ⊆ ker(ψ).
=⇒ F0 ⊆ ker(ψ), as the submodule generated by S is F0.
Therefore, by the fundamental theorem of homomorphism, ψ induces
an R-linear map φ : M ⊗ N = F/F0 −→ L such that φoπ = ψ, where
π : F −→ M ⊗ N is the natural map.
Therefore, (φoπ)(x, y) = ψ(x, y) = θ(x, y).
=⇒ φ(x ⊗ y) = φ((x, y) + F0) = (φoπ)(x, y) = ψ(x, y) = θ(x, y).
Lemma 32 Let M, L, K be R-modules and let θ : M ×L −→ K. Then
the following statements are equivalent:
(a) θ is R-bilinear.
(b) θ(ax + x1, by + y1) = abθ(x, y) + aθ(x, y1) + bθ(x1, y) + θ(x1, y1) for
all x, x1 ∈ M , y, y1 ∈ L and a, b ∈ R.
Proof. (a) =⇒ (b) Let x, x1 ∈ M , y, y1 ∈ L and a, b ∈ R. Then
θ(ax + x1, by + y1) = θ(ax + x1, by) + θ(ax + x1, y1)
= θ(ax, by) + θ(x1, by) + θ(ax, y1) + θ(x1, y1)
= aθ(x, by) + bθ(x1, y) + aθ(x, y1) + θ(x1, y1)
= abθ(x, y) + bθ(x1, y) + aθ(x, y1) + θ(x1, y1)
for all x, x1 ∈ M , y, y1 ∈ L and a, b ∈ R.
(b) =⇒ (a) First note that
(1) θ(x + x1, y) = θ(1.x + x1, 0y + y)
= 1.0 θ(x, y) + 0 θ(x1, y) + 1 θ(x, y) + θ(x1, y)
= θ(x, y) + θ(x1, y) for all x, x1 ∈ M and y ∈ L.
Similarly, θ(x, y + y1) = θ(x, y) + θ(x, y1) for all x ∈ M and y, y1 ∈ L.
(2) θ(ax, y) = θ((a − 1)x + x, 0.y + y)
= (a − 1)0 θ(x, y) + 0 θ(x, y) + (a − 1) θ(x, y) + θ(x, y)
= aθ(x, y) for all a ∈ R, x ∈ M and y ∈ L.
Similarly, θ(x, ay) = aθ(x, y) for all a ∈ R, x ∈ M and y ∈ L.
Therefore, θ is R-bilinear.
27. Properties of Tensor product – I
We shall discuss some properties of tensor products.
∼ M.
Lemma 33 Let M be an R-module. Then R ⊗ M =
Proof. Define φ : M −→ R ⊗ M by setting φ(x) = 1 ⊗ x for all x ∈ M .
=⇒ φ(ax + y) = 1 ⊗ (ax + y) = a(1 ⊗ x) + 1 ⊗ y
= aφ(x) + φ(y) for all a ∈ R and x, y ∈ M .
=⇒ φ is R-linear.
We now define θ : R × M −→ M by setting θ(a, x) = ax for all a ∈ R
and x ∈ M .
Claim. θ is R-bilinear.
Let a, b, α, β ∈ R and x, x1 ∈ M .
Then θ(aα + β, bx + x1) = (aα + β)(bx + x1)
= abαx + aαx1 + bβx + βx1
= abθ(α, x) + aθ(α, x1) + bθ(β, x) + θ(β, x1)
For all a, b, α, β ∈ R and x, x1 ∈ M .
=⇒ θ is R-bilinear.
=⇒ There exists an R-linear map ψ : R ⊗ M −→ M such that
ψ(a ⊗ x) = θ(a, x) = ax.
Before we proceed, we want to see the structure of R ⊗ M .
Let X ∈ R ⊗ M .
=⇒ There exist a1, . . . , an ∈ R and x1, . . . , xn ∈ M such that
Pn
X = i=1 ai ⊗ xi.
Pn
Pn
Pn
Pn
a
⊗x
=
(a
1)⊗x
=
1⊗a
x
=
1⊗
i
i
i i
i=1 i
i=1 i
i=1
i=1 ai xi =
Pn
1 ⊗ x, where x = i=1 aixi.
=⇒ X =
=⇒ Every X ∈ R ⊗ M is of the form 1 ⊗ x for suitable x ∈ M .
Now (ψ oφ)(x) = ψ(φ(x)) = ψ(1 ⊗ x) = 1x = x for all x ∈ M .
=⇒ ψ oφ is the identity map of M .
Also, let X ∈ R ⊗ M .
=⇒ There exist some x ∈ M such that X = 1 ⊗ x.
=⇒ (φoψ)(X) = φ(ψ(X)) = φ(ψ(1 ⊗ x)) = φ(1x) = φ(x) = 1 ⊗ x = X.
=⇒ φoψ is the identity map of R ⊗ M .
=⇒ φ is an isomorphism.
∼ L ⊗ M.
Lemma 34 Let M and L be R-modules. Then M ⊗ L =
Proof.
We shall use universal property of tensor product to con-
struct an R-linear map from M ⊗ L to L ⊗ M .
Define θ : M × L −→ L ⊗ M by θ(x, y) = y ⊗ x for all (x, y) ∈ M × L.
We show that θ is R-bilinear.
Let a, b ∈ R, x, x1 ∈ M and y, y1 ∈ L.
θ(ax + x1, by + y1) = (by + y1) ⊗ (ax + x1)
= by ⊗ (ax + x1) + y1 ⊗ (ax + x1)
= by ⊗ ax + by ⊗ x1 + y1 ⊗ ax + y1 ⊗ x1
= ab(y ⊗ x) + b(y ⊗ x1) + a(y1 ⊗ x) + (y1 ⊗ x1)
= ab θ(x, y) + b θ(x1, y) + a θ(x1, y) + θ(x1, y1)
for all a, b ∈ R, x, x1 ∈ M and y, y1 ∈ L.
Thus, θ is R-bilinear.
By universal property, there exists an R-linear map φ : M ⊗L −→ L⊗M
such that φ(x ⊗ y) = θ(x, y) = y ⊗ x for all x ∈ M and y ∈ L.
Similarly, there exists an R-linear map ψ : L ⊗ M −→ M ⊗ L such that
ψ(y ⊗ x) = x ⊗ y for all x ∈ M and y ∈ L.
=⇒ (ψ oφ)(x ⊗ y) = ψ(φ(x ⊗ y)) = ψ(y ⊗ x) = x ⊗ y for all x ∈ M and
y ∈ L.
We have earlier proved that every element of M ⊗L is a sum of finitely
many elements of the set S = {x ⊗ y | x ∈ M, y ∈ L}. Hence S is a set
of generators of M ⊗ L as an R-module. Since ψ oφ and the identity
map of M ⊗ L agree on S, they are equal.
That is ψ oφ is the identity of M ⊗ L.
Similarly, φoψ is identity of L ⊗ M .
=⇒ φ is an R-linear isomorphism.
Lecture – 12
28. Properties of Tensor product – II
∼
Lemma 35 Let M, L, and K be R-modules. Then (M ⊗ L) ⊗ K =
M ⊗ (L ⊗ K).
Proof.
We shall again use the universal property to construct an
R-linear map from (M ⊗L)⊗K to M ⊗(L⊗K). However, this situation
is a bit more difficult.
Let z ∈ K.
Define θz : M × L −→ M ⊗ (L ⊗ K) by θz (x, y) = x ⊗ (y ⊗ z) for all x ∈ M
and y ∈ L.
We check that θz is R-bilinear.
θz (x + x1, y) = (x + x1) ⊗ (y ⊗ z)
= x ⊗ (y ⊗ z) + x1 ⊗ (y ⊗ z)
= θz (x, y) + θz (x1, y) for all x, x1 ∈ M and y ∈ L.
θz (x, y + y1) = x ⊗ ((y + y1) ⊗ z)
= x ⊗ (y ⊗ z + y1 ⊗ z)
= x ⊗ (y ⊗ z) + x ⊗ (y1 ⊗ z)
= θz (x, y) + θz (x, y1) for all x ∈ M and y, y1 ∈ L.
θz (ax, y) = ax ⊗ (y ⊗ z)
= a(x ⊗ (y ⊗ z))
= aθz (x, y) for all a ∈ R, x ∈ M and y ∈ L.
θz (x, ay) = x ⊗ (ay ⊗ z)
= x ⊗ (a(y ⊗ z))
= a(x ⊗ (y ⊗ z))
= aθz (x, y) for all a ∈ R, x ∈ M and y ∈ L.
=⇒ θz is R-bilinear.
Therefore, there exists an R-linear map φz : M ⊗ N −→ M ⊗ (L ⊗ K)
such that φz (x ⊗ y) = θz (x, y) = x ⊗ (y ⊗ z) for all x ∈ M and y ∈ L.
We now construct an R-bilinear map θ : (M ⊗ L) × K −→ M ⊗ (L ⊗ K).
Define θ(X, z) = φz (X) for all X ∈ M ⊗ L and z ∈ K.
θ(X + X1, z) = φz (X + X1)
= φz (X) + φz (X1)
(as φz is R-linear)
= θ(X, z) + θ(X1, z) for all X ∈ M ⊗ L and z ∈ K.
θ(aX, z) = φz (aX)
= aφz (X)
(as φz is R-linear)
= aθ(X, z) for all a ∈ R, X ∈ M ⊗ L and z ∈ K.
Let X =
Pn
i=1 xi ⊗ yi ∈ M ⊗ L and z, z1 ∈ K. Then
θ(X, z + z1) = φz+z1 (X)
Pn
(as φz+z1 is R-linear)
i=1 φz+z1 (xi ⊗ yi )
Pn
= i=1 xi ⊗ (yi ⊗ (z + z1))
P
= n
i=1 xi ⊗ (yi ⊗ z + yi ⊗ z1 )
Pn
= i=1 (xi ⊗ (yi ⊗ z) + xi ⊗ (yi ⊗ z1))
=
Pn
Pn
x
⊗
(y
⊗
z)
+
i
i=1 i
i=1 xi ⊗ (yi ⊗ z1 )
Pn
Pn
= i=1 φz (xi ⊗ yi) + i=1 φz1 (xi ⊗ yi)
P
Pn
= φz ( n
x
⊗
y
)
+
φ
(
z1
i
i=1 i
i=1 xi ⊗ yi )
=
= φz (X) + φz1 (X)
= θ(X, z) + θ(X, z1) for all X ∈ M ⊗ L and z, z1 ∈ K.
Let a ∈ R, X =
Pn
i=1 xi ⊗ yi ∈ M ⊗ L and z ∈ K. Then
θ(X, az) = φaz (X)
=
=
=
=
Pn
i=1 φaz (xi ⊗ yi )
Pn
i=1 xi ⊗ (yi ⊗ az)
Pn
i=1 xi ⊗ a(yi ⊗ z)
Pn
i=1 a(xi ⊗ (yi ⊗ z))
Pn
= a i=1 xi ⊗ (yi ⊗ z)
P
=a n
i=1 φz (xi ⊗ yi )
Pn
= aφz ( i=1 xi ⊗ yi)
= aφz (X)
= aθ(X, z) for all a ∈ R, X ∈ M ⊗ L and z ∈ K.
Therefore, θ is R-bilinear.
By universal property, there exists an R-linear map φ : (M ⊗ L) ⊗ K −→
M ⊗ (L ⊗ K) such that φ(X ⊗ z) = θ(X, z) = φz (X) for all X ∈ M ⊗ L
and z ∈ K.
=⇒ φ((x ⊗ y) ⊗ z) = φz (x ⊗ y) = x ⊗ (y ⊗ z) for all x ∈ M , y ∈ L and
z ∈ K.
Similarly, there exists an R-linear map ψ : M ⊗ (L ⊗ K) −→ (M ⊗ L) ⊗ K
such that ψ(x ⊗ (y ⊗ z)) = (x ⊗ y) ⊗ z for all x ∈ M , y ∈ L and z ∈ K.
Claim. ψ oφ is identity of (M ⊗ L) ⊗ K.
Pn
First let X = i=1(xi ⊗ yi) ∈ M ⊗ L and z ∈ K.
P
=⇒ (ψ oφ)(X ⊗ z) = (ψ oφ)(( n
i=1 xi ⊗ yi ) ⊗ z)
P
= (ψ oφ)( n
i=1 ((xi ⊗ yi ) ⊗ z)))
P
= n
i=1 (ψ oφ)((xi ⊗ yi ) ⊗ z)
P
= n
i=1 ψ(φ((xi ⊗ yi ) ⊗ z))
Pn
= i=1 ψ(xi ⊗ (yi ⊗ z))
P
= n
i=1 (xi ⊗ yi ) ⊗ z
Pn
= ( i=1 xi ⊗ yi) ⊗ z
=X ⊗z
Now let X ∈ (M ⊗ L) ⊗ K.
=⇒ There exist X1, . . . , Xr ∈ M ⊗ L and z1, . . . , zr ∈ K such that
Pr
X = i=1 Xi ⊗ zi.
P
=⇒ (ψ oφ)(X) = (ψ oφ)( ri=1 Xi ⊗ zi)
Pr
= i=1(ψ oφ)(Xi ⊗ zi)
P
= ri=1 Xi ⊗ zi = X.
=⇒ ψ oφ is the identity of (M ⊗ L) ⊗ K.
Similarly, φoψ is the identity of M ⊗ (L ⊗ K).
=⇒ φ is an isomorphism.
Lemma 36 Let M, L, and K be R-modules.
∼ Hom (M, Hom (L, K)).
L, K) =
R
R
Proof. Let φ ∈ HomR (M ⊗ L, K).
Let x ∈ M .
Note that φ(x ⊗ y) ∈ K for all y ∈ L.
Define φx : L −→ K by setting φx(y) = φ(x ⊗ y).
Claim. φx is R-linear.
Let a ∈ R and y, z ∈ L.
=⇒ φx(ay + z) = φ(x ⊗ (ay + z))
= φ((x ⊗ ay) + (x ⊗ z))
Then HomR (M ⊗
= φ(a(x ⊗ y) + (x ⊗ z))
= aφ(x ⊗ y) + φ(x ⊗ z)
(as φ is R-linear)
= aφx(y) + φx(z) for all a ∈ R and y, z ∈ L.
=⇒ φx is R-linear.
=⇒ φx ∈ HomR (L, K) for all x ∈ M .
Define φ : M −→ HomR (L, K) by setting φ(x) = φx for all x ∈ M .
Claim. φ is R-linear.
Let a ∈ R and x, x1 ∈ M .
=⇒ φ(ax + x1) = φax+x1 .
We shall show that φax+x1 = aφx + φx1 .
Let y ∈ L.
=⇒ φax+x1 (y) = φ((ax + x1) ⊗ y)
= φ(a(x ⊗ y) + (x1 ⊗ y))
= aφ(x ⊗ y) + φ(x1 ⊗ y)
= aφx(y) + φx1 (y)
= (aφx + φx1 )(y) for all y ∈ L.
=⇒ φax+x1 = aφx + φx1 for all a ∈ R and x, x1 ∈ M .
=⇒ φ(ax + x1) = aφ(x) + φ(x1) for all a ∈ R and x, x1 ∈ M .
=⇒ φ is R-linear.
=⇒ φ ∈ HomR (M, HomR (L, K)).
We now define θ : HomR (M ⊗ L, K) −→ HomR (M, HomR (L, K)) by
setting θ(φ) = φ.
Claim. θ is R-linear.
Let a ∈ R and let φ, ψ ∈ HomR (M ⊗ L, K).
=⇒ θ(aφ + ψ) = aφ + ψ.
We shall show that aφ + ψ = aφ + ψ.
Let x ∈ M .
=⇒ (aφ + ψ)(x) = (aφ + ψ)x.
Let y ∈ L.
=⇒ (aφ + ψ)x(y) = (aφ + ψ)(x ⊗ y)
= aφ(x ⊗ y) + φ(x ⊗ y)
= aφx(y) + ψx(y)
= (aφx + ψx)(y) for all y ∈ L.
=⇒ (aφ + ψ)x = aφx + ψx for all x ∈ M .
=⇒ (aφ + ψ)(x) = aφ(x) + ψ(x) for all x ∈ M .
=⇒ (aφ + ψ)(x) = (aφ + ψ)(x) for all x ∈ M .
=⇒ (aφ + ψ) = aφ + ψ.
=⇒ θ(aφ + ψ) = aθ(φ) + θ(ψ) for all a ∈ R and φ, ψ ∈ HomR (M ⊗ L, K).
=⇒ θ is R-linear.
Claim. θ is one-one.
Let φ ∈ ker(θ).
=⇒ θ(φ) = 0 = φ.
Let x ∈ M .
=⇒ φ(x) = 0.
=⇒ φx = 0.
Let y ∈ L.
=⇒ φx(y) = 0.
=⇒ φ(x ⊗ y) = 0 = 0(x ⊗ y) for all x ∈ M and y ∈ L.
As M ⊗ N is generated by {x ⊗ y | x ∈ M, y ∈ N }, we have φ = 0.
=⇒ θ is one-one.
Claim. θ is onto.
Let ψ ∈ HomR (M, HomR (L, K)).
Let x ∈ M and y ∈ L.
=⇒ ψ(x) ∈ HomR (L, K).
=⇒ ψ(x)(y) ∈ K.
Define λ : M × L −→ K by setting λ(x, y) = ψ(x)(y).
Claim. λ is R-bilinear.
Let a, b ∈ R, x, x1 ∈ M and y, y1 ∈ L.
=⇒ λ(ax + x1, by + y1)
= ψ(ax + x1)(by + y1)
= (aψ(x) + ψ(x1))(by + y1)
(as ψ is R-linear)
= aψ(x)(by + y1) + ψ(x1)(by + y1)
= a(bψ(x)(y) + ψ(x)(y1)) + bψ(x1)(y) + ψ(x1)(y1) (as ψ(x) is R-linear)
= abψ(x)(y) + aψ(x)(y1) + bψ(x1)(y) + ψ(x1)(y1)
= abλ(x, y) + aλ(x, y1) + bλ(x1, y) + λ(x1, y1)
all a, b ∈ R, x, x1 ∈ M and y, y1 ∈ L.
=⇒ λ is R-bilinear.
=⇒ There exists an R-linear map f : M ⊗ L −→ K such that f (x ⊗ y) =
λ(x, y) for all x ∈ M and y ∈ L.
=⇒ fx(y) = f (x ⊗ y) = λ(x, y) = ψ(x)(y) for all x ∈ M and y ∈ L.
=⇒ fx = ψ(x) for all x ∈ M .
=⇒ f (x) = fx = ψ(x) for all x ∈ M .
=⇒ f = ψ .
=⇒ θ(f ) = f = ψ .
=⇒ θ is onto and therefore, θ is an isomorphism.
Lecture – 13
29. Exact sequences
Consider the sequence
f
g
M −→ L −→ K,
where M, L, K are R-modules and f, g are R-linear maps. Then this
sequence (of R-modules and R-linear maps) is called exact if Im(f ) =
ker(g).
A sequence
fi−2
fi−1
f
fi+1
i
· · · −→ Mi−1 −→ Mi −→
Mi+1 −→ · · · ,
(1)
of R-modules Mi’s and R-linear maps fi’s is said to be exact at Mi if
fi−1
f
i
Mi−1 −→ Mi −→
Mi+1
is exact (whenever above sequence is possible). The sequence (1) is
said to be exact it is exact at all Mi’s (whenever exactness at such
an Mi makes sense).
Comment. The sequence above may be finite or may extend to infinity in one direction or extend to infinity in both directions.
We now consider a special kind of sequences.
f
(1) 0−→M −→ L. In this sequence 0 is the trivial R-module , the only
possible map from the module 0 to M is the map 0 (therefore, this is
not mentioned). Clearly, the image of the module 0 is the submodule
0 of M . Therefore, this sequence is exact iff ker(f ) = 0. That is, f
is one-one.
g
(2) M −→ L−→0. The only possible map from L to 0 is the map 0.
Therefore, its kernel is the whole of L. The above sequence is exact
iff Im(g) = L. Thas is, g is onto.
f
g
(3) 0 −→ M −→ L −→ K −→ 0.
f
f
g
This sequence is exact iff
g
0 −→ M −→ L, M −→ L −→ K and L −→ K −→ 0 are exact. That
happens iff f is one-one, ker(g) = Im(f ) and g is onto. Such a sequence, if exact, is called a short exact sequence.
f
g
(4) 0 −→ M −→ L −→ K is exact iff f is one-one and ker(g) = Im(f ).
Such a sequence, if exact, is called a left short exact sequence.
f
g
(5) M −→ L −→ K −→ 0 is exact iff g is onto and ker(g) = Im(f ).
Such a sequence, if exact, is called a right short exact sequence.
30. Exactness and Hom – I
f
Let
g
M −→ L −→ K
(1)
be an arbitrary sequence of R-module. Let P be an R-module. Consider
f∗
g∗
g∗
f∗
Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K)
Hom(K, P ) −→ Hom(L, P ) −→ Hom(M, P )
(2)
(3)
We say that (2) is obtained from (1) by apply Hom(P, −) and that
(3) is obtained from (1) by applying Hom(−, P ) .
Assume now that (1) is exact.
There is no guarantee that either (2) or (3) will be exact.
If (2) is always exact for a particular P (and for every arbitrary exact sequence in (1)) then P is called a projective module (or we say
that the functor Hom(P, −) is exact) and if (3) is always exact for a
particular P (and for every arbitrary exact sequence in (1)) then P is
called an injective module (or we say that the functor Hom(−, P ) is
exact).
Comments. (1) The term functor has been used above. This term
comes from homological algebra. We shall not go into details of this
terminology.
(2) There are modules which are projective. We shall learn about
them a little later.
(3) There are modules which are injective. However, their construction is hard and beyond the scope of this course.
We shall learn a bit more about Hom(P, −). First we prove an elementary result.
Lemma 37 Let M, L, K be R-modules. Let φ : M −→ L and ψ : L −→
K be R-linear maps. The ψ oφ = 0 iff Im(φ) ⊆ ker(ψ).
Proof. Assume that ψ oφ = 0.
=⇒ (ψ oφ)(x) = 0 for all x ∈ M .
=⇒ ψ(φ(x)) = 0 for all x ∈ M .
=⇒ φ(x) ∈ ker(ψ) for all x ∈ M .
=⇒ Im(φ) ⊆ ker(ψ).
Conversely, let Im(φ) ⊆ ker(ψ).
=⇒ φ(x) ∈ ker(ψ) for all x ∈ M .
=⇒ ψ(φ(x)) = 0 = (ψ oφ)(x) for all x ∈ M .
=⇒ ψ oφ = 0.
f
g
Lemma 38 Let P be an R-module. Let 0 −→ M −→ L −→ K be
f∗
g∗
an exact sequence. Then so is 0 −→ Hom(P, M ) −→ Hom(P, L) −→
Hom(P, K).
f
g
Proof. 0 −→ M −→ L −→ K is exact.
=⇒ f is one-one and ker(g) = Im(f ).
We need to show that f ∗ is one-one and ker(g ∗) = Im(f ∗).
(1) f ∗ is one-one. Let φ ∈ ker(f ∗).
=⇒ f ∗(φ) = 0 = f oφ.
Note that φ : P −→ M and f oφ : P −→ L.
Let x ∈ P .
=⇒ (f oφ)(x) = 0.
=⇒ f (φ(x)) = 0.
As f one-one φ(x) = 0 for all x ∈ P .
=⇒ φ = 0.
=⇒ f ∗ is one-one.
(2) Im(f ∗) ⊆ ker(g ∗). We have ker(g) = Im(f ).
=⇒ g of = 0.
=⇒ g ∗of ∗ = (g of )∗ = 0∗ = 0.
=⇒ Im(f ∗) ⊆ ker(g ∗).
(2) ker(g ∗) ⊆ Im(f ∗). Let φ ∈ ker(g ∗).
=⇒ g ∗(φ) = 0 = g oφ.
=⇒ Im(φ) ⊆ ker(g) = Im(f ).
We shall define a map ψ : P −→ M in the following way: Let x ∈ P .
Then φ(x) ∈ Im(φ) ⊆ Im(f ). Hence there exists some x ∈ M such
that φ(x) = f (x). As f is one-one this x is unique. We find x for all
x ∈ P and define ψ(x) = x for all x ∈ P .
Claim. f oψ = φ and ψ is R-linear.
Let x ∈ P .
Then φ(x) = f (x) = f (ψ(x)) = (f oψ)(x) for all x ∈ P .
=⇒ φ = f oψ
We now show that ψ is R-linear. Let a ∈ R and x, y ∈ P .
=⇒ φ(ax + y) = aφ(x) + φ(y) = af (x) + f (y) = f (ax + y).
=⇒ ax + y = ax + y (by uniqueness).
=⇒ ψ(ax + y) = aψ(x) + ψ(y) for all a ∈ R and x, y ∈ P .
=⇒ ψ is R-linear.
Also, φ = f oψ = f ∗(ψ) ∈ Im(f ∗) for all φ ∈ ker(g ∗).
=⇒ ker(g ∗) ⊆ Im(f ∗), that is, ker(g ∗) = Im(f ∗).
Lemma 39 Let P be an R-module. Then the following statements
are equivalent
(a) P is projective.
f∗
g∗
(b) 0 −→ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) −→ 0 is exact
f
g
whenever 0 −→ M −→ L −→ K −→ 0 is exact.
f
g
Proof. (a) =⇒ (b) Let 0 −→ M −→ L −→ K −→ 0 be exact.
f
f
g
g
=⇒ 0 −→ M −→ L, M −→ L −→ K and L −→ K −→ 0 are exact.
f∗
Then by (a), 0 −→ Hom(P, M ) −→ Hom(P, L),
f∗
g∗
Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) and
g∗
Hom(P, L) −→ Hom(P, K) −→ 0 are exact.
f∗
g∗
0 −→ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) −→ 0 is exact.
(b) =⇒ (a) We need to show that
g∗
f∗
Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K)
f
g
is exact whenever M −→ L −→ K is exact.
f
g
Let M −→ L −→ K be exact.
=⇒ ker(g) = Im(f ).
=⇒ g of = 0.
=⇒ g ∗of ∗ = (g of )∗ = 0∗ = 0.
=⇒ Im(f ∗) ⊆ ker(g ∗).
Before proceeding further, consider f : M −→ L. Now consider the
map f1 : M −→ Im(f ) and the inclusion map 1 : Im(f ) −→ L, where
f1(x) = f (x) for all x ∈ M . Although f1 is essentially f but we are
using a different symbol f1 as the codomains of f and f1 are not the
same. Note that 1of1 = f .
Furthermore, f1(ax + y) = f (ax + y) = af (x) + f (y) = af1(x) + f1(y)
for all a ∈ R and x, y ∈ M .
=⇒ f1 is R-linear.
Similarly, let g1 : L −→ Im(g) and the inclusion map 2 : Im(g) −→ K,
where g1(x) = g(x) for all x ∈ L. Note that 2og1 = g and that g1 is
R-linear.
Also, let : ker(f ) −→ M be the inclusion and π : K −→ K/Im(g) be
the natural map.
Then we have the following three exact sequences
f1
0 −→ ker(f ) −→ M −→ Im(f ) −→ 0,
g
1
1
0 −→ Im(f ) = ker(g) −→
L −→
Im(g) −→ 0,
2
π
0 −→ Im(g) −→ K −→ K/Im(g) −→ 0.
By (b), the following sequences are also exact
∗
f1∗
0 −→ Hom(P, ker(f )) −→ Hom(P, M ) −→ Hom(P, Im(f )) −→ 0, . . . (1)
∗1
g1∗
∗2
π∗
0 −→ Hom(P, Im(f )) −→ Hom(P, L) −→ Hom(P, Im(g)) −→ 0, . . . . (2)
0 −→ Hom(P, Im(g)) −→ Hom(P, K) −→ Hom(P, K)/Im(g) −→ 0. . (3)
We also have, ∗1of1∗ = (1of1)∗ = f ∗ and ∗2og1∗ = (2og1)∗ = g ∗.
We now prove that ker(g ∗) ⊆ Im(f ∗).
Let φ ∈ ker(g ∗)
=⇒ g ∗(φ) = 0.
=⇒ ∗2(g1∗ (φ)) = (∗2og1∗ )(φ) = 0.
=⇒ g1∗ (φ) = 0 as ∗2 is one-one.
=⇒ φ ∈ ker(g1∗ ) = Im(∗1).
=⇒ There exists ψ ∈ Hom(P, Im(f )) such that φ = ∗1(ψ) = 1oψ.
As f1∗ is onto, there exists θ ∈ Hom(P, M ) such that f1∗(θ) = ψ.
=⇒ φ = 1oψ = 1o(f1∗(θ)) = 1o(f1oθ) = (1of1)oθ = f oθ = f ∗(θ).
=⇒ φ ∈ Im(f ∗) for all φ ∈ ker(g ∗).
=⇒ ker(g ∗) ⊆ Im(f ∗) or ker(g ∗) = Im(f ∗).
f∗
g∗
=⇒ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) is exact.
f
g
As M −→ L −→ K is arbitrary, P is projective.
Lemma 40 Let P be an R-module. Then the following statements
are equivalent
(a) P is projective.
(b) g ∗ : Hom(P, L)−→Hom(P, K) is onto whenever g : L−→K is onto.
Proof. (a) =⇒ (b) Given g : L−→K is onto.
g
=⇒ L −→ K −→ 0 is exact.
g∗
=⇒ Hom(P, L) −→ Hom(P, K) −→ 0 is exact as P is projective.
=⇒ g ∗ is onto.
f
g
(b) =⇒ (a) Let 0 −→ M −→ L −→ K −→ 0 be exact.
We need to show that
f∗
g∗
0 −→ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) −→ 0
is exact.
f
g
As 0 −→ M −→ L −→ K is exact,
f∗
g∗
0 −→ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K)
is exact.
As g is onto g ∗ is also onto (by (b)).
g∗
=⇒ Hom(P, L) −→ Hom(P, K) −→ 0 is exact.
f∗
g∗
=⇒ 0 −→ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) −→ 0 is exact.
f
g
As 0 −→ M −→ L −→ K −→ 0 is arbitrary, P is projective.
Lecture – 14
31. Exactness and Hom – II
Let M be an R-module and N be a submodule of M . Then N is said
to be direct summand of M if there exists a submodule K of M such
that M = N + K and N ∩ K = 0.
Lemma 41 Let M and L be R-modules. Then the following statements are equivalent:
(a) M is isomorphic to a direct summand of L.
(b) There exist R-linear maps f : M −→ L and g : L −→ M such that
g of = IM .
Proof.
(a) =⇒ (b) Assume that M is isomorphic to a direct sum-
mand, say, N of L.
Then there exists a submodule K of L such that L = N + K and
N ∩ K = 0.
Let φ : M −→ N be an R-linear isomorphism and : N −→ L be the
inclusion map.
We shall define g : L −→ M as follows:
Let x ∈ L. Then there exist x1 ∈ N and y1 ∈ K such that x = x1 + y1.
Claim. Given x ∈ L, x1 is unique.
Let x2 ∈ N and y2 ∈ K such that x = x2 + y2 as well.
=⇒ x1 − x2 = y2 − y1 ∈ N ∩ K = 0.
=⇒ x1 − x2 = y2 − y1 = 0, that is, x1 = x2 and y1 = y2.
Define g(x) = φ−1(x1) for all x ∈ L.
Claim. g is R-linear.
Let a ∈ R and x, y ∈ L.
=⇒ There exist x1, x2 ∈ N and y1, y2 ∈ K such that x = x1 + y1 and
y = x2 + y2.
=⇒ ax + y = (ax1 + x2) + (ay1 + y2).
Clearly, ax1 + x2 ∈ N and ay1 + y2 ∈ L.
By uniqueness g(ax + y) = φ−1(ax1 + x2)
= aφ−1(x1) + φ−1(x2) (as φ−1 is R-linear)
= ag(x) + g(y).
As a ∈ R and x, y ∈ L are arbitrary, g is R-linear.
Note that for x ∈ N , x = x + 0 and hence g(x) = φ−1(x).
Let f = oφ.
Let x ∈ M .
=⇒ f (x) = (oφ)(x) = (φ(x)) = φ(x) ∈ N and therefore, g(φ(x)) =
φ−1(φ(x)) = x.
=⇒ (g of )(x) = g(φ(x)) = x for all x ∈ M .
=⇒ (g of ) = IM .
(b) =⇒ (a) Suppose that there exist R-linear maps f : M −→ L and
g : L −→ M such that g of = IM .
Claim. f is one-one.
Let x ∈ ker(f ).
=⇒ x = (g of )(x) = g(f (x)) = g(0) = 0.
=⇒ f is one-one.
=⇒ M is isomorphic to Im(f ).
Put N = Im(f ) and K = ker(g).
Claim. N ∩ K = 0.
Let x ∈ N ∩ K.
=⇒ g(x) = 0 and there exists y ∈ M such that x = f (y).
=⇒ y = (g of )(y) = g(f (y)) = g(x) = 0.
=⇒ x = f (y) = f (0) = 0.
=⇒ N ∩ K = 0.
Claim. L = N + K.
Let x ∈ L.
=⇒ g(x) ∈ M and f (g(x)) ∈ Im(f ) = N .
(as g(x) ∈ M and g of = IM )
=⇒ g(x − f (g(x))) = 0, that is, x − f (g(x)) ∈ ker(g) = K.
=⇒ g(x) = (g of )(g(x)) = g(f (g(x))
=⇒ x = f (g(x)) + (x − f (g(x))) ∈ N + K
As x ∈ L is arbitrary, L ⊆ N + K ⊆ L, that is, L = N + K.
∼ N.
=⇒ N is a direct summand of L and M =
Theorem.
Let P be an R-module. Then the following statements
are equivalent
(a) P is projective.
(b) P is isomorphic to a direct summand of a free R-module.
Proof.
(a) =⇒ (b) Let F be free R-module with an onto R-linear
map π : F −→ P .
=⇒ π ∗ : Hom(P, F ) −→ Hom(P, P ) is onto (as P is projective).
As IP ∈ Hom(P, P ), there exists some f ∈ Hom(P, F ) such that
π ∗(f ) = IP .
=⇒ π of = IP .
So we have R-linear map f : P −→ F and π : F −→ P such that
π of = IP .
=⇒ P is isomorphic to a direct summand of a free R-module F .
(b) =⇒ (a) Suppose that P is isomorphic to a direct summand of a
free module say F .
=⇒ There exist R-linear maps f : P −→ F and g : F −→ P such that
g of = IP .
Let B = {eλ | λ ∈ Λ} be a basis of F .
We need to show that P is projective.
Let t : L −→ K be an onto R-linear map.
We show that t∗ : Hom(P, L) −→ Hom(P, K) is also onto.
Let φ ∈ Hom(P, K), that is, φ : P −→ K is R-linear.
=⇒ φog : F −→ K is R-linear.
=⇒ (φog)(eλ) ∈ K for all λ ∈ Λ.
=⇒ There exists some xλ ∈ L such that t(xλ) = (φog)(eλ), as t is
onto.
Choose and fix one xλ for every λ ∈ Λ.
Let ψ : F −→ L be the R-linear map given by ψ(eλ) = xλ for all λ ∈ Λ.
Such a map does exist as F is free with the basis B.
=⇒ (toψ)(eλ) = t(ψ(eλ)) = t(xλ) = (φog)(eλ) for all λ ∈ Λ.
As B is a set of generators of F and the R-linear maps toψ and φog
agree on all elements of B, we must have toψ = φog.
=⇒ toψ of = φog of = φo(g of ) = φoIP = φ.
=⇒ t∗(ψ of ) = φ.
Note that ψ of : P −→ L, that is, ψ of ∈ Hom(P, L).
=⇒ φ ∈ Im(t∗) for all φ ∈ Hom(P, K).
=⇒ t∗ is onto.
As t : L −→ K is an arbitrary onto R-linear map, P is projective.
Comments. Let M be an R-module. Then M = M +0 and M ∩0 = 0.
Therefore, M is a direct summand of itself. Thus every free R-module
is projective over R.
f
g
Let 0 −→ M −→ L −→ K be a sequence. Then the
Lemma 42
following statements are equivalent:
f
g
(a) 0 −→ M −→ L −→ K is exact.
f∗
g∗
(b) 0 −→ Hom(P, M ) −→ Hom(P, L) −→ Hom(P, K) is exact for all
R-modules P .
Proof. (a) =⇒ (b) Already proved.
(b) =⇒ (a) We shall prove it by suitably choosing P is various cases.
(1) f is one-one. Let x ∈ ker(f ), that is, f (x) = 0.
Take P = R and let φ : R −→ M be given by φ(a) = ax for all a ∈ R.
=⇒ φ ∈ Hom(P, M ) and (f oφ)(a) = f (φ(a)) = f (ax) = af (x) = 0 for
all a ∈ R.
=⇒ f oφ = 0 = f ∗(φ).
As f ∗ is one-one, φ = 0.
=⇒ x = φ(1) = 0.
=⇒ f is one-one.
(2) Im(f ) ⊆ ker(g). We have (g of )∗ = g ∗of ∗ = 0 as Im(f ∗) = ker(g ∗).
Take P = M . Then (g of )∗(IM ) = 0 = g of oIM
=⇒ g of = 0.
=⇒ Im(f ) ⊆ ker(g).
(3) ker(g) ⊆ Im(f ). Let x ∈ ker(g), that is, g(x) = 0.
Take P = R and let φ : R −→ L be given by φ(a) = ax for all a ∈ R.
=⇒ φ ∈ Hom(R, L) and (g oφ)(a) = g(φ(a)) = g(ax) = ag(x) = 0 for
all a ∈ R.
=⇒ g oφ = 0 = g ∗(φ).
=⇒ φ ∈ ker(g ∗) = Im(f ∗).
=⇒ There exists some ψ ∈ Hom(R, M ) such that f ∗(ψ) = φ = f oψ.
=⇒ x = φ(1) = (f oψ)(1) = f (ψ(1)) ∈ Im(f ).
=⇒ x ∈ Im(f ) for all x ∈ ker(g).
=⇒ ker(g) ⊆ Im(f ), that is, ker(g) = Im(f ).
Lemma 43
f
g
Let M −→ L −→ K −→ 0 be a sequence. Then the
following statements are equivalent:
f
g
(a) M −→ L −→ K −→ 0 is exact.
g∗
f∗
(b) 0 −→ Hom(K, Q) −→ Hom(L, Q) −→ Hom(M, Q) is exact for all
R-modules Q.
f
g
Proof. (a) =⇒ (b) Assume that M −→ L −→ K −→ 0 is exact.
Let Q be an R-module.
We need to show that sequence in (b) is exact or equivalently g∗ is
one-one and Im(g∗) = ker(f∗).
(1) g∗ is one-one.
Let φ ∈ ker(g∗). Note that φ : K −→ Q.
=⇒ φog = g∗(φ) = 0.
=⇒ K = Im(g) ⊆ ker(φ) ⊆ K, that is, K = ker(φ).
=⇒ φ(x) = 0 for all x ∈ K.
=⇒ φ = 0.
=⇒ g∗ is one-one.
(2) Im(g∗) ⊆ ker(f∗).
We have f∗og∗ = (g of )∗ = 0∗ = 0.
=⇒ Im(g∗) ⊆ ker(f∗)
(3) ker(f∗) ⊆ Im(g∗).
Let φ ∈ ker(f∗). Note that φ : L −→ Q.
=⇒ φof = f∗(φ) = 0.
=⇒ Im(f ) ⊆ ker(φ).
Note that φ : L −→ Q.
Let y ∈ K. Then there exist some pre-image x ∈ L (not necessarily
unique) of y under g, that is, g(x) = y.
Define ψ : K −→ Q by ψ(y) = φ(x).
Claim. ψ is well defined.
Let x1 ∈ L such that g(x1) = y = g(x).
=⇒ x1 − x = ker(g) = Im(f ) ⊆ ker(φ).
=⇒ φ(x1) = φ(x).
=⇒ ψ(y) does not depend on the choice of x.
=⇒ ψ is well defined.
We now show that ψ is R-linear.
Let a ∈ R and y, y1 ∈ K.
Choose x, x1 ∈ L such that g(x) = y and g(x1) = y1.
=⇒ g(ax + x1) = ag(x) + g(x1) = ay + y1.
=⇒ ψ(y) = φ(x), ψ(y1) = φ(x1) and ψ(ay + y1) = φ(ax + x1).
=⇒ ψ(ay + y1) = φ(ax + x1) = aφ(x) + φ(x1) = aψ(y) + ψ(y1).
As a ∈ R and y, y1 ∈ K are arbitrary, ψ is R-linear.
Claim. ψ og = φ.
Let x ∈ L. Then x is also a pre-image of g(x) under g.
=⇒ ψ(g(x)) = φ(x) for all x ∈ L.
=⇒ ψ og = φ, that is, φ = g∗(ψ) ∈ Im(g∗) for all φ ∈ ker(f∗).
=⇒ ker(f∗) ⊆ Im(g∗), that is, ker(f∗) = Im(g∗).
(b) =⇒ (a) We shall prove is by suitably choosing Q is various cases.
(1) g is onto. We take Q = K/Im(g) and let π : K −→ K/Im(g) be
the natural map.
Let x ∈ L.
=⇒ (π og)(x) = π(g(x)) = g(x) + Im(g) = Im(g) = 0 for all x ∈ L.
=⇒ π og = 0 = g∗(π).
=⇒ π = 0 as g∗ is one-one.
=⇒ K/Im(g) = 0 as π is onto.
=⇒ K = Im(g), that is, g is onto.
(2) Im(f ) ⊆ ker(g). We have (g of )∗ = f∗og∗ = 0 as Im(g∗) = ker(f∗).
Take Q = K. Then (g of )∗(IK ) = 0 = IK og of = g of .
=⇒ Im(f ) ⊆ ker(g).
(3) ker(g) ⊆ Im(f ). We take Q = L/Im(f ) and let π : L −→ L/Im(f )
be the natural map.
=⇒ (π of )(x) = π(f (x)) = f (x) + Im(f ) = Im(f ) = 0 for all x ∈ L.
=⇒ π of = 0 = f∗(π), that is, π ∈ ker(f∗) = Im(g∗).
=⇒ There exists some φ ∈ Hom(K, L/Im(f )) such that g∗(φ) = π.
=⇒ π = φog.
Now let x ∈ ker(g).
=⇒ π(x) = (φog)(x) = φ(g(x)) = φ(0) = 0.
=⇒ x ∈ ker(π) = Im(f ) for all x ∈ ker(g).
=⇒ ker(g) ⊆ Im(f ), that is, ker(g) = Im(f ).
Lecture – 15
32. Exactness and tensor product – I
Let M1, M2, L1, L2 be R-modules and let f : M1 −→ M2 and g : L1 −→
L2 be R-linear maps.
Define a map θ : M1 × L1 −→ M2 ⊗ L2 by setting θ(x, y) = f (x) ⊗ g(y)
for all (x, y) ∈ M1 × L1.
We show that θ is R-bilinear.
Let a, b ∈ R, x, x1 ∈ M1 and y, y1 ∈ L1.
⇒ θ(ax + x1, by + y1) = f (ax + x1) ⊗ g(by + y1)
= (af (x) + f (x1)) ⊗ (bg(y) + g(y1)
= ab (f (x) ⊗ g(y)) + a (f (x) ⊗ g(y1))
+ b (f (x1) ⊗ g(y)) + f (x1) ⊗ g(y1)
= ab θ(x, y) + a θ(x, y1) + b θ(x1, y) + θ(x1, y1)
Therefore, θ is R-bilinear.
Hence there exists an R-linear map φ : M1 ⊗ L1 −→ M2 ⊗ L2 such that
φ(x ⊗ y) = θ(x, y) = f (x) ⊗ g(y) for all x ∈ M1 and y ∈ L1.
We denote this map by f ⊗ g.
We look at some properties of this map.
(1) Suppose that f = 0. Then (f ⊗g)(x⊗y) = f (x)⊗g(y) = 0⊗g(y) =
0 for all x ∈ M1 and y ∈ L1. As {x ⊗ y | x ∈ M1, y ∈ L1} is a set of
generators of M1 ⊗ L1, 0 ⊗ g = 0.
(2) Similarly, f ⊗ 0 = 0.
(3) Let M1, M2, M3, L1, L2, L3 be R-modules and f : M1 −→ M2, f1 : M2 −→
M3, g : L1 −→ L2, g1 : L2 −→ L3 be R-linear maps. Then f ⊗ g : M1 ⊗
L1 −→ M2 ⊗ L2 and f1 ⊗ g1 : M2 ⊗ L2 −→ M3 ⊗ L3. Also
((f1 ⊗ g1)o(f ⊗ g))(x ⊗ y) = (f1 ⊗ g1)((f ⊗ g))(x ⊗ y))
= (f1 ⊗ g1)((f (x) ⊗ g(y)))
= f1(f (x)) ⊗ g1(g(y))
= (f1of )(x) ⊗ (g1og)(y)
= ((f1of ) ⊗ (g1og))(x ⊗ y)
for all x ∈ M1 and y ∈ L1.
As {x ⊗ y | x ∈ M1, y ∈ L1} is a set of generators of M1 ⊗ L1 and
(f1 ⊗ g1)o(f ⊗ g) and (f1of ) ⊗ (g1og) agree on this set, we have
(f1 ⊗ g1)o(f ⊗ g) = (f1of ) ⊗ (g1og).
Now let M, L, K be R-modules and let f : L −→ K be an R-linear map.
Then we have IM ⊗ f : M ⊗ L −→ M ⊗ K. We write fe for IM ⊗ f .
Then we have
(1) f = 0 =⇒ fe = 0.
(2) Let T be an R-module and g : K −→ T be an R-linear map. Then
eofe.
gg
of = g
Then (IM ⊗ IL)(x ⊗ y) =
IM (x) ⊗ IL(y) = x ⊗ y for all x ∈ M and y ∈ L. As {x ⊗ y | x ∈ M, y ∈ L}
(3) Let f be the identity map of L.
is a set of generators of M ⊗ L and (IM ⊗ IL) and IM ⊗L agree on this
set, we have If
L = IM ⊗ IL = IM ⊗L . Loosely, this means, identity gives
rise to identity.
−1 = f^
(4) Suppose that f is an isomorphism. Then feofg
of −1 = IM ⊗K
e
−1 ofe = f^
−1 of = I
and fg
M ⊗L . Hence f is an isomorphism with inverse
−1 .
fg
We now define flat modules. Let
f
g
M −→ L −→ K
(1)
be an arbitrary sequence of R-module. Let P be an R-module. Consider
fe
ge
P ⊗ M −→ P ⊗ L −→ P ⊗ K
(2)
We say that (2) is obtained from (1) by apply P ⊗ −.
Assume now that (1) is an arbitrary exact sequence.
There is no guarantee that (2) will be exact.
If (2) is always exact for a particular P whenever (1) is an arbitrary
exact sequence then P is called a flat R-module (or we say that the
functor P ⊗ − is exact).
There are some flat modules and there are some ring over which some
modules are not flat.
We aim to understand flat modules. We first prove a lemma.
f
g
Lemma 44 Let P be an R-module and let M −→ L −→ K −→ 0 be
fe
ge
an exact sequence. Then P ⊗ M −→ P ⊗ L −→ P ⊗ K −→ 0 is exact.
Proof. We need to show that ge is onto and ker(ge) = Im(fe).
(1) ge is onto. Let X ∈ P ⊗ K.
=⇒ There exist x1, . . . , xn ∈ P and y1, . . . , yn ∈ K such that X =
Pn
i=1 xi ⊗ yi .
As g is onto, there exist z1, . . . , zn ∈ L such that g(zi) = yi for all
i = 1, . . . , n.
P
Pn
Pn
=⇒ ge( n
x
⊗
z
)
=
x
⊗
g(z
)
=
i
i
i=1 xi ⊗ yi = X, that is,
i=1 i
i=1 i
X ∈ Im(ge).
As X ∈ P ⊗ K is arbitrary, ge is onto.
(2) Im(fe) ⊆ ker(ge). As Im(f ) = ker(g), we have g of = 0.
e = 0.
=⇒ geofe = gg
of = 0
=⇒ Im(fe) ⊆ ker(ge).
(3) ker(ge) ⊆ Im(fe). We define θ : P × K −→ (P ⊗ L)/Im(fe) as follows:
Let x ∈ P and y ∈ K. As g is onto y has a pre-image under g, say,
z ∈ L, that is, g(z) = y. Define θ(x, y) = (x ⊗ z) + Im(fe).
Claim. θ is well defined.
The map θ appears to depend upon z which is not unique.
Let z1 ∈ L such that g(z1) = y = g(z).
=⇒ z − z1 ∈ ker(g) = Im(f ).
=⇒ There exists t ∈ M such that f (t) = z − z1.
=⇒ x ⊗ z − x ⊗ z1 = x ⊗ (z − z1) = x ⊗ f (t) = fe(x ⊗ t) ∈ Im(fe).
=⇒ (x ⊗ z) + Im(fe) = (x ⊗ z1) + Im(fe).
Therefore, θ is well defined.
Claim. θ is R-bilinear.
Let a, b ∈ R, x, x1 ∈ P and y, y1 ∈ K.
Choose pre-images z, z1 in L of y, y1 under g.
=⇒ bz + z1 is also a pre-image of by + y1 in L under g.
=⇒ θ(ax + x1, by + y1) = ((ax + x1) ⊗ (bz + z1)) + Im(fe)
= (ab (x ⊗ z) + a(x ⊗ z1) + b(x1 ⊗ z) + (x1 ⊗ z1)) + Im(fe)
e
e
e
= ab (x ⊗ z) + Im(f ) + a (x ⊗ z1) + Im(f ) + b (x1 ⊗ z) + Im(f )
+ (x1 ⊕ z1) + Im(fe) .
= ab θ(x, y) + a θ(x, y1) + b θ(x1, y) + θ(x1, y1).
As a, b ∈ R, x, x1 ∈ P and y, y1 ∈ K are arbitrary, θ is R-bilinear.
=⇒ There exists an R-linear map φ : P ⊗ K −→ (P ⊗ L)/Im(fe) such
that φ(x ⊗ y) = θ(x, y) for all x ∈ P and y ∈ K.
Now let X ∈ ker(ge). Then there exist x1, . . . , xn ∈ P and z1, . . . , zn ∈ L
P
such that X = n
i=1 xi ⊗ zi .
As X ∈ ker(ge), we have
Pn
i=1 xi ⊗ g(zi ) = 0.
Note that zi is a pre-image in L of g(zi) for all i = 1, . . . , n.
P
Pn
Therefore, 0 = φ(0) = φ( n
x
⊗
g(z
))
=
φ(xi ⊗ g(zi)) =
i
i
i=1
i=1
P
Pn n
e
e) = X + Im(fe).
(x
⊗
z
)
+
Im(
f
)
=
x
⊗
z
+
Im(
f
i
i
i
i
i=1
i=1
=⇒ X ∈ Im(fe) for all X ∈ ker(ge).
=⇒ ker(ge) ⊆ Im(fe).
Lemma 45 Let P be an R-module. Then the following statements
are equivalent:
(a) P is flat.
(b) Whenever f : M −→ L is one-one then so is fe: P ⊗ M −→ P ⊗ L.
f
g
(c) Whenever 0 −→ M −→ L −→ K −→ 0 is exact then so is
fe
ge
0 −→ P ⊗ M −→ P ⊗ L −→ P ⊗ K −→ 0.
Proof. (a) =⇒ (b) Assume that f : M −→ L is one-one.
f
=⇒ 0 −→ M −→ L is exact.
fe
=⇒ 0 −→ P ⊗ M −→ P ⊗ L is exact as P is flat.
=⇒ fe: P ⊗ M −→ P ⊗ L is one-one.
f
g
(b) =⇒ (c) Let 0 −→ M −→ L −→ K −→ 0 be exact.
f
g
f
=⇒ M −→ L −→ K −→ 0 is exact and f : M −→ L is one-one.
fe
ge
=⇒ P ⊗ M −→ P ⊗ L −→ P ⊗ K −→ 0 is exact and fe: P ⊗ M −→ P ⊗ L
is one-one (by statement (b)).
fe
ge
=⇒ 0 −→ P ⊗ M −→ P ⊗ L −→ P ⊗ K −→ 0 is exact.
f
g
(c) =⇒ (a) Let M −→ L −→ K be exact.
fe
ge
We need to show that P ⊗ M −→ P ⊗ L −→ P ⊗ K is exact or
ker(ge) = Im(fe).
e = 0, we have Im(fe) ⊆ ker(g
e).
(1) Im(fe) ⊆ ker(ge). As geofe = gg
of = 0
f
g
(2) ker(ge) ⊆ Im(fe). From M −→ L −→ K, we get the following three
exact sequences
f
1
0 −→ ker(f ) −→ M −→
Im(f ) −→ 0,
1
g1
0 −→ Im(f ) = ker(g) −→ L −→ Im(g) −→ 0,
π
2
0 −→ Im(g) −→
K −→ K/Im(g) −→ 0,
where, , 1, 2 are inclusions, π is the natural map, f1(x) = f (x) for
all x ∈ M and g1(x) = g(x) for all x ∈ L. Furthermore, 1of1 = f ,
2og1 = g; f1 and g1 are R-linear (see Lemma 39).
By (c), these sequences are also exact
e
f
f
1
0 −→ P ⊗ ker(f ) −→ P ⊗ M −→ P ⊗ Im(f ) −→ 0, . . . . . . . . . . . . . . . . (1)
e
gf
1
1
P ⊗ Im(g) −→ 0, . . . . . (2)
0 −→ P ⊗ Im(f ) = P ⊗ ker(g) −→
P ⊗ L −→
e
π
e
2
0 −→ P ⊗ Im(g) −→
P ⊗ K −→ P ⊗ (K/Im(g)) −→ 0. . . . . . . . . . . . . (3)
Let x ∈ ker(ge).
e(x) = 0.
=⇒ (f2ogf1)(x) = ^
2 og1 (x) = g
=⇒ f2(gf1(x)) = 0.
=⇒ gf1(x) = 0
=⇒ x ∈ ker(gf1) = Im(f1),
(as (3) is exact, f2 is one-one).
(as (2) is exact).
=⇒ There exist some y ∈ P ⊗ Im(f ) such that f1(y) = x.
=⇒ There exist some z ∈ P ⊗ M such that ff1(z) = y (as ff1 is onto).
f
f
=⇒ fe(z) = ^
1 of1 (z) = (f
1 of1 )(z) = f
1 (f1 (z)) = f
1 (y) = x.
=⇒ x ∈ Im(fe) for all x ∈ ker(ge).
=⇒ ker(ge) ⊆ Im(fe).
Lecture – 16
33. Exactness and tensor product – II
We want to prove next that a projective module is flat. For that we
need the following lemma.
Lemma 46 Let M be an R-module and let F be a free R-module
with a basis B = {eλ | λ ∈ Λ}. Let X ∈ F ⊗ M . Then there exist some
unique xλ ∈ M such that X =
P
λ∈Λ eλ ⊗ xλ and xλ = 0 for all but
finitely many values of λ.
There exist y1, . . . , yn ∈ F and z1, . . . , zn ∈ M such that
Pn
X = i=1 yi ⊗ zi.
Proof.
=⇒ For every i ∈ {1, . . . , n}, there exist aiλ ∈ R such that aiλ = 0 for
P
all but finitely many values of λ and yi = λ∈Λ aiλeλ.
=⇒ yi ⊗ zi = ( λ∈Λ aiλeλ) ⊗ zi
P
= λ∈Λ(aiλeλ ⊗ zi)
P
P
= λ∈Λ(eλ ⊗ aiλzi) for all i = 1, . . . , n.
Pn
i=1 yi ⊗ zi
P
P
= n
i=1 λ∈Λ eλ ⊗ aiλ zi
P
P
= λ∈Λ n
i=1 eλ ⊗ aiλ zi
P
P
= λ∈Λ eλ ⊗ ( n
i=1 aiλ zi )
P
= λ∈Λ eλ ⊗ xλ,
P
where xλ = n
i=1 aiλ zi for all λ ∈ Λ.
=⇒ X =
Claim. xλ = 0 for all but finitely many values of λ.
Let Si = {λ ∈ Λ | aiλ 6= 0}, for all i = 1, . . . , n.
Then Si is finite for all i = 1, . . . , n.
=⇒ ∪n
i=1 Si is finite.
Let λ ∈ Λ \ ∪n
i=1 Si .
=⇒ aiλ = 0 for all i = 1, . . . , n.
=⇒ xλ =
Pn
n
i=1 aiλ zi = 0 for all λ ∈ Λ \ ∪i=1 Si .
Uniqueness. Suppose, if possible, that there exist xλ’s and tλ’s in M
P
P
such that X = λ∈Λ eλ ⊗ xλ and X = λ∈Λ eλ ⊗ tλ, where xλ’s and tλ’s
are 0 for all but finitely many values of λ ∈ Λ.
Fix some µ ∈ Λ.
Let φ : F −→ R be the R-linear map such that φ(eµ) = 1 and φ(eλ) = 0
for all λ ∈ Λ \ {µ}.
Then φe : F ⊗ M −→ R ⊗ M .
Let ψ : R ⊗ M −→ M be the isomorphism, that is, ψ(a ⊗ x) = ax for
all a ∈ R and x ∈ M .
e
e
=⇒ (ψ oφ)(X)
= ψ(φ(X))
e
= ψ(φ(
λ∈Λ eλ ⊗ xλ ))
P
e
= ψ( λ∈Λ φ(e
λ ⊗ xλ ))
P
= ψ( λ∈Λ φ(eλ) ⊗ xλ)
P
= ψ(1 ⊗ xµ) = xµ.
e
Similarly, (ψ oφ)(X)
= tµ.
=⇒ xµ = tµ for all µ ∈ Λ.
Theorem. Every free R-module is flat over R.
Proof. Let F be a free R-module with a basis B = {eλ | λ ∈ Λ}.
Let M and N be R-modules and let f : M −→ L be a one-one R-linear
map.
We need to show that fe: F ⊗ M −→ F ⊗ L is one-one.
Let X ∈ ker(fe).
P
=⇒ There exists unique xλ’s in M such that X = λ∈Λ eλ ⊗ xλ, where
xλ’s are 0 for all but finitely many values of λ.
P
=⇒ fe( λ∈Λ eλ ⊗ xλ) = 0.
=⇒
P
P
e
⊗
f
(x
)
=
0
=
λ
λ∈Λ λ
λ∈Λ eλ ⊗ 0.
By uniqueness in F ⊗ L, f (xλ) = 0 for all λ ∈ Λ.
=⇒ xλ = 0 for all λ ∈ Λ as f is one-one.
P
P
=⇒ X = λ∈Λ eλ ⊗ xλ = λ∈Λ eλ ⊗ 0 = 0.
=⇒ fe is one-one.
As M, L and f are arbitrary, F is flat.
Theorem. Every projective R-module is flat over R.
Proof. Let P be a projective R-module.
Then P is isomorphic to a direct summand of a free R-module, say,
F.
=⇒ There exist R-linear maps f : P −→ F and g : F −→ P such that
g of = IP .
Let M and N be R-modules and let φ : M −→ L be a one-one R-linear
map.
We need to show that IP ⊗ φ : P ⊗ M −→ P ⊗ L is one-one.
The following diagram may be useful in understanding the proof.
f ⊗I
g⊗I
M
M
P⊗
M
−
−
−
−
→
F
⊗
M
−
−
−
−
→ P⊗


M

IP ⊗φy
f ⊗I

yIF ⊗φ

IP ⊗φy
g⊗I
L
L
P ⊗L −
−−−
→
F ⊗ L −−−→
P ⊗L
Note that (1) (g ⊗ IM )o(f ⊗ IM ) = (g of ) ⊗ IM = IP ⊗ IM = IP ⊗M .
(2) IF ⊗ φ : F ⊗ M −→ F ⊗ L is one-one as F is free and hence flat.
(3) (IF ⊗ φ)o(f ⊗ IM ) = (IF of ) ⊗ (φoIM ) = f ⊗ φ.
(4) (f ⊗ IL)o(IP ⊗ φ) = (f oIP ) ⊗ (ILoφ) = f ⊗ φ.
=⇒ (f ⊗ IL)o(IP ⊗ φ) = (IF ⊗ φ)o(f ⊗ IM ). . . . . . . . . . . . . . . . . . . . . . (A)
Let X ∈ ker(IP ⊗ φ).
=⇒ ((IF ⊗ φ)o(f ⊗ IM ))(X) = ((f ⊗ IL)o(IP ⊗ φ))(X) (by (A) above)
= (f ⊗ IL)((IP ⊗ φ)(X))
= (f ⊗ IL)(0) = 0.
=⇒ (IF ⊗ φ)((f ⊗ IM )(X)) = 0.
=⇒ (f ⊗ IM )(X) = 0 as IF ⊗ φ is one-one
(by (2) above).
=⇒ ((g⊗IM )o(f ⊗IM ))(X) = (g⊗IM )((f ⊗IM )(X)) = (g⊗IM )(0) = 0.
=⇒ IP ⊗M (X) = 0
=⇒ X = 0, that is, IP ⊗ φ is one-one.
(by (1) above).
34. Tensor product of algebras
Let A and B be R-algebras. This means A and B are rings and there
exist ring homomorphisms φ : R −→ A and ψ : R −→ B such that A
and B acquire R-module structures via φ and ψ respectively.
We want to show that the R-module A ⊗ B = A ⊗R B also has a
natural ring structure.
we would like to define

n
X

i=1

ai ⊗ bi 
m
X
j=1

cj ⊗ d j  =
n X
m
X
aicj ⊗ bidj
i=1 j=1
but the well definedness of this multiplication is not clear.
We discuss the theoretically correct way of introducing this multiplication in A ⊗ B.
Let a ∈ A and b ∈ B.
Define fa : A −→ A by setting fa(x) = ax for all x ∈ A.
Then fa is R-linear.
Similarly, gb : B −→ B defined by gb(y) = by for all y ∈ B is R-linear.
=⇒ fa ⊗ gb : A ⊗ B −→ A ⊗ B is R-linear.
Put hab = fa ⊗ gb.
=⇒ hab(x ⊗ y) = ax ⊗ by for all x ∈ A and y ∈ B.
Note that hab ∈ Hom(A ⊗ B, A ⊗ B).
We now define a map λ : A × B −→ Hom(A ⊗ B, A ⊗ B) by setting
λ(a, b) = hab for all a ∈ A and b ∈ B.
Claim. λ is R-bilinear.
Let α, β ∈ R, a, a1 ∈ A and b, b1 ∈ B.
=⇒ λ(αa + a1, βb + b1) = h(αa+a1)(βb+b1).
Let x ∈ A and y ∈ B.
=⇒ h(αa+a1)(βb+b1)(x ⊗ y) = (αa + a1)x ⊗ (βb + b1)y
= αβ(ax ⊗ by) + α(ax ⊗ b1y) + β(a1x ⊗ by) + a1x ⊕ b1y
= αβhab(x ⊗ y) + αhab1 (x ⊗ y) + βha1b(x ⊗ y) + ha1b1 (x ⊗ y)
= (αβhab + αhab1 + βha1b + ha1b1 )(x ⊗ y)
for all x ∈ A and y ∈ B.
As S = {x ⊗ y | x ∈ A, y ∈ B} generates A ⊗ B as an R-module and
the maps h(αa+a1)(βb+b1) and αβhab + αhab1 + βha1b + ha1b1 agree on
S, they are equal, that is,
h(αa+a1)(βb+b1) = αβhab + αhab1 + βha1b + ha1b1 .
=⇒ λ(αa + a1, βb + b1) = αβλ(a, b) + αλ(a, b1) + βλ(a1, b) + λ(a1, b1)
for all α, β ∈ R, a, a1 ∈ A and b, b1 ∈ B.
=⇒ λ is R-bilinear.
=⇒ There exists an R-linear map µ : A ⊗ B −→ Hom(A ⊗ B, A ⊗ B)
such that µ(a ⊗ b) = λ(a, b) = hab for all a ∈ A and b ∈ B.
Pn
i=1 ai ⊗ bi ∈ A ⊗ B.
Pn
Pn
Pn
=⇒ µ(X) = µ( i=1 ai ⊗ bi) = i=1 µ(ai ⊗ bi) = i=1 haibi .
P
Let Y = m
j=1 xj ⊗ yj ∈ A ⊗ B.
Let X =
Pn
=⇒ µ(X)(Y ) = ( i=1 haibi )(Y )
P
= n
i=1 hai bi (Y )
Pn
Pm
= i=1 haibi ( j=1 xj ⊗ yj )
Pm
P
= n
i=1 j=1 hai bi (xj ⊗ yj )
Pn Pm
= i=1 j=1(aixj ⊗ biyj ).
We define multiplication in A ⊗ B by X.Y = µ(X)(Y ).
Then
(1) . is commutative. Let X, Y ∈ A ⊗ B, where X =
Y =
Pn
i=1 ai ⊗ bi and
Pm
j=1 xj ⊗ yj .
Pn Pm
=⇒ X.Y = µ(X)(Y ) = i=1 j=1(aixj ⊗ biyj ) = µ(Y )(X) = Y.X
for all X, Y ∈ A ⊗ B.
Pn
(2) . is Associative. Let X, Y, Z ∈ A ⊗ B, where X = i=1 ai ⊗ bi,
P
Pl
Y = m
x
⊗
y
and
Z
=
j
j=1 j
k=1 sk ⊗ tk .
Pm Pl
P
Clearly, (X.Y ).Z = n
i=1 j=1 k=1 ai xj sk ⊗ bi yj tk .
Pn Pm Pl
Similarly, X.(Y.Z) = i=1 j=1 k=1 aixj sk ⊗ biyj tk .
=⇒ (X.Y ).Z = X.(Y.Z) for all X, Y, Z ∈ A ⊗ B.
Pn
(3) Unity. Let E = 1 ⊗ 1. Let X = i=1 ai ⊗ bi ∈ A ⊗ B.
P
=⇒ E.X = n
i=1 1ai ⊗ 1bi = X = X.E.
=⇒ 1 ⊗ 1 is the unity of A ⊗ B.
(4) Distributivity. Let X, Y, Z ∈ A ⊗ B.
=⇒ X.(Y + Z) = µ(X)(Y + Z) = µ(X)(Y ) + µ(X)(Z) as µ(X) is Rlinear.
=⇒ X.(Y + Z) = X.Y + X.Z and by commutativity (X + Y ).Z =
X.Z + Y.Z for all X, Y, Z ∈ A ⊗ B.
Lecture – 17
35. Restriction and extension of scalars
Let A be ring and let φ : R −→ A be aring homomorphism so that A
becomes an R-algebra via φ. More precisely, R-module structure of
A is given by ax = φ(a)x for all a ∈ R and x ∈ A.
Let N be an A-module. Then N also becomes an R-module in the
same way, that is, ax = φ(a)x for all a ∈ R and x ∈ N . When N is
treated as an R-module in this manner we say that N is an R-module
by restriction of scalars.
Now let M is an R-module.
Treat N as an R-module by restriction of scalars.
Then we can construct R-modules M ⊗R N and N ⊗R M .
We shall introduce A-module structure on both M ⊗R N and N ⊗R M .
However, the idea is the same so we do it only for M ⊗R N .
Let a ∈ A and let fa : N −→ N be given by fa(y) = ay for all y ∈ N .
Claim. fa is R-linear.
Let α ∈ R and y, y1 ∈ N .
=⇒ fa(αy + y1) = fa(φ(α)y + y1)
= a(φ(α)y + y1)
= φ(α)ay + ay1
= φ(α)fa(y) + fa(y1)
= αfa(y) + fa(y1).
=⇒ fa is R-linear.
We construct fa for all a ∈ A.
Let ffa = IM ⊗ fa : M ⊗R N −→ M ⊗R N .
We define a.X = ffa(X) for all a ∈ A and X ∈ M ⊗R N .
Note that ffa(x ⊗ y) = x ⊗ fa(y) = x ⊗ ay.
Pn
Pn
Pn
Pn
f
f
=⇒ a.( i=1 xi ⊗yi) = fa( i=1 xi ⊗yi) = i=1 fa(xi ⊗yi) = i=1 xi ⊗ayi.
Claim. M ⊗R N is an A-module with respect to the above.
Pn
i=1 xi ⊗ yi .
P
Pn
=⇒ 1.X = ff1(X) = n
x
⊗1y
=
i
i=1 i
i=1 xi ⊗yi = X for all X ∈ M ⊗R N .
(1) Let X =
(2) a.(X + Y ) = ffa(X + Y ) = ffa(X) + ffa(Y ) = a.X + a.Y for all a ∈ A
and X, Y ∈ M ⊗R N .
P
(3) Let X = n
i=1 xi ⊗ yi .
=⇒ (a + b).X = fg
a+b (X)
P
= n
i=1 xi ⊗ (a + b)yi
Pn
i=1 xi ⊗ (ayi + byi )
P
= n
i=1 (xi ⊗ ayi + xi ⊗ byi )
P
Pn
= n
x
⊗
ay
+
i
i=1 i
i=1 xi ⊗ byi
=
= a.X + b.X
for all a, b ∈ A and X ∈ M ⊗R N .
Pn
(4) Let a, b ∈ A and X = i=1 xi ⊗ yi ∈ M ⊗R N .
=⇒ (ab).X = fg
ab (X)
P
= n
i=1 xi ⊗ abyi
P
= n
i=1 xi ⊗ abyi
Pn
= a.( i=1 xi ⊗ byi)
= a.(b.X)
for all a, b ∈ A and X ∈ M ⊗R N .
Therefore, M ⊗R N is an A-module.
Comments. (1) If M and N both are A-modules then M ⊗R N becomes A-module via the first factor M or via the second factor N .
In such case we need to specify the way it is being considered as an
A-module.
(2) We can construct M ⊗R A and A ⊗R M . The module M ⊗R A is an
A-module via the second factor and A ⊗R M becomes an A-module
via the first factor. They are isomorphic as R-modules by the isomorphism φ : M ⊗R A −→ A ⊗R M , given by φ(x ⊗ a) = a ⊗ x. However, now
φ is also A-linear and hence M ⊗R A and A ⊗R M are also isomorphic
as A-modules.
(3) When we consider M ⊗R A (or A ⊗R M ) as an A-module, we say
that it is being treated as an A-module by extension of scalar.
Comments. Let I be an ideal of R.
(1) Since R is a module over itself and I is a submodule of R, R/I is
a quotient module of R. The R-module structure of R/I in this case
is given by a.(x + I) = ax + I.
(2) R/I is also a ring and we have the natural map π : R −→ R/I.
The map π is also a ring homomorphism.
R-algebra.
Therefore, R/I is an
The R-module structure of R-algebra R/I is given by
a.(x + I) = π(a)(x + I) = (a + I)(x + I) = ax + I.
These two R-module structures of R/I are identical.
(3) R/I is a module over itself and the R/I-module structure of R/I is
given by the multiplication of R/I. Therefore, R/I-module structure
of R/I is given by (a + I).(x + I) = ax + I.
(4) Let M be an R-module. Then the R-module R/I ⊗R M is an
R/I-module by extension of scalars. The R/I-modules structure of
R/I ⊗R M is given by (a + I).((b + I) ⊗ x) = ((a + I)(b + I)) ⊗ x =
(ab + I) ⊗ x.
(5) Note that I ⊆ annR (M/IM ) and hence M/IM is an R/I-module.
The R/I-module structure of M/IM is given by (a + I).(x + IM ) =
a(x + IM ) = ax + IM .
Lemma 47 Let M be an R-module and let I be an ideal of R. Then
∼ R/I ⊗ M as an R/I-module.
M/IM =
R
Proof. Note that R/I ⊗R M is an R/I-module by extension of scalers.
Furthermore, I ⊆ annR (M/IM ) and hence M/IM is also an R/Imodule.
We shall first construct an R-linear map from M/IM to R/I ⊗R M .
Define φ : M/IM −→ R/I ⊗R M by setting φ(x + IM ) = (1 + I) ⊗ x for
all x ∈ M .
Claim. φ is well defined.
Let x + IM = y + IM .
=⇒ x − y ∈ IM .
=⇒ There exists a1, . . . , an ∈ I and x1, . . . , xn ∈ M such that x − y =
Pn
i=1 ai xi .
Pn
i=1 ai xi )
P
+ (1 + I) ⊗ n
i=1 ai xi )
P
+ n
i=1 (1 + I) ⊗ ai xi
P
+ n
i=1 (ai (1 + I)) ⊗ xi
Pn
+ i=1 (ai + I ) ⊗ xi
P
+ n
i=1 0 ⊗ xi
=⇒ (1 + I) ⊗ x = (1 + I) ⊗ (y +
= (1 + I) ⊗ y
= (1 + I) ⊗ y
= (1 + I) ⊗ y
= (1 + I) ⊗ y
= (1 + I) ⊗ y
= (1 + I) ⊗ y.
=⇒ φ is well defined.
Claim. φ is R-linear.
Let a ∈ R and x + IM , y + IM ∈ M/IM .
=⇒ φ(a(x + IM ) + (y + IM ))
= φ((ax + y) + IM )
= (1 + I) ⊗ (ax + y)
= a((1 + I) ⊗ x) + ((1 + I) ⊗ y)
= aφ(x + IM ) + φ(y + IM )
for all a ∈ R and x + IM , y + IM ∈ M/IM .
=⇒ φ is R-linear.
We now define θ : R/I × M −→ M/IM by setting θ(a + I, x) = ax + IM
for all a ∈ R and x ∈ M .
Claim. θ is well defined.
Let a + I = b + I.
=⇒ a − b ∈ I.
=⇒ (a − b)x ∈ IM for all x ∈ M .
=⇒ ax + IM = bx + IM for all x ∈ M .
=⇒ θ is well defined.
Claim. θ is R-bilinear.
Let a, b, α, β ∈ R and x, x1 ∈ M . Then
θ(a(α + I) + (β + I), bx + x1)
= θ((aα + β) + I, bx + x1)
= ((aα + β)(bx + x1)) + IM
= (abαx + aαx1 + bβx + βx1) + IM
= (abαx + IM ) + (aαx1 + IM ) + (bβx + IM ) + (βx1 + IM )
= ab(αx + IM ) + a(αx1 + IM ) + b(βx + IM ) + (βx1 + IM )
= ab θ(α + I, x) + a θ(α + I, x1) + b θ(β + I, x) + θ(β + I, x1)
for all a, b, α, β ∈ R and x, x1 ∈ M .
=⇒ θ is R-bilinear.
=⇒ There exists an R-linear map ψ : R/I ⊗R M −→ M/IM such that
ψ((a + I) ⊗ x) = θ(a + I, x) = ax + IM .
Before we proceed, we want to see the structure of R/I ⊗R M .
Let X ∈ R/I ⊗R M .
=⇒ There exist a1, . . . , an ∈ R and x1, . . . , xn ∈ M such that
P
X= n
i=1 (ai + I) ⊗ xi .
Pn
=⇒ X = i=1(ai(1 + I)) ⊗ xi
P
= n
i=1 (1 + I) ⊗ ai xi
= (1 + I) ⊗
Pn
i=1 ai xi
Pn
= (1 + I) ⊗ x, where x = i=1 aixi.
=⇒ Every X ∈ R/I ⊗R M is of the form (1 + I) ⊗ x for suitable x ∈ M .
Now (ψ oφ)(x + IM ) = ψ(φ(x + IM )) = ψ((1 + I) ⊗ x)) = 1x + IM =
x + IM for all x ∈ M .
=⇒ ψ oφ is the identity map of M/IM .
Also, let X ∈ R/I ⊗R M .
=⇒ There exist some x ∈ M such that X = (1 + I) ⊗ x.
=⇒ (φoψ)(X) = φ(ψ(X)) = φ(ψ((1 + I) ⊗ x))
= φ(1x + IM ) = φ(x + IM ) = (1 + I) ⊗ x = X.
=⇒ φoψ is the identity map of R/I ⊗R M .
=⇒ φ is an isomorphism.
Claim. φ is R/I-linear.
Let a ∈ R and x ∈ M .
=⇒ φ((a + I)(x + M )) = φ(ax + M )
= (1 + I) ⊗ ax
= (a(1 + I)) ⊗ x
= (a + I) ⊗ x
= ((a + I)(1 + I)) ⊗ x
= (a + I)((1 + I) ⊗ x)
= (a + I)φ((x + IM ) for all a ∈ R and x ∈ M .
=⇒ φ is R/I-linear.
36. Commutative diagram
Let M, P1, . . . , Pn−1, Q1, . . . , Qm−1, N be R-modules. Let f0 : M −→ P1,
f1 : P1 −→ P2, ......, fn−1 : Pn−1 −→ Pn, fn : Pn −→ N , g0 : M −→ Q1,
g1 : Q1 −→ Q2, ......, gm−1 : Qm−1 −→ Qm and gm : Qm −→ N be Rlinear maps such that fnofn−1o . . . of1of0 = gmogm−1o . . . og1og0. Then
we say that the diagram
f
f
f
fn−1
0
1
2
−
→
P
−
→
. . . −−−→ M
−
→
P
M
2
1

n
g0 
y

yfn
g1
g2
g3
gm
Q1 −→ Q2 −→ Q3 −→ . . . −−→ N
is commutative. Although the diagram drawn here is of rectangular
shape but it can be of any arbitrary shape.
Commutative diagram are useful in visualising and understanding
given data. They also help in the construction of the proof.
We consider the following situation:
f
g
φ
ψ
Let M1 −→ M2 −→ M3 and L1 −→ L2 −→ L3 be two sequences (not
necessarily exact). Let αi : Mi −→ Li be isomorphisms such that the
squares in the diagram
f
g
M
1 −→ M
2 −→ M
3
α1 
y
α2 
y
φ
α3 
y
ψ
L1 −→ L2 −→ L3
are commutative.
This means α2of = φoα1 and α3og = ψ oα2.
We shall now show that the first sequence is exact iff the second
sequence is exact. Assume that the first sequence is exact.
(1) Im(φ) ⊆ ker(ψ). We shall solve it on diagram in this way.
Put x below L2 which is in Im(φ).
Put x1 below L1 which goes to x under φ.
Put x2 around M1 which goes to x1 under α1 (exists as α1 is onto).
Put x3 around M2, the image of x2 under f .
As the first square is commutative, x3 goes to x under α2.
As Im(f ) = ker(g), x3 goes to x4 = 0 under g, put it around M3.
This x4 = 0 goes to x5 = 0 under α3, put is around L3.
As the second square is commutative, x must go to x5 = 0 under ψ.
That is, x ∈ ker(ψ).
(2) ker(ψ) ⊆ Im(φ). We do it similarly.
Put x below L2 which is in ker(ψ).
Then x goes to x1 = 0 under ψ, put it around L3.
Put x2 around M2 which goes to x under α2 (exists as α2 is onto).
Put x3 around M3, the image of x2 under g.
As the second square is commutative, x3 goes to x1 = 0 under α3.
As α3 is an isomorphism, x3 = 0.
Therefore, x2 ∈ ker(g) = Im(f ).
Put x4 around M1 which goes to x2 under f .
Put x5 around L1, the image of x4 under α1.
As the first square is commutative, x4 must go to x under φ.
That is, x ∈ Im(φ).
Solving on the diagram does not appear to be very rigorous but a
rigorous proof can be constructed from these steps.
Similarly, if the second sequence is exact we prove that the first sequence is also exact.
As an application of this result we look at tensoring. We had proved
∼ N ⊗ M given by x ⊗ y goes to y ⊗ x for all x ∈ M and
that M ⊗ N =
y ∈ N.
f
g
Let M1 −→ M2 −→ M3 be an exact sequence and let P be an Rmodule.
Let αi : P ⊗ Mi −→ Mi ⊗ P be the isomorphism, i = 1, 2, 3.
Then the squares in the following diagram are commutative.
I ⊗f
I ⊗g
P ⊗M1 −−P−−→ P ⊗M2 −
−P−−
→ P ⊗M3
α1 
y
α2 
y
φ⊗I
α3 
y
ψ⊗I
P
P
M1 ⊗ P −−−−→
M2 ⊗ P −−−−→
M3 ⊗ P
Therefore, the upper sequence is exact iff the lower sequence is exact.
This means, for determining the projectivity of P , it does not matter
whether we tensor the given exact sequence by P from left or from
right. It also means that P ⊗ − is exact iff − ⊗ P is exact.
Lecture – 18
37. Rings and Modules of fractions
We shall be discussing a very important topic in commutative algebra
known as construction of rings and modules of fractions.
Let S ⊆ R. Then S is called a multiplicative set if 1 ∈ S and S is
closed under multiplication, that is, ab ∈ S for all a, b ∈ S.
We look at some examples of multiplicative sets.
(1) S = {1}.
(2) Let a ∈ R and S = {1, a, a2, . . .}.
(3) S = the set of all nonzero divisors of R.
(4) S = the set of all units of R.
(5) Let P be a prime ideal of R and let S = R \ P .
As P 6= R, 1 6∈ P , that us 1 ∈ S.
Let a, b ∈ S.
=⇒ a 6∈ P and b 6∈ P .
=⇒ ab 6∈ P .
=⇒ ab ∈ S for all a, b ∈ S.
(6) Intersection of multiplicative sets is multiplicative.
(7) Let I be an ideal of R and let S = {1 + a | a ∈ I}. Then S is
multiplicative.
Let s, t ∈ S.
=⇒ There exist a, b ∈ I such that s = 1 + a and t = 1 + b.
=⇒ st = 1 + a + b + ab ∈ S as a + b + ab ∈ I.
=⇒ S is multiplicative.
Let M be an R-module and let S ⊆ R be multiplicative.
Consider M × S.
We shall construct an abelian group from M × S by defining an equivalence relation in M × S and then defining a binary operation in the
set of all equivalence classes of this relation.
In M × S, define a relation ∼ by (x, s) ∼ (y, t) iff there exists some
u ∈ S such that
u(tx − sy) = 0 or utx = usy.
We shall show that ∼ is an equivalence relation.
(1) Reflexivity. Let (x, s) ∈ M × S.
=⇒ 1(sx − sx) = 0 for all (x, s) ∈ M × S .
=⇒ As 1 ∈ S, we are done.
(2) Symmetry. Let (x, s), (y, t) ∈ M × S such that (x, s) ∼ (y, t).
=⇒ There exists u ∈ S such that u(tx − sy) = 0.
=⇒ u(sy − tx) = 0.
=⇒ (y, t) ∼ (x, s).
=⇒ ∼ is symmetric.
(3) Transitivity. Let (x, s), (y, t), (z, r) ∈ M ×S such that (x, s) ∼ (y, t)
and (y, t) ∼ (z, r).
=⇒ There exist u, v ∈ S such that utx = usy and vry = vtz.
=⇒ uvrtx = uvrsy = usvry = usvtz or (uvt)rx = (uvt)sz.
As S is multiplicative and u, v, t ∈ S, we have uvt ∈ S.
=⇒ (x, s) ∼ (z, r).
Thus ∼ is an equivalence relation.
Therefore, ∼ partitions M × S into equivalence classes.
x
s
for all (x, s) ∈ M × S. Furthermore, the family of all the equivalence
We shall denote the equivalence class of (x, s) ∈ M × S under ∼ by
classes in M × S under ∼ shall be denoted by S −1M .
Comments. (1) In the definition of ∼, u appears to be superfluous.
However, it is required for the transitivity of ∼.
x
and s
(2) Let (x, s) ∈ M × S. Then we call x as the numerator of
s
x
as the denominator of
.
s
(3) Let (x, s) ∈ M × S and u ∈ S.
=⇒ 1(us(x)) = 1(s(ux)).
ux
x
for all (x, s) ∈ M × S and u ∈ S.
=⇒ (x, s) ∼ (ux, us) or =
s
us
x
This means, if we multiply the numerator and the denominator of
s
by the same u ∈ S then the equivalence class remains the same.
s0
0
0
= for all s ∈ S.
(4) =
s
s1
1
In S −1M , we define a binary operation ⊕ as follows: Let X, Y ∈ S −1M .
x
y
=⇒ There exist (x, s), (y, t) ∈ M × S such that X = and Y = .
s
t
As s, t ∈ S, st ∈ S.
tx + sy
Define ⊕ by setting X ⊕ Y =
.
st
So ⊕ is defined by choosing elements in X and Y . Therefore it does
not appear to be well defined. We shall prove that ⊕ is indeed well
defined.
x
y
Suppose that X = 1 and Y = 1 as well.
s1
t1
t1x1 + s1y1
tx + sy
We need to show that
=
.
st
s1 t1
x
x
y
y
As = 1 and = 1 , we have (x, s) ∼ (x1, s1) and (y, t) ∼ (y1, t1).
s
s1
t
t1
=⇒ There exist u, v ∈ S such that us1x = usx1 and vt1y = vty1.
tx + sy
uvs1t1(tx + sy)
=⇒
=
st
uvs1t1st
uvs1t1tx + uvs1t1sy
uvs1t1st
vt t(us1x) + uss1(vt1y)
= 1
uvs1t1st
vt t(usx1) + uss1(vty1)
= 1
uvs1t1st
uvst(t1x1) + uvst(s1y1)
=
uvs1t1st
t1x1 + s1y1
.
=
s1 t1
=⇒ ⊕ is well defined.
=
Before we proceed further we make the following observations.
(1) Let x, y ∈ M and s ∈ S.
x
y
sx + sy
s(x + y)
x+y
+ =
=
=
.
s
s
ss
ss
s
In other words, if two elements of S −1M have same denominator then
=⇒
in their sum the numerator is sum of their respective numerators and
the denominator is their common denominator.
(2) Let X1, . . . , Xn ∈ S −1M .
xi
=⇒ There exist xi ∈ M and si ∈ S such that Xi = , i = 1, . . . , n.
si
s1 . . . si−1si+1 . . . snxi
yi
=⇒ Xi =
= ,
s1 . . . si−1si+1 . . . snsi
s
where yi = s1 . . . si−1si+1 . . . snxi and s = s1 . . . sn.
In other words, all X1, . . . , Xn may be assumed to have the same denominator.
We shall now prove that S −1M is an abelian group.
Let X, Y, Z ∈ S −1M .
x
y
Then there exist s ∈ S and x, y, z ∈ M such that X = , Y =
and
s
s
z
Z= .
s
x y
x+y
y+x
y
x
(1) ⊕ is commutative. X ⊕ Y = ⊕ =
=
= ⊕ =
s
s
s
s
s
s
−1
Y ⊕ X for all X, Y ∈ S M .
z
x y
(2) ⊕ is Associative. (X ⊕ Y ) ⊕ Z =
⊕
⊕
s
s
s
x+y z
=
⊕
s
s
(x + y) + z
=
s
x + (y + z)
=
s
x y+z
= ⊕
s
s
= X ⊕ (Y ⊕ Z)
for all X, Y, Z ∈ S −1M .
0
(3) Identity. Note that ∈ S −1M .
1
0
x 0
1x + s0
x
0
And X ⊕ = ⊕ =
= =X = ⊕X
1
s 1
s1
s
1
for all X ∈ S −1M .
x
−x
(4) Inverses. For X = , let X 0 =
.
s
s
x −x
x−x
0
0
0
=
= = = X 0 ⊕ X.
Then X ⊕ X = ⊕
s
s
s
s
1
Thus X 0 is the inverse of X with respect to ⊕.
Therefore, S −1M is an abelian group under ⊕.
Henceforth, we shall use + instead of ⊕ and use standard notations
in the abelian group S −1M .
In particular, we can construct S −1R.
We shall define a multiplication in S −1R so that it becomes a ring.
We define this multiplication
as follows:
b
a
Let X = and Y = be in S −1R.
s
t
ab
Define X Y = .
st
We need to check well definedness of .
a0
b0
Let X = 0 and Y = 0 .
s
t
=⇒ There exist u, v ∈ S such that
uvs0t0ab
(us0a)(vt0b)
ab
=
=
=⇒
0
0
st
uvs t st
uvs0t0st
uvsta0b0
(usa0)(vtb0)
=
=
=
0
0
0
0
uvs t st
uvs t st
=⇒
is well defined.
We shall show that (S −1R, ⊕,
us0a = usa0 and vt0b = vtb0.
a0b0
.
0
0
st
) is a commutative ring with unity.
Let X, Y, Z ∈ S −1R.
a
b
=⇒ There exist s ∈ S and x, y, z ∈ R such that X = , Y =
and
s
s
c
Z= .
s
a b
ab
ba
b a
(1) Commutativity. X Y =
=
=
=
=Y
X for
s s
ss
ss
s s
all X, Y ∈ S −1R.
(2) Associativity. (X
Y)
(ab)c
ab c
=
ss s
sss
a(bc)
a bc
=
=
sss
s ss
= X (Y Z) for all X, Y, Z ∈ S −1R.
Z=
1
(3) Identity. We have ∈ S −1R.
1
1
a 1
a1
a
1
Also, X
=
=
= =X=
X for all X ∈ S −1R.
1
s 1
s1
s
1
a
b
c
a b+c
(4) Distributivity. X (Y + Z) =
+
=
s
s
s
s
s
a(b + c)
(ab + ac)
=
=
ss
ss
ab
ac
a b
a
+
=
+
ss
ss
s s
s
=X Y +X Z
=
c
s
for all X, Y, Z ∈ S −1R.
By commutativity, we also have
(X + Y )
Z=X
Z+Y
Therefore, (S −1R, +,
Z for all X, Y, Z ∈ S −1R.
) is a commutative ring with unity.
Hence-
forth, we shall use the standard notations in S −1R and suppress
.
We now show that S −1M is an S −1R-module.
a
x
−1
Let ∈ S R and ∈ S −1M .
s
t
a x
ax
Define · =
.
s t
st
We need to show that the above is well defined.
a
a0
x0
x
Let = 0 and = 0 .
s
s
t
t
=⇒ There exist u, v ∈ S such that us0a = usa0 and vt0x = vtx0.
uvs0t0ax
(us0a)(vt0x)
(usa0)(vtx0)
ax
=
=
=
=⇒
0
0
0
0
st
uvs t st
uvs t st
uvs0t0st
uvsta0x0
a0x0
=
= 0 0.
0
0
uvs t st
st
=⇒ · is well defined.
We now check that S −1M has S −1R-module structure.
a a0
x x0
−1
Let , 0 ∈ S R and , 0 ∈ S −1M .
s s
t t
1x
x
x
1 x
= for all ∈ S −1M .
(1) · =
1 t
1t
t
t
!
a
a0
s0a + sa0 x
(s0a + sa0)x
x
(2)
+ 0 · =
· =
s
s
t
ss0
t
ss0t
a a0
for all , 0
s s
s0ax
sa0x
s0ax + sa0x
=
= 0 + 0
0
ss t
ss t
ss t
ax
a0x
a x
a0 x
=
+ 0 = · + 0·
st
st
s t
s t
x
∈ S −1R and ∈ S −1M .
t
x0
a
x
(3) ·
+ 0
s
t
t
!
a t0x + tx0
a(t0x + tx0)
= ·
=
0
s
tt
stt0
at0x + atx0
t0ax
ta0x
=
=
+
0
0
stt
stt
stt0
a0x
a x0
ax
a x
+ 0 = · + · 0
=
st
st
s t
s t
a
x x0
−1
for all ∈ S R and , 0 ∈ S −1M .
s
t t
(4)
a a0
s s0
!
x
aa0 x
aa0x
· = 0· =
t
ss t
ss0t
a0x
a
a
= · 0 = ·
s st
s
a0
x
·
0
s t
!
x
a a0
for all , 0 ∈ S −1R and ∈ S −1M .
s s
t
Lecture – 19
Lemma 48 Let S be a multiplicative set in R and let M be an Rmodule. Let x ∈ M and s ∈ S.
x
(1) Then = 0 in S −1M iff there exists some t ∈ S such that tx = 0.
s
(2) If 0 ∈ S then S −1M = 0.
(3) If S −1R = 0 then 0 ∈ S.
x
0
= 0 = in S −1M .
s
1
=⇒ there exists some t ∈ S such that t1x = ts0 = 0, or tx = 0.
Proof. (1) Let
Conversely, suppose that there exists some t ∈ S such that tx = 0.
tx
0
ts0
0
x
=
=
= = 0.
=⇒ =
s
ts
ts
ts1
1
(2) Let X ∈ S −1M . Then there exist y ∈ M and u ∈ S such that
y
X= .
u
As 0y = 0, by (1), X = 0.
1
−1
(3) Assume that S R = 0. Then = 0.
1
By (1), there exists t ∈ S such that t1 = 0 = t, that is, 0 ∈ S.
Let M and L be R-modules and let f : M −→ L be R-linear.
Let S ⊆ R be multiplicative.
Define θ : S −1M −→ S −1L by setting θ
s ∈ S.
x
s
f (x)
=
for all x ∈ M and
s
We shall show that θ is well defined.
y
x
Let = .
s
t
=⇒ There exists u ∈ S such that utx = usy.
=⇒ f (utx) = f (usy) or utf (x) = usf (y).
f (x)
f (y)
=⇒
=
.
s
t
=⇒ θ is well defined.
Claim. θ is S −1R-linear.
Let X, Y ∈ S −1M and α ∈ S −1R.
x
y
=⇒ There exist x, y ∈ M , a ∈ R and s, t, u ∈ S such that X = , Y =
s
t
and α =
a
u
a x
y
ax
y
=⇒ θ(αX + Y ) = θ
+
=θ
+
u s
t
us
t
tax + usy
f (tax + usy)
=θ
=
ust
ust
taf (x) + usf (y)
taf (x)
usf (y)
=
=
+
ust
ust
ust
af (x)
f (y)
a f (x)
f (y)
+
=
+
=
us
t
u s
t
= αθ(X) + θ(Y ) for all X, Y ∈ S −1M and α ∈ S −1R.
=⇒ θ is S −1R-linear.
We denote θ by S −1f .
We look at some properties of this map.
(1) Let f : M −→ L be the map 0.
f (x)
0
x
x
=
= = 0 for all ∈ S −1M .
=⇒ (S −1f )
s
s
s
s
=⇒ S −1f = 0.
(2) Let f : M −→ M be the identity map IM .
x
f (x)
x
−1
=⇒ (S f )
=
= .
s
s
s
=⇒ S −1IM = IS −1M .
(3) Let f : M −→ L and g : L −→ K be R-linear.
!
x
x
f (x)
−1
−1
−1
−1
−1
= (S g) S f
=S g
=⇒ ((S g)o(S f ))
s
s
s
=
g(f (x))
s
!
x
(g of )(x)
−1
=
= (S (g of ))
s
s
x
for all ∈ S −1M .
s
=⇒ (S −1g)o(S −1f ) = (S −1(g of ).
(4) Let f : M −→ L be an isomorphism. Let g = f −1. Then g of = IM
and f og = IL.
=⇒ S −1g oS −1f = S −1(g of ) = S −1IM = IS −1M and S −1f oS −1g =
S −1(f og) = S −1IL = IS −1L. Hence S −1f is an isomorphism with inverse S −1(f −1).
We now prove a theorem.
f
g
Theorem. Let M −→ L −→ K be an exact sequence of R-modules
and R-linear maps. Then
exact.
S −1 f
S −1 g
−1
−1
S M −
−−−
→ S L −−−→ S −1K
Proof. We need to show that Im(S −1f ) = ker(S −1g).
(1) Im(S −1f ) ⊆ ker(S −1g). We are given ker(g) = Im(f ).
=⇒ g of = 0.
=⇒ (S −1g)o(S −1f ) = S −1(g of ) = 0.
=⇒ Im(S −1f ) ⊆ ker(S −1g).
(2) ker(S −1g) ⊆ Im(S −1f ). Let
x
∈ ker(S −1g).
s
is also
g(x)
x
=
=0
s
s
=⇒ There exists some t ∈ S such that tg(x) = 0 = g(tx).
=⇒ S −1g
=⇒ tx ∈ ker(g) = Im(f ).
=⇒ There exists some y ∈ M such that tx = f (y).
x
tx
f (y)
y
=⇒ =
=
= S −1f
.
s
ts
ts
ts
x
x
−1
=⇒ ∈ Im(S f ) for all ∈ ker(S −1g).
s
s
=⇒ ker(S −1g) ⊆ Im(S −1f ).
Comments. (1) In the language of homological algebra we say that
S −1− is an exact functor.
a
−1
(2) We have natural map R : R −→ S R given by a −→ . Clearly,
1
−1
R is a ring homomorphism. Therefore, S R is an R-algebra and so
an R-module. Note that
a
a
b a
ba
b · = R (b) =
=
for all b, a ∈ R and s ∈ S.
s
s
1 s
s
(3) Let M be an R-module. We treat S −1M as an R-module by restriction of scalars. Note that
x
x
a x
ax
a. = R (a) =
=
for all a ∈ R, x ∈ M and s ∈ S.
s
s
1 s
s
x
−1
(4) We again have a natural map M : M −→ S M given by x −→ .
1
Note that M is R-linear.
We now prove a very useful result.
Theorem. Let M be an R-module and let S ⊆ R be a multiplicative
∼ S −1R ⊗ M as S −1R-modules.
set. Then S −1M =
R
Proof. Note that S −1R ⊗R M is an S −1R-module by the first factor.
However, treating S −1R as an R-module we construct S −1R ⊗R M .
We also treat S −1M as an R-module by restriction of scalars.
a
ax
Define θ : S −1R × M −→ S −1M by setting θ
, x =
for all a ∈ R,
s
s
s ∈ S and x ∈ M .
Claim. θ is well defined.
a
b
Let = .
s
t
=⇒ There exists u ∈ S such that uta = usb
ax
utax
usbx
bx
=⇒
=
=
=
s
uts
uts
t
=⇒ θ is well defined.
Claim. θ is R-bilinear.
a b
Let α, β ∈ R, ,
∈ S −1R and x, y ∈ M .
s s
a
b
=⇒ θ α. + , βx + y
s
s
αa + b
, βx + y
=θ
s
(αa + b)(βx + y)
=
s
αβax + αay + βbx + by
=
s
αβax
αay
βbx
by
=
+
+
+
s
s
s
s
ay
bx
by
ax
+ α. + β. +
= (αβ).
s
s
s
s
a
a
b
b
= (αβ).θ
, x + α.θ
, y + β.θ , x + θ , y
s
s
s
s
a b
As α, β ∈ R, ,
∈ S −1R and x, y ∈ M are arbitrary, θ is R-bilinear.
s s
Therefore there exists an R-linear map ψ : S −1R ⊗R M −→ S −1M such
a
a
ax
that ψ
⊗x =θ
, x =
for all a ∈ R, s ∈ S and x ∈ M .
s
s
s
Claim. ψ is onto.
x
∈ S −1M .
s x
1
⊗x = .
=⇒ ψ
s
s
x
=⇒ ∈ Im(ψ).
s
x
As ∈ S −1M is arbitrary, ψ is onto.
s
Let
Before we proceed further, we look at the form of elements of
S −1R ⊗R M . Let X ∈ S −1R ⊗R M .
=⇒ There exist a1, . . . , an ∈ R, s ∈ S and x1, . . . , xn ∈ M such that
ai
Pn
i=1 s ⊗ xi .
ai 1
1
Pn
Pn
=⇒ X = i=1
⊗ xi = i=1 ai ·
⊗ xi
1s
s


n
1 X
1
1
P
Pn
aixi = ⊗ x, where x = n
aixi.
= i=1 ⊗ aixi = ⊗
i=1
s
s
s
i=1
X=
1
−1
Therefore every element of S R ⊗R M is of the form ⊗ x for some
s
suitable s ∈ S and x ∈ M .
Claim. ψ is one-one.
Let X ∈ ker(ψ).
=⇒ There exist some s ∈ S and x ∈ S −1M such that X =
1
⊗ x.
s
x
Also, ψ (X ) = 0 = .
s
=⇒ There exists some t ∈ S such that tx = 0.
1
1
t
1
1
=⇒ X = ⊗ x =
⊗x=
⊗x= t·
⊗ tx =
⊗ 0 = 0.
s
ts
ts
ts
ts
=⇒ ψ is one-one.
Claim. ψ is S −1R-linear.
Let X ∈ S −1R ⊗R M . Then X =
a 1
a
=⇒ ψ
.X = ψ
.
⊗x
t
t s
a
1
ax
=ψ
⊗x =ψ
⊗ ax =
ts
ts
ts
a x
a
= ψ(X).
=
t s
t
1
⊗ x for some s ∈ S and x ∈ M .
s
a
∈ S −1R and X ∈ S −1R ⊗R M are arbitrary, ψ is S −1R-linear.
t
−1 , is also an S −1 R-linear isomorphism and is given by
Note that
ψ
x
1
ψ −1
= ⊗ x.
s
s
As
Lecture – 20
38. Flatness of the ring of fractions
As a corollary to above theorem we have the following theorem.
Theorem. The ring S −1R is flat as an R-module.
f
g
Let M1 −→ M2 −→ M3 be an exact sequence of R-modules
and R-linear maps.
x
=
Let αi : S −1Mi −→ S −1R ⊗R Mi be the isomorphism given by αi
s
Proof.
1
⊗ x, i = 1, 2, 3.
s
Then the squares in the following diagram are commutative.
S −1M1
α1 
y
S −1R ⊗R M1
S −1 f
S −1M2
α2 
y
−−−−−−−−−→
IS −1 R ⊗f
−−−−−−→
S −1R ⊗
R M2
S −1 g
S −1M3
−−−−−−−−−→
IS −1 R ⊗g
−−−−−−→
α3 
y
S −1R ⊗R M3
As the top sequence is exact so is the bottom sequence.
=⇒ S −1R is a flat R-module.
Comments. If an R-algebra is flat as an R-module then we call it a
flat R-algebra.
39. An Isomorphism
We shall now prove a theorem.
Theorem. Let M and N be R-modules and let S be a multiplicative
∼ S −1(M ⊗ N ) as S −1R-modules.
set. Then S −1M ⊗ −1 S −1N =
S
R
R
−1 M × S −1 N −→ S −1 (M ⊗ N ) by setting
Proof.
We
define
θ
:
S
R
x y
x⊗y
θ
,
=
for all x ∈ M , y ∈ N and s, t ∈ R.
s t
st
Claim. θ is well defined.
x
y
y
x
= 1 and = 1 .
s
s1
t
t1
=⇒ There exist u, v ∈ S such that us1x = usx1 and vt1y = vty1.
x⊗y
uvs1t1(x ⊗ y)
=⇒
=
st
uvs1t1st
(us1x) ⊗ (vt1y)
=
uvs1t1st
(usx1) ⊗ (vty1)
=
uvs1t1st
usvt(x1 ⊗ y1)
=
uvs1t1st
x ⊗ y1
= 1
s1 t1
Therefore, θ is well defined.
Let
Claim. θ is S −1R-bilinear.
a b
x
x
y
y
Let , ∈ S −1R, 1 , 2 ∈ S −1M and 1 , 2 ∈ S −1N .
s t
s1 s 2
t1 t2
!
!
x2 b y1
y2
as2x1 + ss1x2 bt2y1 + tt1y2
a x1
+
,
+
=θ
,
=⇒ θ
s s1
s2
t t1
t2
ss1s2
tt1t2
(as2x1 + ss1x2) ⊗ (bt2y1 + tt1y2)
=
ss1s2tt1t2
as bt (x ⊗ y1) + as2tt1(x1 ⊗ y2) + ss1bt2(x2 ⊗ y1) + ss1tt1(x2 ⊗ y2)
= 2 2 1
ss1s2tt1t2
as2bt2(x1 ⊗ y1) as2tt1(x1 ⊗ y2) ss1bt2(x2 ⊗ y1) ss1tt1(x2 ⊗ y2)
=
+
+
+
ss1s2tt1t2
ss1s2tt1t2
ss1s2tt1t2
ss1s2tt1t2
ab(x1 ⊗ y1)
a(x1 ⊗ y2)
b(x2 ⊗ y1)
x2 ⊗ y2
+
+
+
=
ss1tt1
ss1t2
s2tt1
s2t2
a
s
a
=
s
=
b x1 ⊗ y1
t s1 t1
x1
b
θ
,
t
s1
a x1 ⊗ y2
b x2 ⊗ y1
x ⊗ y2
+
+ 2
s s1 t2
t s2t1
s2 t2
!
!
!
x1 y2
x2 y1
y1
a
b
x2
+ θ
,
+ θ
,
+θ
,
t1
s
s1
t2
t
s2
t1
s2
+
y2
t2
!
Therefore, θ is S −1R-bilinear.
−1 M ⊗
−1 N −→
Hence there exists an S −1
R-linear
map
φ
:
S
S
−1
S R
y
x
y
x
⊗
y
x
⊗
=θ
,
=
.
S −1(M ⊗R N ) such that φ
s
t
s
t
st
Our aim now is to construct the inverse of φ. This is a bit longish.
x y
We define λ : M × N −→ S −1M ⊗S −1R S −1N by setting λ(x, y) = ⊗
1 1
We treat S −1R-module S −1M ⊗S −1R S −1N as an R-module by restric-
tion of scalars.
Claim. λ is R-bilinear.
Let a, b ∈ R, x, x1 ∈ M and y, y1 ∈ N .
ax + x1 by + y1
⊗
=⇒ λ(ax + x1, by + y1) =
1
1
a x
x1
b y
y1
⊗
=
+
+
1 1
1
1 1
1
a x y1
b x1 y
x
a b x y
y
+
+
+ 1⊗ 1
=
⊗
⊗
⊗
1 1 1 1
1 1
1
1 1
1
1
1
ab
a
b
λ (x, y ) + λ (x, y1) + λ (x1, y ) + λ (x1, y1)
=
1
1
1
= (ab) · λ (x, y ) + a · λ (x, y1) + b · λ (x1, y ) + λ (x1, y1)
=⇒ λ is R-bilinear.
Therefore, there exists an R-linear map σ : M ⊗R N −→ S −1M ⊗S −1R
S −1N such that σ(x ⊗ y) = λ(x, y) =
x y
⊗
1 1
We now define ψ : S −1(M ⊗R N ) −→ S −1M ⊗S −1R S −1N by setting
1
X
= σ(X), for all X ∈ M ⊗R N and s ∈ S.
ψ
s
s
Claim. ψ is well defined.
X
Y
Let
= .
s
t
=⇒ There exists some u ∈ S such that utX = usY .
=⇒ (ut) · σ(X) = (us) · σ(Y ).
ut
us
=⇒
σ(X) =
σ(Y ).
1
1
1 ut
1 us
=⇒
σ(X) =
σ(Y ).
ust 1
ust 1
us
ut
σ(X) =
σ(Y ).
=⇒
ust
ust
1
1
=⇒
σ(X) = σ(Y ).
s
t
Therefore, ψ is well defined.
Claim. ψ is S −1R-linear.
a
Let ∈ S −1R and X, Y ∈ S −1(M ⊗R N ).
s
X1
=⇒ There exists t ∈ S, X1, Y1 ∈ M ⊗R N such that X =
and
t
Y1
Y =
.
t
a
a X1
Y1
aX1
Y1
aX1 + sY1
=⇒ ψ
X +Y =ψ
+
=ψ
+
=ψ
s
s t
t
st
t
st
1
1
=
σ(aX1 + sY1) =
(a · σ(X1) + s · σ(Y1))
st
st
a
1
1
σ(X1) + σ(Y1)
=
st 1
s
1 a
1 s
=
σ(X1) +
σ(Y1))
st 1
st 1
a
s
=
σ(X1) +
σ(Y1))
st
st
1
a 1
σ(X1) + σ(Y1))
=
s t
t
a
= ψ (X ) + ψ (Y )
s
Therefore, ψ is S −1R-linear.
Claim. φ is an isomorphism.
x y
x y
x⊗y
1
ψ oφ
⊗
=ψ φ
⊗
=ψ
= (σ(x ⊗ y))
s
t
s
t
st
st
1 x y
1 x 1 y
x y
=
⊗
=
⊗
=
⊗
st 1 1
s 1
t 1
s
t
x y
As
⊗ | x ∈ M, y ∈ N, s, t ∈ S generates S −1M ⊗S −1R S −1N as an
s
t
−1
S R-module, and ψ oφ and the identity of S −1M ⊗S −1R S −1N agree
on this set, we have ψ oφ = IS −1M ⊗
S
S −1 R
−1 N .
Let X ∈ S −1(M ⊗R N ).
=⇒ There exist x1, . . . , xn ∈ M , y1, . . . , yn ∈ N and s ∈ S such that
Pn
xi ⊗ yi
i=1
X=
s
!
Pn
xi ⊗ yi
i=1
=⇒ φoψ (X ) = φoψ
s
xi ⊗ yi
xi ⊗ yi
Pn
Pn
=φ ψ
= i=1 φ ψ
i=1
s
s
1 xi yi
1
xi yi
Pn
Pn
= i=1 φ
⊗
= i=1 φ
⊗
s 1
1
s
1
1
Pn
xi ⊗ yi
1 xi ⊗ yi
xi ⊗ yi
Pn
Pn
i=1
= i=1
= i=1
=
=X
s
1
s
s
Therefore, φoψ is the identity of S −1(M ⊗R N ).
=⇒ φ is an S −1R-linear isomorphism.
Let S be a multiplicative set of ring R. Then S −1R has the following
universal property.
Theorem. Let S be a multiplicative set of ring R. Let A be a ring
and let φ : R −→ A be a ring homomorphism such that φ(s) is a unit
of A for all s ∈ S. Then there exists a unique ring homomorphism
ψ : S −1R −→ A such that ψ o = φ, where : R −→ S −1R is natural
a
map (given by (a) = ).
1
Proof.
Define ψ : S −1R −→ A by setting ψ
a
s
= φ(s)−1φ(a) for all
a ∈ R and s ∈ S.
Claim. ψ is well defined.
b
a
Let = .
s
t
=⇒ There exists some u ∈ S such that uta = usb.
=⇒ φ(u)φ(t)φ(a) = φ(uta) = φ(usb) = φ(u)φ(s)φ(b).
=⇒ φ(u)−1φ(t)−1φ(s)−1φ(u)φ(t)φ(a)
= φ(u)−1φ(t)−1φ(s)−1φ(u)φ(s)φ(b)
=⇒ φ(s)−1φ(a) = φ(t)−1φ(b)
Therefore, ψ is well defined.
Claim. ψ is a ring homomorphism.
(as φ(u), φ(s), φ(s) are units of A).
a b
Let , ∈ S −1R.
s t
a
b
ta + sb
=⇒ ψ
+
=ψ
s
t
st
= φ(st)−1φ(ta + sb) = (φ(s)φ(t))−1(φ(t)φ(a) + φ(s)φ(b))
= φ(s)−1φ(t)−1φ(t)φ(a) + φ(s)−1φ(t)−1φ(s)φ(b)
a
b
−1
−1
= φ(s) φ(a) + φ(t) φ(b) = ψ
+ψ
.
s
t
ab
a b
=ψ
= φ(st)−1φ(ab) = (φ(s)φ(t))−1φ(ab)
Also, ψ
s t
st
= φ(s)−1φ(t)−1φ(a)φ(b)
a
b
= φ(s)−1φ(a)φ(t)−1φ(b) = ψ
ψ
.
s
t
1
And ψ
= φ(1)−1φ(1) = 1.
1
Therefore, ψ is a ring homomorphism.
a
= φ(1)−1φ(a) = 1−1φ(a) =
We also have, ψ o(a) = ψ((a)) = ψ
1
φ(a) for all a ∈ R.
Therefore, ψ o = φ.
Uniqueness. Let ψ 0 : S −1R −→ A be a ring homomorphism such that
ψ 0o = φ.
a
Let ∈ S −1R.
s
1
s
1
s
1
=⇒ 1 = ψ 0
= ψ0
= ψ0
ψ0
1
1 s
1
s
1
1
1
0
0
0
0
0
= ψ ((s)) ψ
= ψ o (s) ψ
= φ(s)ψ
.
s
s
s
1
= φ(s)−1.
s
a
1
1
a
a
0
0
0
0
=⇒ ψ
=ψ
=ψ
ψ
= φ(s)−1ψ 0 ((a))
s
s 1
s
1
a
= φ(s)−1ψ 0o(a) = φ(s)−1φ(a) = ψ
for all a ∈ R and s ∈ S.
s
Therefore, ψ 0 = ψ.
=⇒ ψ 0
Quotient field of an integral domain. Let R be an integral domain
and let S = R \ {0}. As product of any two non-zero elements of R is
again non-zero, S is a multiplicative set of R. Put K = S −1R.
Claim. K is a field.
a
Let X ∈ K and X 6= 0. Let a ∈ R and s ∈ S such that X = .
s
As X 6= 0, a can not be 0. Therefore, a ∈ S
s
=⇒ ∈ K.
a
as
1
a s
=
= .
=⇒
s a
sa
1
a
=⇒
is a unit of K.
s
=⇒ K is a field.
This field K is called the quotient field of R.
a
Let : R −→ K be the natural map, that is, (a) = for all a ∈ R.
1
Claim. is one-one.
a
Let a ∈ ker(). =⇒ = (a) = 0.
1
=⇒ There exists some s ∈ S such that sa = 0.
As R is integral domain and s 6= 0, we have a = 0.
Let L be a field and let φ : R −→ L be a one-one ring homomorphism.
Let a ∈ R be non-zero.
Then φ(a) 6= 0 (as φ is one-one) and hence φ(a) is a unit of L.
Then by the previous result, there exists a ring homomorphism ψ : K −→
a
= φ(s)−1φ(a). We also have φ = ψ o.
L given by ψ
s
Claim. ψ is one-one.
a
Let ∈ ker(ψ).
s a
= 0 = φ(s)−1φ(a).
=⇒ ψ
s
=⇒ φ(s)φ(s)−1φ(a) = 0.
=⇒ φ(a) = 0.
=⇒ a = 0 (as φ is one-one).
a
0
=⇒ = = 0.
s
s
=⇒ ψ is one-one.
As φ = ψ o, we have φ(R) ⊆ ψ(K) ⊆ L.
So we conclude the following: If a field L contains an isomorphic copy
of R then L also contains an isomorphic copy of K and the isomorphic
copy of K in L contains the given isomorphic copy of R.
Lecture – 21
40. Submodules and modules of fractions
Comments. (1) Let S be a multiplicative set in R. Let M be an
R-module and let N be a submodule of M . Let : N −→ M be the
inclusion map. Then 0 −→ N −→ M is exact.
S −1 ()
−1
=⇒ 0 −−−−→ S N −−−−−→ S −1M
=⇒ S −1() is one-one.
is exact.
=⇒ S −1N is isomorphic to its image in S −1M under S −1().
x
x
x
−1
−1
Let ∈ S N . Then notationally, S ()
= .
s
s
s
=⇒ The image of S −1N in S −1M under S −1() is precisely the set
x
| x ∈ N, s ∈ S . We also denote this set by S −1N .
s
Therefore, S −1N denotes two objects now
(a) the S −1R-module when N is treated as an R-module and S −1N
is constructed.
(b) The image of S −1R-module S −1N in S −1M under the map S −1().
Since S −1() is S −1R-linear, this image is a S −1R-submodule of S −1M .
This may cause some ambiguity occasionally.
(2) Let x ∈ N and s ∈ S. Then the equivalence class of (x, s) in N × S
is contained in the equivalence class of (x, s) in M × S. Hence the
equivalence class of (x, s) in N × S is smaller than the equivalence
x
class of (x, s) in M × S. Notationally, both are denoted by .
s
Prime ideals of S −1R
We discuss prime ideals of S −1R.
We prove a very elementary result.
Lemma 49 Let P be a prime ideal of R and let S be a multiplicative
a
set of R such that P ∩ S = ∅. Let a ∈ R and s ∈ S such that ∈ S −1P .
s
Then a ∈ P .
a
b
Proof. There exist b ∈ P and t ∈ S such that = .
s
t
=⇒ There exists some u ∈ S such that uta = usb ∈ P
(as b ∈ P ).
=⇒ Either ut ∈ P or a ∈ P .
But ut ∈ S and P ∩ S = ∅ and thus ut 6∈ P .
=⇒ a ∈ P .
Lemma 50 Let P be a prime ideal of R and let S be a multiplicative
set of R. If P ∩ S = ∅ then S −1P is a prime ideal of S −1R.
Proof. As P is an ideal of R, it is also a submodule of R. Hence S −1P
is an S −1R-submodule of S −1R. Therefore, S −1P is an ideal of S −1R.
Claim. S −1P is a proper ideal of S −1R.
Suppose, if possible, that S −1P = S −1R.
1
∈ S −1P .
1
=⇒ BY Lemma 49, 1 ∈ P , that is, P = R, a contradiction.
=⇒
=⇒ S −1P 6= S −1R.
Claim. S −1P is a prime ideal of S −1R.
Let X, Y ∈ S −1R such that XY ∈ S −1P .
=⇒ There exist a, b ∈ R and s, t ∈ S such that X =
a b
ab
−1
Therefore,
∈ S P or
∈ S −1P .
s t
st
=⇒ By Lemma 49, ab ∈ P .
As P is a prime ideal either a ∈ P or b ∈ P .
a
b
and Y = .
s
t
Assume that a ∈ P .
a
=⇒ X = ∈ S −1P .
s
=⇒ S −1P is a prime ideal of S −1R.
Lemma 51 Let S be multiplicative set in R such that 0 6∈ S. Let
: R −→ S −1R be the natural map. Let Q be a prime ideal of S −1R.
Then P = −1(Q) is is a prime ideal of R such that P ∩ S = ∅ and
S −1P = Q.
Proof. As 0 6∈ S, S −1R 6= 0. Hence S −1R has prime ideals.
By an earlier result P = −1(Q) is a prime ideal of R.
Claim. P ∩ S = ∅.
Suppose, if possible, that P ∩ S 6= ∅.
Let s ∈ P ∩ S.
s
=⇒ = (s) ∈ Q.
1
s 1
s
1
s
But
= = , that is
is a unit of S −1R.
1 s
s
1
1
=⇒ Q = S −1R, a contradiction.
Therefore, P ∩ S = ∅.
Claim. S −1P = Q.
a
Let X ∈ Q. Then there exist a ∈ R and s ∈ S such that X = .
s
s a
a
∈Q
=⇒ (a) = =
1
1 s
a
=⇒ a ∈ P and hence X = ∈ S −1P for all X ∈ Q.
s
=⇒ Q ⊆ S −1P .
Conversely, let X ∈ S −1P .
a
=⇒ There exist some a ∈ P and s ∈ S such that X = .
s
a
=⇒ = (a) ∈ Q.
1
1 a
=⇒ X =
∈ Q for all X ∈ S −1P .
s 1
=⇒ S −1P ⊆ Q or Q = S −1P .
Theorem. Let S be multiplicative set in R such that 0 6∈ S. Let
: R −→ S −1R be the natural map. Let
X = {Q ⊆ S −1R | Q is a prime ideal of S −1R}, and
Y = {P ⊆ R | P is a prime ideal of R and P ∩ S = ∅}.
Then the map θ : X −→ Y given by θ(Q) = −1(Q) is a bijection.
Proof.
Let Q ∈ X. Then Q is a prime ideal of S −1R and −1(Q) is
a prime ideal of R such that −1(Q) ∩ S = ∅.
=⇒ θ is well defined.
Claim. θ is one-one.
Let Q1, Q2 ∈ X such that θ(Q1) = θ(Q2). Pt P = θ(Q1) = θ(Q2).
=⇒ Q1 = S −1P = Q2, by Lemma 51.
=⇒ θ is one-one.
Claim. θ is onto.
Let P ∈ Y . Then P is a prime ideal of R such that P ∩ S = ∅.
=⇒ By Lemma 50, Q = S −1P is a prime ideal of S −1R.
Put P1 = −1(Q).
Let a ∈ P .
a
=⇒ ∈ S −1P = Q.
1
=⇒ (a) ∈ Q and hence a ∈ P1 for all a ∈ P .
=⇒ P ⊆ P1.
Let a ∈ P1.
a
= (a) ∈ Q = S −1P .
1
=⇒ By Lemma 49, a ∈ P for all a ∈ P1.
=⇒
=⇒ P1 ⊆ P or P = P1 = −1(Q) = θ(Q).
=⇒ θ is onto.
Let P be a prime ideal of R and let S = R \ P .
Let M be an R-module.
Then we write MP instead of S −1M and RP instead of S −1R.
We call RP and MP as localisation of R and M respectively, at P .
Furthermore, if M and L are R-modules and φ : M −→ N is an R-linear
map then we write φP instead of the map S −1φ.
Lemma 52 Let P be a prime ideal of R. Then RP is a local ring
with maximal ideal PP .
Proof. Clearly, 0 6∈ R \ P . Hence RP 6= 0.
=⇒ RP has maximal ideals.
Let m be a maximal ideal of RP .
Let : R −→ RP be the natural map and let Q = −1(m).
Then, by Lemma 51, Q is a prime ideal of R, Q ∩ (R \ P ) = ∅ and
QP = m.
=⇒ Q ⊆ P (as Q ∩ (R \ P ) = ∅).
=⇒ m = QP ⊆ PP .
As P ∩ (R \ P ) = ∅, by Lemma 50, PP is a prime ideal of RP . Hence
PP 6= RP . =⇒ m = PP .
In other words, RP has only one maximal ideal, namely, PP . Therefore, RP is local.
Lecture – 22
41. Local properties
Definition. A property X is called a local property of ring R (or an
R-module M ) if R (or M ) has X iff RP (or MP ) has X for every prime
ideal P of R. We shall prove some local properties.
Lemma 53
Let M be an R-module and let x ∈ M .
following statements are equivalent
(a) x = 0.
x
(b) = 0 in MP for all prime ideals P of R.
1
x
(c) = 0 in Mm for all maximal ideals m of R.
1
Then the
Proof. (a) =⇒ (b) is obvious and (b) =⇒ (c) is true as every maximal ideal is also a prime ideal.
(c) =⇒ (a) Put I = {a ∈ R | ax = 0}.
Then I is an ideal of R (it is called the annihilator of element x and
is denoted by annR (x)).
Claim. I = R.
Suppose, if possible, that I 6= R.
=⇒ There exists a maximal ideal m of R such that I ⊆ m.
x
By (c), = 0 in Mm.
1
=⇒ There exists some s ∈ R \ m such that sx = 0.
=⇒ s ∈ I ∩ (R \ m) ⊆ m ∩ (R \ m) = ∅, a contradiction.
=⇒ I = R, that is 1 ∈ I.
=⇒ x = 1x = 0.
Lemma 54
Let M and N be R-modules and let φ : M −→ N be
R-linear. Then the following statements are equivalent
(a) φ is one-one.
(b) φP is one-one for all prime ideals P of R.
(c) φm is one-one for all maximal ideals m of R.
Proof. (a) =⇒ (b) Follows by Theorem after Lemma 48.
(b) =⇒ (c) True as every maximal ideal is also a prime ideal.
(c) =⇒ (a) Let x ∈ ker(φ). Let m be a maximal ideal of R.
φ(x)
0
x
=
= = 0.
=⇒ φm
1
1
1
x
=⇒ = 0 in Mm for all maximal ideals m of R
1
By Lemma 53, x = 0.
(as φm is one-one).
=⇒ φ is one-one.
Lemma 55
Let M and N be R-modules and let φ : M −→ N be
R-linear. Then the following statements are equivalent
(a) φ is onto.
(b) φP is onto for all prime ideals P of R.
(c) φm is onto for all maximal ideals m of R.
Proof. (a) =⇒ (b) Let P be a prime ideal of R.
φP
φ
As M −→ N −→ 0 is exact, so is MP −→ NP −→ 0.
=⇒ φP is onto.
(b) =⇒ (c) Obvious as every maximal ideal is also a prime ideal.
(c) =⇒ (a) Let m be a maximal ideal of R. Let L = N/Im(φ) and let
π : N −→ L be the natural map. Consider the sequence
φ
π
M −−−−−→ N −−−−−→ L −−−−−→ 0.
As ker(π) = Im(φ) and π is onto, this sequence is exact.
π
=⇒ Mm −−−φ−m
−−→ Nm −−−−m
−−→ Lm −−−−→ 0 . . . . . . . . . . . . . . . . (1)
is also exact.
=⇒ ker(πm) = Im(φm) and πm is onto.
By (c), φm is onto and therefore, Im(φm) = Nm.
=⇒ ker(πm) = Im(φm) = Nm.
=⇒ πm is the map 0.
=⇒ Lm = Im(πm) = 0 (as πm is onto).
Let x ∈ L.
x
=⇒ = 0 in Lm (as Lm = 0).
1
x
=⇒ = 0 in Lm for every maximal ideal m of R.
1
Therefore, by Lemma 53, x = 0 for all x ∈ L.
=⇒ L = 0, that is, N = Im(φ).
=⇒ φ is onto.
Before we come to the next result we state a result without proof.
Theorem. Let M be flat R-module and let A be an R-algebra. Then
A ⊗R M is a flat A-module.
Now the result.
Lemma 56 Let M be an R-module. Then the following statements
are equivalent
(a) M is flat over R.
(b) MP is flat over RP for all prime ideals P of R.
(c) Mm is flat over Rm for all maximal ideals m of R.
Proof. (a) =⇒ (b) By the Theorem (before this Lemma), RP ⊗R M
is flat over RP .
As MP is isomorphic to RP ⊗R M as an RP -module, MP is also flat
as an RP -module.
(b) =⇒ (c) Obvious as every maximal ideal is also a prime ideal.
(c) =⇒ (a) Let X and Y be R-modules and let φ : X −→ Y be a
one-one R-linear map.
We need to show that φe = IM ⊗φ : M ⊗R X −→ M ⊗R Y is also one-one.
e X
f = M ⊗ X and Ye = M ⊗ Y .
Put ψ = φ,
R
R
Let m be a maximal ideal of R.
f
Let α be the Rm-linear isomorphism from X
m = (M ⊗R X)m to
Mm ⊗Rm Xm.
z⊗x
1 z
x
Then α
=
⊗
for all z ∈ M , x ∈ X, and s ∈ R \ m.
s
s 1 1
Let β be the Rm-linear isomorphism from Yem = (M ⊗R Y )m to
Mm ⊗Rm Ym.
z⊗y
1 z
y
Then β
=
⊗
for all z ∈ M , y ∈ Y and s ∈ R \ m.
s
s 1 1
As φ : X −→ Y is a one-one R-linear map, φm : Xm −→ Ym is also a
one-one Rm-linear map.
By (c), Mm is a flat Rm-module. Therefore
IMm ⊗ φm : Mm ⊗Rm Xm −→ Mm ⊗Rm Ym is also a one-one Rm-linear
map.
f −→ Ye is R -linear ( induced by ψ).
Note that ψm : X
m
m
m
Consider the following diagram
ψm
−−−−
−→
f
X
m

αy
IM ⊗φm
Yem

βy
Mm ⊗Rm Xm −
−−m
−−−−
→ Mm ⊗Rm Ym
Note that
z⊗x
z⊗x
= β ψm
(β oψm)
s
s !
ψ(z ⊗ x)
=β
s
=β
z ⊗ φ(x)
s
!
φ(x)
z
⊗
1
1
1
=
s
!
And
IMm ⊗ φm oα
z ⊗ x
s
z ⊗ x = IMm ⊗ φm
α
s
1 z
x
⊗
s 1 1
= IMm ⊗ φm
1 z
x
=
⊗ φm
s 1
1
!
1 z
φ(x)
=
⊗
s 1
1
z⊗x
= (β oψm)
s
f = (M ⊗ X) .
Now let Z ∈ X
m
R
=⇒ There exist z1, . . . , zn ∈ M , x1, . . . , xn ∈ X and s ∈ R \ m such that
Pn
z i ⊗ xi
Z = i=1
.
s
!
Pn z ⊗ x
i
i=1 i
=⇒ IMm ⊗ φm oα (Z ) = IMm ⊗ φm oα
s
P
z
⊗
x
i
i
n
= IMm ⊗ φm oα
i=1
s
n z ⊗x X
i
i
=
IMm ⊗ φm oα
s
i=1
n
X
zi ⊗ xi
=
(β oψm)
s
i=1

= (β oψm)

n
X zi ⊗ xi


i=1
s
= (β oψm)
!
Pn
i=1 zi ⊗ xi = (β oψ ) (Z ).
m
s
Therefore, IMm ⊗ φm oα = β oψm
Claim. ψm is one-one.
Let Z ∈ ker(ψm).
=⇒
IMm ⊗ φm oα (Z) = (β oψm) (Z) = β (ψm(Z)) = β(0) = 0
As IMm ⊗ φm and α are both one-one, so is IMm ⊗ φm oα.
=⇒ Z = 0, that is, ψm is one-one for every maximal ideal m of R.
Therefore, by Lemma 54, ψ = φe is one-one.
=⇒ M is flat.
Assignment – I
MTH611A – 2020-21
1. Let R and A be rings and φ : R −→ A be a ring homomorphism.
P
Pn
n
i
i.
e
e
a
x
=
φ(a
)x
Define φ : R[x] −→ A[x] by φ
i
i
i=0
i=0
that φe is a ring homomorphism.
e
We have φ(1)
= φ(1) = 1.
Let f, g ∈ R[x], f =
Pn
i and g = Pm b xj .
a
x
j=0 j
i=0 i
Assume that m ≥ n and put an+1 = · · · = am = 0.
Pm
Then f + g = i=0(ai + bi)xi.
Show
Pm
Pm
i
e
=⇒ φ(f + g) = i=0 φ(ai + bi)x = i=0(φ(ai) + φ(bi))xi
P
i + Pm φ(b )xi = φ(f
e ) + φ(g).
e
= m
φ(a
)x
i
i
i=0
i=0
Pn Pm
i+j .
i=0 j=0 ai bj x
Pn Pm
Pn Pm
i+j )
i+j
e
e
e
) = i=0 j=0 φ(a
=⇒ φ(f g) = φ( i=0 j=0 aibj x
i bj x
Pm
Pm
P
i+j
i+j = Pn
φ(a
)
φ(b
)x
φ(a
b
)x
= n
i
j
i
j
i=0
i=0 j=0
j=0
P
P
m φ(b )xj = φ(f
n
i
e )φ(g).
e
=
j
j=0
i=0 φ(ai )x
Clearly, f g =
2. (a) Let I be an ideal of R. Put
J = {f (x) ∈ R[x] | all coefficients of f (x) are in I}.
Show that J is an ideal of R[x], where R[x] is the polynomial ring
in one variable x over R. We denote this ideal as I[x].
(b) Show that if I is prime then so is I[x].
∼ A[x] as rings.
(c) Let A = R/I. Show that R[x]/I[x] =
(a) Easy.
(b) Let f, g ∈ R[x], f =
Pn
i and g = Pm b xj such that
a
x
i=0 i
j=0 j
f g ∈ I[x].
Assume that f 6∈ I[x].
=⇒ At least one of the coefficients of f is not in I.
Let k ≥ 0 be the smallest integer such that ak 6∈ I.
=⇒ ai ∈ I for all i = 0, . . . , k − 1.
Suppose, if possible, that g 6∈ I[x].
Let l ≥ 0 be the smallest integer such that bl 6∈ I.
=⇒ bj ∈ I for all j = 0, . . . , l − 1.
As f g ∈ I[x], all coefficients of f g are in I.
In particular, coefficient of xk+l ∈ I.
P
i+j=k+l ai bj ∈ I.
P
P
=⇒ i+j=k+l,i<k aibj + ak bl + i+j=k+l,j<l aibj ∈ I.
P
In i+j=k+l,i<k aibj as i < k, ai ∈ I and hence all terms of this
This means,
summation are in I.
=⇒
P
P
a
b
∈
I
and
similarly,
i+j=k+l,i<k i j
i+j=k+l,j<l ai bj ∈ I.
=⇒ ak bl ∈ I, a contradiction, as I is prime and ak 6∈ I, bl 6∈ I.
=⇒ g ∈ I[x].
Therefore, I[x] is prime.
(c) Let π : R −→ A be the natural map, which is a ring homomorphism.
e : R[x] −→ A[x] is a ring homomorphism. Clearly, π
e is onto.
=⇒ π
Pn
Let f = i=0 aixi ∈ R[x].
e ) ⇐⇒ π
e (f ) = 0 =
f ∈ ker(π
Pn
i
π(a
)x
i
i=0
⇐⇒ π(ai) = 0 for all i = 0, 1, . . . , n
⇐⇒ ai ∈ I for all i = 0, 1, . . . , n
⇐⇒ f ∈ I[x].
e ) = I[x].
=⇒ ker(π
Therefore, by fundamental theorem of homomorphism for rings,
∼ A[x] as rings.
R[x]/I[x] =
3. Let a, b ∈ R such that a is a unit and b is nilpotent. Show that
a + b is a unit.
Suppose, if possible, that a + b is a nonunit.
=⇒ There exists a maximal ideal m of R such that a + b ∈ m.
But b is nilpotent, therefore, it is in every prime ideal and hence
b ∈ m.
=⇒ a ∈ m, that is, m = R.
A contradiction, and therefore, a + b is a unit.
Pn
i
4. Let f =
i=0 ai x ∈ R[x].
Then f is said to be primitive if
(a0, . . . , an) = R . Let f, g ∈ R[x]. Show that f and g are primitive
iff f g is primitive.
Pn
Pm
i
Assume that f = i=0 aix and g = i=0 bixi are primitive.
Assume, if possible, that f g is not primitive.
Let f g =
Pn+m
i
i=0 ci x .
Then (c0, c1, . . . , cn+m) 6= R.
Let m be a maximal ideal such that (c0, c1, . . . , cn+m) ⊆ m.
=⇒ f g ∈ m[x].
As m[x] is prime, either f ∈ m[x] or g ∈ m[x].
Assume that f ∈ m[x].
=⇒ a0, a1, . . . , an ∈ m.
=⇒ R = (a0, a1, . . . , an) ⊆ m, a contradiction.
Therefore, f g is primitive.
Conversely, let f g be primitive.
Assume, if possible, f is not primitive.
=⇒ (a0, a1, . . . , an) 6= R.
Let m be a maximal ideal such that (a0, a1, . . . , an) ⊆ m.
=⇒ ck = i+j=k aibj ∈ m for all k = 0, 1, . . . , n + m.
=⇒ (c0, . . . , cn+m) ⊆ m, a contradiction.
P
=⇒ f is primitive.
Similarly, g is primitive.
5. Let f =
Pn
i ∈ R[x]. Show that f is a unit iff a is a unit and
a
x
0
i
i=0
a1, . . . , an are nilpotents.
Let f be a unit.
=⇒ There exists a g =
Pm
i ∈ R[x] such that f g = 1.
b
x
i
i=0
=⇒ a0b0 = 1 and hence a0 is a unit.
We shall now prove that a1, . . . , an are nilpotents.
If n = 0 then there is nothing to prove.
Assume that n ≥ 1.
Let P be a prime ideal.
Claim. ai ∈ P for all i = 1, . . . , n.
Let A = R/P and let π : R −→ A be the natural map.
e : R[x] −→ A[x] and ker(π
e) =
Then we have the ring homomorphism π
P [x].
e (f g) = π
e (1) = 1 = 1 + P .
As f g = 1, we have π
e (f )π
e (g) = 1.
=⇒ π
e (f )π
e (g)) = deg(1) = 0.
=⇒ deg (π
e (f )) + deg (π
e (g)) = deg(1) = 0 (as A is an integral
=⇒ deg (π
domain).
e (f )) = 0 = deg (π
e (g)).
=⇒ deg (π
=⇒ π(ai) = 0 for all i = 1, . . . , n.
=⇒ ai ∈ P for all i = 1, . . . , n.
As P is arbitrary, ai ∈ nil(R) for all i = 1, . . . , n.
Therefore, ai is nilpotent for all i = 1, . . . , n.
Conversely, let a0 be a unit of R and a1, . . . , an be nilpotents.
=⇒ a1x, a2x2, . . . , anxn are nilpotents of R[x].
=⇒ a1x + a2x2 + · · · + anxn is a nilpotent of R[x].
=⇒ f = a0 + a1x + a2x2 + · · · + anxn is a unit of R[x].
6. Show that nil(R[x]) = J(R[x]).
We know that nil(R[x]) ⊆ J(R[x]).
Pn
Let f = i=0 aixi ∈ J(R[x]).
=⇒ 1 + xf is a unit of R[x].
=⇒ a0, . . . , an are nilpotents.
=⇒ a0, a1x, a2x2, . . . , anxn are nilpotents of R[x].
Pn
=⇒ f = i=0 aixi is a nilpotent of R[x] for all f ∈ J(R[x]).
=⇒ J(R[x]) ⊆ nil(R[x]).
7. An element e ∈ R is called an idempotent if e2 = e.
If in R
every ideal which is not contained in the nil(R) contains a nonzero
idempotent then show that J(R) = nil(R).
Suppose, if possible, that J(R) 6⊆ nil(R).
=⇒ There exists a nonzero idempotent e ∈ R such that e ∈ J(R).
As e2 = e, we have (1 − e)e = 0 and as e ∈ J(R), 1 − e is a unit.
=⇒ e = 0, a contradiction.
8. A ring R has the property that for every element x ∈ R there
exists some n > 1 such that xn = x. Show that every prime ideal
is maximal in R.
Let P be a prime ideal of R.
As P is proper ideal, there exists a maximal ideal m such that
P ⊆ m.
Claim. P = m.
Suppose, if possible, that P 6= m.
Let x ∈ m such that x 6∈ P .
=⇒ There exists some n > 1 such that xn = x, that is,
x(xn−1 − 1) = 0 ∈ P .
=⇒ xn−1 − 1 ∈ P ⊆ m (as P is prime and x 6∈ P ).
=⇒ 1 ∈ m and therefore, m = R, a contradiction.
=⇒ P = m, that is, P is maximal.
9. Show that the following statements are equivalent:
(a) R has exactly one prime ideal.
(b) Every element of R is either a unit or a nilpotent.
(c) R/nil(R) is a field.
(a) =⇒ (b) Let P be the unique prime ideal of R. Then nil(R) = P
and P is also a maximal ideal of R. Furthermore, P is the only
maximal ideal of R.
Let a ∈ R. If a is a unit then we there is nothing to prove.
Let a be a non-unit.
=⇒ a is contained in a maximal ideal of R and hence a ∈ P .
=⇒ a is nilpotent.
(b) =⇒ (c) Let X ∈ R/nil(R) be a nonzero element.
Let a ∈ R such that X = a + nil(R). Then a 6∈ nil(R).
By (b), a must be a unit of R.
=⇒ There exists b ∈ R such that ab = 1.
=⇒ X(b + nil(R)) = 1 + nil(R) and hence X is a unit of R/nil(R).
As X is arbitrary, R/nil(R) is a field.
(c)=⇒ (a) By (c), nil(R) is maximal ideal.
Let P be a prime ideal of R.
=⇒ nil(R) ⊆ P and hence nil(R) = P .
=⇒ nil(R) is the only prime ideal of R.
10. Let e1, . . . , en ∈ R be idempotents and let x ∈ R. Show that
(a) There exists an idempotent e ∈ R such that (e1, . . . , en) = (e).
(b) (e1, . . . , en, x) = (y) for some y ∈ R.
(a) Induction on n. For n = 1 there is nothing to prove.
Let n ≥ 2.
By induction there exists an idempotent a of R such that
(e1, . . . , en−1) = (a).
=⇒ (e1, . . . , en) = (a, en).
Let e = a + (1 − a)en.
Then e2 = a2 + 2 a(1 − a)en + a2(1 − en)2
= a + 2 (a − a2)en + a2(1 − 2en + e2
n ) = a + a(1 − en ) = e.
Also ae = a2 + (a − a2)en = a and ene = en.
=⇒ (a, en) ⊆ (e) ⊆ (a, en), that is, (e1, . . . , en) = (a, en) = (e).
(b) Put I = (e1, . . . , en, x).
By (a), there exists an idempotent e such that (e1, . . . , en) = (e).
=⇒ I = (e, x). Put y = e + (1 − e)x.
Then ey = e2 + (e2 − e)x = e and x = y − e + ex = y − ey + eyx =
(1 − e + ex)y.
=⇒ I ⊆ (y) and as y ∈ I, (y) ⊆ I.
=⇒ I = (y).
11. Show that if I is an ideal of R which is free as an R-module then
I is principal.
If I = (0) then we are done.
Assume that I 6= (0). Let B be a basis of I.
=⇒ I(B) = I and hence B 6= ∅.
We claim that B is singleton.
Suppose, if possible, that |B| ≥ 2.
Let x, y ∈ B such that x 6= y.
As yx − xy = 0, B is not LI, a contradiction.
Therefore, B is singleton, that is, B = {x}.
=⇒ I = (x). Further note that x is a nonzero divisor of R.
12. Let I be an ideal of R. If I is a direct summand of R then show
that I is generated by an idempotent.
There exists an ideal J of R such that I + J = R and I ∩ J = (0).
=⇒ There exist x ∈ I and y ∈ J such that x + y = 1.
As xy ∈ I ∩ J, xy = 0, that is, x(1 − x) = 0.
=⇒ x2 = x.
Let a ∈ I. Then a = ax + ay and ay ∈ I ∩ J.
=⇒ ay = 0 and thus a = ax.
=⇒ I = (x).
13. Let I be an ideal of R generated by an idempotent. Show that I
is a direct summand of R. Hence I is projective as an R-module.
Show that if R contains an idempotent e such that e 6= 0, 1 then
I = (e) is a projective R-module which is not free.
Let I = (e), where e2 = e. Let J = (1 − e).
Then 1 ∈ I + J and hence I + J = R.
Let x ∈ I ∩ J.
=⇒ As x ∈ I, x = ae for some a ∈ R.
=⇒ xe = ae2 = ae = x.
As x ∈ J, x = b(1 − e) for some b ∈ R.
=⇒ x = xe = b(1 − e)e = 0.
=⇒ I ∩ J = (0).
Therefore, I is a direct summand of R.
=⇒ I is projective as an R-module.
Let e ∈ R be an idempotent, e 6= 0, 1. Let I = (e).
Then I 6= (0).
Suppose, if possible, that I is free. Let B be a basis of I.
Then B is singleton, that is, B = {x} and x 6= 0.
As x ∈ I, x = ae for some a ∈ R.
=⇒ (1 − e)x = 0.
As B is LI, 1 − e = 0, that is e = 1, a contradiction.
Therefore, I is not free.
So there are projective modules which are not free.
14. Let I be the ideal generated by 6 in Z. Then e = 3 + I is an
idempotent of R = Z/I and e 6= 0, 1.
15. Let R = Z[x]. Show that the ideal (n, x) is not principal, where
n ≥ 2 is an integer.
Suppose, if possible, that (n, x) is principal.
=⇒ (n, x) = (f (x)) for some f (x) ∈ Z[x].
=⇒ There exist g1(x), g2(x) ∈ Z[x] such that n = g1(x)f (x) and
x = g2(x)f (x).
=⇒ deg(n) = 0 = deg(g1(x)) + deg(f (x)) and
deg(x) = 1 = deg(g2(x)) + deg(f (x)).
=⇒ deg(f (x)) = 0 = deg(g1(x)) and deg(g2(x)) = 1.
=⇒ f (x) = m, g1(x) = l and g2(x) = a + bx, where m, l, a, b ∈ Z,
b 6= 0. Furthermore, n = ml.
As x = g2(x)f (x), we have 1 = bm. Therefore, m = ±1.
Also, there exist h1(x), h2(x) ∈ Z[x] such that ±1 = m = h1(x)n+
h2(x)x.
Let h1(x) =
Ps
i and h (x) = Pt
i , where a , b ∈ Z.
a
x
b
x
2
i
i
i
j
i=0
i=0
=⇒ ±1 = a0n, that is, n = 1, a contradiction.
16. Let P be a projective module and let M be an R-module. Let
f : M −→ P be an onto R-linear map. Show that there exists a
submodule N of M such that M = N + ker(f ), N ∩ ker(f ) = 0 and
∼ N.
P =
Since P is projective and f : M −→ P is onto, f ∗ : Hom(P, M ) −→
Hom(P, P ) is onto.
=⇒ There exists g ∈ Hom(P, M ) such that f ∗(g) = IP or f og = IP .
=⇒ g is one-one.
∼ N , M = N +ker(f ) and N ∩ker(f ) = 0.
Let N = Im(g). Then P =
17. Let I be an ideal of R such that R/I is projective. Show that I is
a direct summand of R.
Previous problem.
18. Show that Q is not projective over Z.
Suppose, if possible, Q is projective over Z.
=⇒ Q is isomorphic to a direct summand of a free Z-module, say,
F.
=⇒ There exist Z-linear maps f : Q −→ F and g : F −→ Q such
that g of = IQ.
Let B = {eλ | λ ∈ Λ} be a basis of F .
P
Let x ∈ Q. Then f (x) = λ∈Λ aλeλ for aλ ∈ Z such that aλ’s are 0
for all but finitely many values of λ.
Let n ∈ N. Then f (x) = f (n(x/n)) = nf (x/n).
P
Let f (x/n) = λ∈Λ bλeλ for bλ ∈ Z such that bλ’s are 0 for all but
finitely many values of λ.
=⇒ aλ = nbλ for all λ ∈ Λ.
=⇒ n | |aλ| for all λ ∈ Λ and n ∈ N.
=⇒ |aλ| = 0 for all λ ∈ Λ.
=⇒ aλ = 0 for all λ ∈ Λ.
=⇒ f (x) = 0 for all x ∈ Q, a contradiction, as f is one-one.
19. An R-module M is called torsion-free over R if every nonzero
divisor of R is also a nonzero divisor of M (element a ∈ R is said
to be a nonzero divisor of M if ax 6= 0 for every nonzero element
x of M ). Show that if M is flat then M is also torsion-free.
Let a ∈ R be a nonzero divisor of R.
=⇒ The R-linear map φ : R −→ R given by φ(x) = ax is one-one.
=⇒ φe : M ⊗R R −→ M ⊗R R is one-one.
Let f : M −→ M be the R-linear map given by f (x) = ax.
We need to show that f is one-one.
Let x ∈ ker(f ).
Let ψ : M −→ M ⊗R R be the isomorphism. Note that ψ(y) = y ⊗ 1
for all y ∈ M .
e
Now, φ(x
⊗ 1) = x ⊗ φ(1) = x ⊗ a = ax ⊗ 1 = 0 ⊗ 1 = 0.
=⇒ x ⊗ 1 = 0 as φe is one-one.
=⇒ ψ(x) = 0.
=⇒ x = 0 as ψ is one-one.
=⇒ ker(f ) = 0, that is, f is one-one.
=⇒ Every nonzero divisor of R is a nonzero divisor of M .
Therefore, M is torsion free.
20. Let I 6= R be an ideal of R and let
F = {P ⊆ R | P is a prime ideal of R and I ⊆ P }.
Show that F contains minimal elements.
Note that as I 6= R, F is nonempty.
We apply Zorn’s Lemma to the set G = {R \ P | P ∈ F }.
Order G by inclusion.
Let T be a chain in G. Then T = {R \ Pλ | λ ∈ Λ}.
Let X = ∪λ∈ΛR \ Pλ = R \ (∩λ∈ΛPλ).
Claim. ∩λ∈ΛPλ is a prime ideal.
Let a, b ∈ R such that a, b 6∈ ∩λ∈ΛPλ.
=⇒ There exist i, j ∈ Λ such that a 6∈ Pi and b 6∈ Pj .
As T is a chain, either, R \ Pi ⊆ R \ Pj or R \ Pj ⊆ R \ Pi.
Assume that R \ Pi ⊆ R \ Pj .
=⇒ Pj ⊆ Pi.
=⇒ a, b 6∈ Pj .
As Pj is a prime ideal ab 6∈ Pj .
=⇒ ab 6∈ ∩λ∈ΛPλ.
=⇒ ∩λ∈ΛPλ is a prime ideal and I ⊆ ∩λ∈ΛPλ.
=⇒ ∩λ∈ΛPλ ∈ F .
=⇒ R \ ∩λ∈ΛPλ ∈ G and R \ Pk ⊆ R \ ∩λ∈ΛPλ for all k ∈ Λ.
Therefore, by Zorn’s Lemma, G has a maximal element, say, R \ P
for some P ∈ F .
Claim. P is a minimal element of F .
Let Q ∈ F such that Q ⊆ P .
=⇒ R \ P ⊆ R \ Q and R \ Q ∈ G.
As R \ P is maximal in G, R \ P = R \ Q, that is, P = Q.
21. Let A be an R-algebra and let M and N be an R-modules. Let
f : M −→ N be an R-linear map. Show that
(a) fe: A ⊗R M −→ A ⊗R N is A-linear.
(b) if M is generated by S = {xλ | λ ∈ Λ} as an R-module then
A ⊗R M is generated by S 0 = {1 ⊗ xλ | λ ∈ Λ} as an A-module.
(c) if M is finitely generated as an R-module then A ⊗R M is
finitely generated as an A-module.
(a) As fe is R-linear, it respects addition.
Pn
R M.
i=1 ai ⊗ xi ∈ A ⊗
P
Pn
n
e
e
e
Then f (aX) = f (a i=1 ai ⊗ xi) = f
i=1 aai ⊗ xi
P
Pn
= n
aa
⊗
f
(x
)
=
a
i
i
i=1
i=1 ai ⊗ f (xi )
= afe(X) for all a ∈ A and X ∈ A ⊗R M .
=⇒ fe is A-linear.
Let a ∈ A and X =
(b) Let T = {1 ⊗ x | x ∈ M } and let L be the A-submodule of
A ⊗R M generated by T .
Claim. L = A ⊗R M .
Pn
1=1 ai ⊗ xi ∈ A ⊗R M .
Pn
Pn
=⇒ X = 1=1 ai ⊗ xi = 1=1 ai(1 ⊗ xi) ∈ L for all X ∈ A ⊗R M .
Let X =
=⇒ A ⊗R M = L.
Let K be the A-submodule of A ⊗R M generated by S 0.
Let x ∈ M .
Then there exist aλ’s in R such that x =
P
λ∈Λ aλ xλ and aλ = 0
for all but finitely many values of λ ∈ Λ.
P
P
=⇒ 1 ⊗ x = 1 ⊗ λ∈Λ aλxλ = λ∈Λ(1 ⊗ aλxλ)
P
P
= λ∈Λ(aλ.1 ⊗ xλ) = λ∈Λ(f (aλ)1) ⊗ xλ
P
= λ∈Λ f (aλ)(1 ⊗ xλ) ∈ K for all x ∈ M .
=⇒ T ⊆ K.
=⇒ L ⊆ K and hence K = A ⊗R M .
(c) Follows from (b).
22. Let A be an R-algebra and let M be a free R-module. Show that
A ⊗R M is a free A-module.
Let B = {eλ | λ ∈ Λ} be a basis of M .
Let X ∈ A ⊗R M .
By previous problem, the set B 0 = {1 ⊗ eλ | λ ∈ Λ} generates
A ⊗R M as an A-module.
We need to show that B 0 is LI.
Let aλ’s be elements of A such that aλ = 0 for all but finitely
P
many values of λ ∈ Λ and λ∈Λ aλ(1 ⊗ eλ) = 0.
P
P
=⇒ λ∈Λ aλ ⊗ eλ = 0 = λ∈Λ 0 ⊗ eλ.
By uniqueness, aλ = 0 for all λ ∈ Λ.
=⇒ B 0 is a basis of A ⊗R M as an A-module.
23. Let A be an R-algebra and let M be a projective R-module. Show
that A ⊗R M is a projective A-module.
As M is projective R-module, it is isomorphic to a direct summand
of a free module, say, F . Therefore, there exist R-linear maps
f : M −→ F and g : F −→ M such that g of = IM .
=⇒ fe: A ⊗R M −→ A ⊗R F and ge : A ⊗R F −→ A ⊗R M are A-linear.
Also geofe = gg
of = IA⊗R M .
=⇒ A⊗R M is isomorphic to a direct summand of A-module A⊗R F .
As A ⊗R F is free as an A-module, A ⊗R M is projective as an Amodule.
Assignment – II
MTH611A – 2020-21
1. Let I and J be ideals of R and let M be an R-module. Show that
J (M/IM ) = (JM + IM )/IM .
Let X ∈ J (M/IM ).
=⇒ There exist a1, . . . , an ∈ J and x1, . . . , xn ∈ M such that X =
Pn
i=1 ai (xi + IM ).
Pn
Pn
=⇒ X = i=1(aixi + IM ) = ( i=1 aixi) + IM ∈ (JM + IM )/IM .
=⇒ J (M/IM ) ⊆ (JM + IM )/IM .
Conversely, let X ∈ (JM + IM )/IM .
=⇒ There exists some x ∈ JM + IM such that X = x + IM .
=⇒ There exists some y ∈ JM and z ∈ IM such that x = y + z.
=⇒ X = (y + z) + IM = (y + IM ) + (z + IM ) = y + IM .
Also, as y ∈ JM , there exist a1, . . . , an ∈ J and x1, . . . , xn ∈ M such
P
that y = n
i=1 ai xi .
Pn
Pn
=⇒ X = i=1 aixi + IM = i=1 ai(xi + IM ) ∈ J(M/IM ).
=⇒ (JM + IM )/IM ⊆ J (M/IM ), that is, J (M/IM ) = (JM +
IM )/IM .
2. (a) Let M and N be R-modules and let I = ann(M ) and J =
ann(N ). If I + J = R then show that M ⊗ N = 0.
(b) Let I and J be ideals of R such that I + J = R. Show that
(R/I ) ⊗ (R/J ) = 0.
(c) Let m, n be relatively prime integers. Let I = (m) and J = (n).
Show that (Z/I ) ⊗ (Z/J ) = 0.
(a) There exist a ∈ I and b ∈ J such that a + b = 1.
Let x ∈ M and y ∈ N .
=⇒ x ⊗ y = (a + b)(x ⊗ y) = a(x ⊗ y) + b(x ⊗ y)
= (ax ⊗ y) + (x ⊗ by) = (0 ⊗ y) + (x ⊗ 0) = 0 + 0 = 0.
As every element of M ⊗ N is a sum of elements of the form x ⊗ y,
M ⊗ N = 0.
(b) Follows from (a) as I = ann(R/I) and J = ann(R/J).
(c) Follows from (b) as I + J = Z.
3. Let A be an R-algebra and let M and N be A-modules.
Let
f : M −→ N be an A-linear map. Treat M and N as R-modules
by restriction of scalars. Show that
(a) f is R-linear.
(b) every A-submodule of M is also an R-submodule of M .
(a) As f is A-linear, it respect addition.
Let a ∈ R and x ∈ M .
Let φ : R −→ A be the given ring homomorphism.
=⇒ f (a.x) = f (φ(a)x) = φ(a)f (x) = a.f (x). for all a ∈ R and
x ∈ M.
=⇒ f is R-linear.
(b) Let K be an A-submodule of M . Then K is a subgroup of M .
Let a ∈ R and x ∈ K.
=⇒ a.x = φ(a)x ∈ K for all a ∈ R and x ∈ K.
=⇒ K is an R-submodule of M .
4. Let A be an R-algebra, M be an R-module and L, K be A-modules.
Show that (M ⊗R L) ⊗A K is naturally isomorphic M ⊗R (L ⊗A K)
to as an A-module.
Note that L ⊗A K is an A-module.
We treat it as an R-module vis restriction of scalars and construct
M ⊗R (L ⊗A K).
Also, M ⊗R L is an A module via second factor.
Hence we construct (M ⊗R L) ⊗A K.
Note that M ⊗R (L ⊗A K) is an A-module via the second factor.
The proof of this is nearly identical to the proof of Lemma 35 of
the lectures.
Let z ∈ K.
Define θz : M × L −→ M ⊗R (L ⊗A K) by θz (x, y) = x ⊗ (y ⊗ z) for
all x ∈ M and y ∈ L.
We check that θz is R-bilinear.
θz (x + x1, y) = (x + x1) ⊗ (y ⊗ z)
= x ⊗ (y ⊗ z) + x1 ⊗ (y ⊗ z)
= θz (x, y) + θz (x1, y) for all x, x1 ∈ M and y ∈ L.
θz (x, y + y1) = x ⊗ ((y + y1) ⊗ z)
= x ⊗ (y ⊗ z + y1 ⊗ z)
= x ⊗ (y ⊗ z) + x ⊗ (y1 ⊗ z)
= θz (x, y) + θz (x, y1) for all x ∈ M and y, y1 ∈ L.
θz (ax, y) = ax ⊗ (y ⊗ z)
= a(x ⊗ (y ⊗ z))
= aθz (x, y) for all a ∈ R, x ∈ M and y ∈ L.
θz (x, ay) = x ⊗ (a.y ⊗ z)
= x ⊗ (φ(a)y ⊗ z)
= x ⊗ (φ(a)(y ⊗ z))
= x ⊗ (a.(y ⊗ z))
= a(x ⊗ (y ⊗ z))
= aθz (x, y) for all a ∈ R, x ∈ M and y ∈ L.
=⇒ θz is R-bilinear.
Therefore, there exists an R-linear map φz : M ⊗R L −→ M ⊗R (L⊗A
K) such that φz (x ⊗ y) = θz (x, y) = x ⊗ (y ⊗ z) for all x ∈ M and
y ∈ L.
We now construct an A-bilinear map θ : (M ⊗R L) × K −→ M ⊗R
(L ⊗A K).
Define θ(X, z) = φz (X) for all X ∈ M ⊗ L and z ∈ K.
θ(X + X1, z) = φz (X + X1)
= φz (X) + φz (X1)
(as φz is R-linear)
= θ(X, z) + θ(X1, z) for all X ∈ M ⊗ L and z ∈ K.
Let X =
Pn
i=1 xi ⊗ yi ∈ M ⊗ L and z, z1 ∈ K. Then
θ(X, z + z1) = φz+z1 (X)
Pn
i=1 φz+z1 (xi ⊗ yi )
Pn
= i=1 xi ⊗ (yi ⊗ (z + z1))
=
(as φz+z1 is R-linear)
=
=
=
=
=
Pn
i=1 xi ⊗ (yi ⊗ z + yi ⊗ z1 )
Pn
i=1 (xi ⊗ (yi ⊗ z) + xi ⊗ (yi ⊗ z1 ))
Pn
Pn
x
⊗
(y
⊗
z)
+
i
i=1 i
i=1 xi ⊗ (yi ⊗ z1 )
Pn
Pn
φ
(x
⊗
y
)
+
i
i=1 z i
i=1 φz1 (xi ⊗ yi )
P
Pn
φz ( n
x
⊗
y
)
+
φ
(
z1
i
i=1 i
i=1 xi ⊗ yi )
= φz (X) + φz1 (X)
= θ(X, z) + θ(X, z1) for all X ∈ M ⊗ L and z, z1 ∈ K.
P
Let a ∈ A, X = n
i=1 xi ⊗ yi ∈ M ⊗ L and z ∈ K. Then
P
=⇒ θ(aX, z) = φz (a( n
i=1 (xi ⊗ yi ))
P
= φz ( n
i=1 a(xi ⊗ yi ))
Pn
= φz ( i=1(xi ⊗ ayi))
=
=
=
=
=
=
Pn
i=1 xi ⊗ (ayi ⊗ z)
Pn
i=1 xi ⊗ (a(yi ⊗ z)) (as in L ⊗A K tensoring is over A)
Pn
i=1 a(xi ⊗ (yi ⊗ z))
Pn
a i=1(xi ⊗ (yi ⊗ z))
P
a n
i=1 φz ((xi ⊗ yi )).
P
aφz ( n
i=1 (xi ⊗ yi )).
= aφz (X).
= aθ(X, z) for all a ∈ A, X ∈ M ⊗R L and z ∈ K.
θ(X, az) = φaz (X)
Pn
= i=1 φaz (xi ⊗ yi)
P
= n
i=1 xi ⊗ (yi ⊗ az)
=
=
=
=
=
Pn
i=1 xi ⊗ a(yi ⊗ z) (as in L ⊗A K tensoring is over A)
Pn
i=1 a(xi ⊗ (yi ⊗ z))
P
a n
i=1 xi ⊗ (yi ⊗ z)
Pn
a i=1 φz (xi ⊗ yi)
P
aφz ( n
i=1 xi ⊗ yi )
= aφz (X)
= aθ(X, z) for all a ∈ R, X ∈ M ⊗ L and z ∈ K.
Therefore, θ is A-bilinear.
By universal property, there exists an A-linear map φ : (M ⊗R L)⊗A
K −→ M ⊗R (L ⊗A K) such that φ(X ⊗ z) = θ(X, z) = φz (X) for
all X ∈ M ⊗ L and z ∈ K.
=⇒ φ((x ⊗ y) ⊗ z) = φz (x ⊗ y) = x ⊗ (y ⊗ z) for all x ∈ M , y ∈ L
and z ∈ K.
Similarly, there exists an A-linear map ψ : M ⊗R (L ⊗A K) −→
(M ⊗R L) ⊗A K such that ψ(x ⊗ (y ⊗ z)) = (x ⊗ y) ⊗ z for all
x ∈ M , y ∈ L and z ∈ K.
Claim. ψ oφ is identity of (M ⊗R L) ⊗A K.
Pn
First let X = i=1(xi ⊗ yi) ∈ M ⊗R L and z ∈ K.
P
=⇒ (ψ oφ)(X ⊗ z) = (ψ oφ)(( n
i=1 xi ⊗ yi ) ⊗ z)
Pn
= (ψ oφ)( i=1((xi ⊗ yi) ⊗ z)))
P
= n
i=1 (ψ oφ)((xi ⊗ yi ) ⊗ z)
Pn
= i=1 ψ(φ((xi ⊗ yi) ⊗ z))
Pn
= i=1 ψ(xi ⊗ (yi ⊗ z))
P
= n
i=1 (xi ⊗ yi ) ⊗ z
P
=( n
i=1 xi ⊗ yi ) ⊗ z
=X ⊗z
Now let X ∈ (M ⊗R L) ⊗A K.
=⇒ There exist X1, . . . , Xr ∈ M ⊗R L and z1, . . . , zr ∈ K such that
P
X = ri=1 Xi ⊗ zi.
P
=⇒ (ψ oφ)(X) = (ψ oφ)( ri=1 Xi ⊗ zi)
Pr
i=1 (ψ oφ)(Xi ⊗ zi )
P
= ri=1 Xi ⊗ zi = X.
=
=⇒ ψ oφ is the identity of (M ⊗R L) ⊗A K.
Similarly, φoψ is the identity of M ⊗R (L ⊗A K).
=⇒ φ is an A-linear isomorphism.
5. Let A be an R-algebra. Let M be a flat R-module. Show that
A ⊗R M is flat as an A-module.
Let L and K be A-module and φ : L −→ K be a one-one A-linear
map.
Treating K and L as R-modules by restriction of scalars.
Then φ is also R-linear.
Also, IA ⊗ φ : A ⊗A L −→ A ⊗A K is one-one A-linear map.
=⇒ IA ⊗ φ : A ⊗A L −→ A ⊗A K is one-one R-linear map.
=⇒ IM ⊗ (IA ⊗ φ) : M ⊗R (A ⊗A L) −→ M ⊗R (A ⊗A K) is one-one
R-linear map.
Let α : (M ⊗R A) ⊗A L −→ M ⊗R (A ⊗A L) and
β : (M ⊗R A) ⊗A K −→ M ⊗R (A ⊗A K)
be the isomorphisms as discussed in previous problem.
=⇒ β −1o(IM ⊗ (IA ⊗ φ))oα : (M ⊗R A) ⊗A L −→ (M ⊗R A) ⊗A K is
a one-one map.
Let x ∈ M , a ∈ A and y ∈ L.
Then (β −1o(IM ⊗ (IA ⊗ φ))oα)((x ⊗ a) ⊗ y)
= (β −1o(IM ⊗ (IA ⊗ φ))α(((x ⊗ a) ⊗ y))
= (β −1o(IM ⊗ (IA ⊗ φ))(x ⊗ (a ⊗ y)
= β −1((IM ⊗ (IA ⊗ φ))(x ⊗ (a ⊗ y))
= β −1(x ⊗ (a ⊗ φ(y)) = (x ⊗ a) ⊗ φ(y)
= (IM ⊗A ⊗ φ)((x ⊗ a) ⊗ y).
Let X ∈ M ⊗R A and y ∈ L. Then there exist x1, . . . , xn ∈ M and
Pn
a1, . . . , an ∈ A such that X = i=1 xi ⊗ ai.
=⇒ (β −1o(IM ⊗ (IA ⊗ φ)oα)(X ⊗ y)
Pn
−1
= (β o(IM ⊗ (IA ⊗ φ))oα)(( i=1 xi ⊗ ai) ⊗ y)
P
= (β −1o(IM ⊗ (IA ⊗ φ))oα)( n
i=1 (xi ⊗ ai ) ⊗ y)
Pn
= i=1(β −1o(IM ⊗ (IA ⊗ φ))oα)((xi ⊗ ai) ⊗ y)
P
= n
i=1 (IM ⊗A ⊗ φ)((xi ⊗ ai ) ⊗ y)
= (IM ⊗A ⊗ φ)(X ⊗ y).
As every element of (M ⊗R A) ⊗A L is a sum of elements of the
form X ⊗ y, we have β −1o(IM ⊗ (IA ⊗ φ))oα = IM ⊗A ⊗ φ.
=⇒ IM ⊗A ⊗ φ is one-one.
=⇒ M ⊗R A is flat as an A-module.
=⇒ A ⊗R M is flat as an A-module.
6. Let {Mλ | λ ∈ Λ} be a family of R-modules and N be an R-module.
Show that (⊕λ∈ΛMλ) ⊗ N is isomorphic to ⊕λ∈Λ (Mλ ⊗ N ).
Put M = ⊕λ∈ΛMλ.
To avoid the ambiguity, we shall use ≺ xλ instead of (xλ) to
write the elements of a direct sum.
Define θ : M × N −→ ⊕λ∈Λ (Mλ ⊗ N ) by setting θ(≺ xλ , y) =≺
xλ ⊗ y for all ≺ xλ ∈ M and y ∈ N .
Note that θ is well defined as xλ’s are 0 for all but finitely many
values of λ ∈ Λ and hence so are xλ ⊗ y’s.
Claim. θ is R-bilinear.
Let a, b ∈ R, ≺ xλ , ≺ x0λ ∈ M and y, y 0 ∈ N .
=⇒ θ(a ≺ xλ + ≺ x0λ , by + y 0)
= θ(≺ axλ + x0λ , by + y 0)
=≺ (axλ + x0λ) ⊗ (by + y 0) =≺ ab(xλ ⊗ y) + a(xλ ⊗ y 0) + b(x0λ ⊗ y) + x0λ ⊗ y 0 = ab ≺ xλ ⊗ y +a ≺ xλ ⊗ y 0 +b ≺ x0λ ⊗ y + ≺ x0λ ⊗ y 0 = abθ(≺ xλ , y) + aθ(≺ xλ , y 0) + bθ(≺ x0λ , y) + θ(≺ x0λ , y 0)
=⇒ θ is R-bilinear.
Therefore, there exists an R-linear map φ : M ⊗N −→ ⊕λ∈Λ (Mλ ⊗ N )
such that φ(≺ xλ ⊗y) = θ(≺ xλ , y) =≺ xλ ⊗ y .
For λ ∈ Λ, let λ : Mλ −→ M be the inclusion and pλ : M −→ Mλ
be the projection.
Then we have R-linear maps fλ : Mλ ⊗ N −→ M ⊗ N and pfλ : M ⊗
N −→ Mλ ⊗ N .
Define ψ : ⊕λ∈Λ (Mλ ⊗ N ) −→ M ⊗ N by setting ψ(≺ Xλ ) =
P
λ (Xλ ).
λ∈Λ f
Claim. ψ oφ = IM ⊗N .
P
ψ oφ((≺ xλ ⊗y) = ψ(≺ xλ ⊗ y ) = λ∈Λ fλ(xλ ⊗ y)
P
P
= λ∈Λ λ(xλ) ⊗ y = ( λ∈Λ λ(xλ)) ⊗ y.
P
Let λ∈Λ λ(xλ) =≺ yλ .
Let µ ∈ Λ.
=⇒ yµ = pµ(≺ yλ ) = pµ( λ∈Λ λ(xλ)) = λ∈Λ pµ(λ(xλ)) =
P
λ∈Λ (pµ oλ )(xλ )) = xµ as pµ oµ = IMµ and pµ oλ = 0 if λ 6= µ.
=⇒ yµ = xµ for all µ ∈ Λ.
P
=⇒
P
P
λ∈Λ λ (xλ ) =≺ xλ .
=⇒ ψ oφ((≺ xλ ⊗y) =≺ xλ ⊗y for all ≺ xλ ∈ M and y ∈ N .
As every element of M ⊗ N is a sum of elements of the form
≺ xλ ⊗y, we must have ψ oφ = IM ⊗N .
We now want to show that φoψ is the identity of ⊕λ∈Λ (Mλ ⊗ N ).
But we shall need the following
Claim. φ(X) =≺ pfλ(X) for all X ∈ M ⊗ N .
First let X =≺ xλ ⊗y.
Put Z =≺ xλ .
=⇒ φ(X) =≺ xλ ⊗ y =≺ pλ(Z) ⊗ y =≺ pfλ(Z ⊗ y) =≺ pfλ(X) .
P
Now let X = n
i=1 Zi ⊗ yi , where Zi ∈ M and yi ∈ N .
Pn
Pn
=⇒ φ(X) = φ( i=1 Zi ⊗ yi) = i=1 φ(Zi ⊗ yi)
P
Pn
f
f
= n
≺
p
(Z
⊗
y
)
=≺
i
λ i
λ (Zi ⊗ yi ) i=1
i=1 p
Pn
f
=≺ pλ( i=1 Zi ⊗ yi) =≺ pfλ(X) for all X ∈ M ⊗ N .
Claim. φoψ is the identity of ⊕λ∈Λ (Mλ ⊗ N ).
Let ≺ Xλ ∈ ⊕λ∈Λ (Mλ ⊗ N ).
=⇒ φoψ(≺ Xλ ) = φ(ψ(≺ Xλ ))
P
= φ( λ∈Λ fλ(Xλ))
P
λ (Xλ ))
λ∈Λ φ(f
P
= λ∈Λ ≺ pfµ(fλ(Xλ) P
= λ∈Λ ≺ (pfµofλ)(Xλ) P
= λ∈Λ ≺ p^
µ oλ (Xλ ) P
=≺ λ∈Λ p^
µ oλ (Xλ ) =
=≺ Xλ as p^
µ oλ = 0 if λ 6= µ and p^
µ oµ = IMµ ⊗N .
=⇒ φoψ is the identity of ⊕λ∈Λ (Mλ ⊗ N ).
=⇒ φ is an isomorphism.
7. Let {Mλ | λ ∈ Λ} be a family of R-modules. Show that ⊕λ∈ΛMλ is
flat iff each Mλ is flat.
Put M = ⊕λ∈ΛMλ.
Let P, Q be R-modules and let f : P −→ Q be an R-linear one-one
map.
Let α : M ⊗ P −→ ⊕λ∈Λ (Mλ ⊗ P ) and β : M ⊗ Q −→ ⊕λ∈Λ (Mλ ⊗ Q)
be the isomorphisms.
Let gλ = IMλ ⊗ f : Mλ ⊗ P −→ Mλ ⊗ Q.
Define φ : ⊕λ∈Λ (Mλ ⊗ P ) −→ ⊕λ∈Λ (Mλ ⊗ Q)
φ(≺ Xλ ) =≺ gλ(Xλ) .
Then φ is R-linear (check).
by
setting
Let g = IM ⊗ f : M ⊗ P −→ M ⊗ Q.
Consider
0
 −→

y
M
⊗P
g
−→

yα
M
⊗Q

yβ
φ
0 −→ ⊕λ∈Λ (Mλ ⊗ P ) −→ ⊕λ∈Λ (Mλ ⊗ Q)
Claim. The squares in above diagram are commutative.
First square is commutative as both compositions are 0 maps.
Let ≺ xλ ∈ M and y ∈ P .
Then (β og)(≺ xλ ⊗y) = β(g(≺ xλ ⊗y))
= β(≺ xλ ⊗f (y)) =≺ xλ ⊗ f (y) and (φoα)(≺ xλ ⊗y) = φ(α(≺ xλ ⊗y))
= φ(≺ xλ ⊗ y )
=≺ gλ(xλ ⊗ y) =≺ xλ ⊗ f (y) = (β og)(≺ xλ ⊗y).
As every element of M ⊗ P is a sum of elements of the form
≺ xλ ⊗y we must have β og = φoα.
=⇒ Second square is also commutative.
Suppose that M is flat.
=⇒ g is one-one.
=⇒ β og is one-one as β is an isomorphism.
=⇒ φoα is one-one.
=⇒ φ is one-one as α is an isomorphism.
Let µ ∈ Λ. We want to show that gµ is one-one for all µ ∈ Λ.
Let X ∈ ker(gµ).
Let ≺ Xλ ∈ ⊕λ∈Λ (Mλ ⊗ P ), where Xµ = X and Xλ = 0 for λ 6= µ.
=⇒ φ(≺ Xλ ) =≺ gλ(Xλ) = 0.
=⇒ ≺ Xλ = 0 as φ is one-one.
=⇒ Xλ = 0 for all λ ∈ Λ.
=⇒ Xµ = 0 or gµ is one-one for all µ ∈ Λ.
=⇒ Mµ is flat for all µ ∈ Λ.
Conversely, assume that all Mλ’s are flat.
=⇒ gλ is one-one for all λ ∈ Λ.
Claim. φ is one-one.
Let ≺ Xλ ∈ ker(φ).
=⇒ ≺ gλ(Xλ) = 0.
=⇒ gλ(Xλ) = 0 for all λ ∈ Λ.
=⇒ Xλ = 0 for all λ ∈ Λ as gλ’s are one-one.
=⇒ ≺ Xλ = 0 or φ is one-one.
=⇒ The lower sequence in the above diagram is exact.
=⇒ The upper sequence in the above diagram is exact.
=⇒ g is one-one.
=⇒ M is flat.
8. Let A be an R-algebra, M and N be A-modules. Show that there
exists a natural R-linear map φ : M ⊗R N −→ M ⊗A N . Furthermore,
φ is onto. Also show that M is isomorphic to a direct summand
of A ⊗R M .
We treat M , N and M ⊗A N as R-modules by restriction of scalars
and construct M ⊗R N .
Define θ : M × N −→ M ⊗A N by setting θ(x, y) = x ⊗ y for all x ∈ M
and y ∈ N .
Claim.
θ is R-bilinear.
Let a, b ∈ R, x, x0 ∈ M and y, y 0 ∈ N .
Let f : R −→ A be the given ring homomorphism.
=⇒ θ(a.x + x0, b.y + y 0)
= θ(f (a)x + x0, f (b)y + y 0)
= (f (a)x + x0) ⊗ (f (b)y + y 0)
= f (a)f (b)(x ⊗ y) + f (a)(x ⊗ y 0) + f (b)(x0 ⊗ y) + (x0 ⊗ y 0)
= f (ab)(x ⊗ y) + f (a)(x ⊗ y 0) + f (b)(x0 ⊗ y) + (x0 ⊗ y 0)
= (ab).θ(x, y) + a.θ(x, y 0) + b.θ(x0, y) + θ(x0, y 0) for all a, b ∈ R,
x, x0 ∈ M and y, y 0 ∈ N .
=⇒ θ is R-bilinear.
Hence there exists an R-linear map φ : M ⊗R N −→ M ⊗A N such
that φ(x ⊗ y) = θ(x, y) = x ⊗ y for all x ∈ M and y ∈ N .
Clearly, φ is onto.
Taking M = R and N = M , we have R-linear onto map φ : A ⊗R
M −→ A ⊗A M .
Let α : A ⊗A M −→ M be the isomorphism.
Then α is A-linear and hence also R-linear.
=⇒ ψ = αoφ : A ⊗R M −→ M is an onto R-linear map.
We also have β : M −→ A⊗R M given by β(x) = 1⊗x for all x ∈ M .
Let a ∈ R and x, y ∈ M .
=⇒ β(a.x + y) = 1 ⊗ (a.x + y) = a.(1 ⊗ x) + (1 ⊗ y)
= a.β(x) + β(y) for all a ∈ R and x, y ∈ M .
=⇒ β is R-linear.
Also, (ψ oβ)(x) = ψ(β(x)) = (αoφ)(1 ⊗ x) = α(φ(1 ⊗ x))
= α(1 ⊗ x)) = x = IM (x) for all x ∈ M .
=⇒ ψ oβ = IM .
=⇒ M is isomorphic to a direct summand of A ⊗R M (as an Rmodule).
9. Let R be a local ring and let M, N be finitely generated R-modules.
If M ⊗R N = 0 then show that either M = 0 or N = 0.
Assume that neither M = 0 nor N = 0.
Let m be the unique maximal ideal of the local ring R.
Put F = R/m. Then F is a field.
Let α : M −→ M/mM be the natural map.
Then α is R-linear.
By Nakayama Lemma M/mM 6= 0.
(If M/mM = 0, then M = mM and hence M = 0, by Nakayama
Lemma as M is fg.)
As M/mM is a nonzero vector-space over F , there exists an onto
F -linear map f : M/mM −→ F .
Then f is R-linear.
=⇒ f 0 = f oα : M −→ F is R-linear and onto.
Similarly, there exists an onto R-linear map g 0 : N −→ F .
=⇒ f 0 ⊗ g 0 : M ⊗R N −→ F ⊗R F is an onto R-linear map.
=⇒ F ⊗R F = 0 as M ⊗R N = 0.
By previous problem, there exists an onto R-linear map h : F ⊗R
F −→ F ⊗F F .
=⇒ F ⊗F F = 0.
Let φ : F ⊗F F −→ F be the F -linear isomorphism.
Therefore, F = 0.
=⇒ R = m, a contradiction.
Therefore, either M = 0 or N = 0.
10. Let A be an R-algebra which is flat as an R-module. Let N be a
flat A-module. Show that N is a flat R-module.
Let K and L be R-modules and let f : L −→ K be a one-one R-
linear map.
As A is flat as an R-module, f ⊗IA : L⊗R A −→ K ⊗R A is a one-one
R-linear map.
Note that L ⊗R A and K ⊗R A are A-modules by second factor.
Claim. f ⊗ IA is A-linear.
As f ⊗ IA is R-linear, it respects addition.
P
Let a ∈ A and X = n
i=1 xi ⊗ ai ∈ L ⊗R A.
=⇒ (f ⊗ IA)(aX) = (f ⊗ IA)(a n
i=1 xi ⊗ ai )
P
= (f ⊗ IA)( n
i=1 xi ⊗ aai )
P
Pn
i=1 f (xi ) ⊗ aai
Pn
= a i=1 f (xi) ⊗ ai = a(f ⊗ IA)(X)
=
for all a ∈ A and X ∈ L ⊗R A. Thus (f ⊗ IA) is A-linear.
=⇒ (f ⊗ IA) ⊗ IN : (L ⊗R A) ⊗A N −→ (K ⊗R A) ⊗A N is a one-one
A-linear map (as N is flat as an A-module).
Treating, N as R-module, by restriction of scalars.
Then we have R-linear map f ⊗ IN : L ⊗R N −→ K ⊗R N .
We want to show that f ⊗ IN is one-one.
Let α0 : (L ⊗R A) ⊗A N −→ L ⊗R (A ⊗A N ) be the A-linear isomorphism as discussed in a previous problem.
Also let α00 : A ⊗A N −→ N be the A-linear isomorphism.
=⇒ α0 and α00 are R-linear isomorphism.
=⇒ IL ⊗ α00 : L ⊗R (A ⊗A N ) −→ L ⊗R N is an R-linear isomorphism.
Then α = (IL ⊗α00)oα0 : (L⊗R A)⊗A N −→ L⊗R N is also an R-linear
isomorphism.
Note that α((x ⊗ a) ⊗ y) = ((IL ⊗ α00)oα0)((x ⊗ a) ⊗ y)
= (IL ⊗ α00)(x ⊗ (a ⊗ y))
= x ⊗ α00(a ⊗ y)
= x ⊗ ay, for all x ∈ L, a ∈ A and y ∈ L.
Similarly, there exists an R-linear isomorphism β : K ⊗R (A ⊗A
N ) −→ K ⊗R N .
Consider
(f ⊗I )⊗I
A
N
−
−
−
−
−
−
−
−
→
(K ⊗R A)
⊗A N
0
−
→
(L
⊗
A)
⊗
N
R 
A



y
0 −→

yα
L ⊗R N
f ⊗IN
−−−−→

yβ
K ⊗R N
Claim. The squares in above diagram are commutative.
The first square is commutative as compositions are 0 maps.
Let x ∈ L, a ∈ A and y ∈ N .
=⇒ (β o((f ⊗ IA) ⊗ IN ))((x ⊗ a) ⊗ y)
= β(((f ⊗ IA) ⊗ IN )((x ⊗ a) ⊗ y))
= β((f (x) ⊗ a) ⊗ y) = f (x) ⊗ ay.
Similarly, ((f ⊗ IN )oα)((x ⊗ a) ⊗ y) = f (x) ⊗ ay.
As every element of (L ⊗R A) ⊗A N is a finite sum of elements of
the form (x⊗a)⊗y (check), we have β o((f ⊗IA)⊗IN ) = (f ⊗IN )oα,
and hence the second square is also commutative.
=⇒ The lower sequence is exact.
=⇒ f ⊗ IN is one-one.
11. Let M and N be flat R-modules. Show that so is M ⊗ N .
Let P and Q be R-modules and f : P −→ Q be a one-one R-linear
map.
=⇒ f 0 = f ⊗ IM : P ⊗ M −→ Q ⊗ M a one-one R-linear map.
=⇒ f 00 = f 0 ⊗ IN : (P ⊗ M ) ⊗ N −→ (Q ⊗ M ) ⊗ N a one-one R-linear
map.
We also have an R-linear map g = f ⊗ IM ⊗N : P ⊗ (M ⊗ N ) −→
Q ⊗ (M ⊗ N ).
We want to show that g is one-one.
Let α : (P ⊗ M ) ⊗ N −→ P ⊗ (M ⊗ N ) be the isomorphism.
Note that α((x ⊗ y) ⊗ z) = x ⊗ (y ⊗ z) for all x ∈ P , y ∈ M and
z ∈ N.
Let β : (Q ⊗ M ) ⊗ N −→ Q ⊗ (M ⊗ N ) be the isomorphism.
Note that β((x ⊗ y) ⊗ z) = x ⊗ (y ⊗ z) for all x ∈ Q, y ∈ M and
z ∈ N.
Consider
f 00
0
−
→ (Q ⊗ M
 −→ (P ⊗ M
)⊗N −
)⊗N

y

yα
g

yβ
0 −→ P ⊗ (M ⊗ N ) −→ Q ⊗ (M ⊗ N )
Then the squares in the above diagram are commutative.
As the upper sequence is exact, so is the lower sequence.
=⇒ g is one-one.
12. Let f : M −→ N be an R-linear map. Let I be an ideal of R. Show
that
(a) f induces an R-linear map f¯: M/IM −→ N/IN .
(b) if I is contained in J(R), f¯ is onto and N is finitely generated
then f is also onto.
(a) Define f¯: M/IM −→ N/IN by setting f¯(x + IM ) = f (x) + IN
for all x ∈ M .
Claim. f¯ is well defined and R-linear.
Let x + IM = y + IM .
=⇒ x − y ∈ IM .
=⇒ There exist a1, . . . , an ∈ I and x1, . . . , xn ∈ M such that x − y =
Pn
i=1 ai xi .
Pn
=⇒ f (x − y) = i=1 aif (xi) ∈ IN .
=⇒ f (x) + IN = f (y) + IN and hence f¯ is well defined.
Let a ∈ R, x, y ∈ M .
=⇒ f¯(a(x + IM ) + (y + IM )) = f¯((ax + y) + IM ))
= f (ax + y) + IN
= (af (x) + f (y)) + IN
= a(f (x) + IN ) + (f (y) + IN )
= af¯(x + IM ) + f¯(y + IM )
for all a ∈ R, x, y ∈ M .
=⇒ f¯ is R-linear.
(b) Let y ∈ N .
As f¯ is onto, there exist x ∈ M such that
f¯(x + IM ) = y + IN or f (x) + IN = y + IN .
=⇒ y − f (x) ∈ IN .
=⇒ y ∈ Im(f ) + IN for all y ∈ N .
=⇒ N = Im(f ) + IN .
As N is finitely generated and I ⊆ J(R), we have N = Im(f ).
That is, f is onto.
13. Let m, n ∈ N and let f : Rm −→ Rn be R-linear. Show that
(a) if f is onto then n ≤ m.
(b) if f is an isomorphism then n = m.
(c) if f is one-one then m ≤ n.
(a) Let m be a maximal ideal of R. Put F = R/m. Then F is an
R-algebra. Note that Rm is a free R-module and there is a basis
of Rm which contains exactly m elements. Note that Rm ⊗R F is
an F -module via the second factor. Furthermore, by a previous
problem, there is a basis of Rm ⊗R F as a module over F , which
contains exactly m elements.
As F is a field, dimF (Rm ⊗R F ) = m.
Similarly, dimF (Rn ⊗R F ) = n.
As f is onto so is f ⊗ IF .
Also, as both Rm ⊗R F and Rn ⊗R F are F -modules via the second
factor, f ⊗ IF is F -linear.
=⇒ m ≥ n, by rank-nullity theorem of linear algebra.
(b) In this case f ⊗ IF is an F -linear isomorphism. Therefore,
again by rank-nullity theorem m = n.
(c) This result is hard and can not be proved with the theory
developed so far.
f
g
14. Let M −→ N −→ P −→ 0 be an exact sequence of R-modules.
Show that if M and P are fg (finitely generated) then so is N .
Also if N is fg then so is P .
Assume that M is generated by S = {x1, . . . , xm} and P is generated by T = {y1, . . . , yn}.
Let x ∈ N .
Pn
As g(x) ∈ P , there exist a1, . . . , an ∈ R such that g(x) = i=1 aiyi.
Let s1, . . . , sn ∈ N such that g(si) = yi for all i = 1, . . . , n.
Pn
Pn
Pn
=⇒ g(x − i=1 aisi) = g(x) − i=1 aig(si) = g(x) − i=1 aiyi = 0.
=⇒ x −
Pn
i=1 ai si ∈ ker(g) = Im(f ).
Pn
=⇒ There exist z ∈ M such that f (z) = x − i=1 aisi. Also, there
P
exist b1, . . . , bm ∈ R such that z = m
j=1 bj xj .
P
Pm
=⇒ x − n
a
s
=
f
(z)
=
i=1 i i
j=1 bj f (xj ).
Put X = {s1, . . . , sn, f (x1), . . . , f (xm)}.
f for all x ∈ N .
Then x ∈ X
f As X is finite, N is fg.
=⇒ N = X.
Second part has been done in the lectures.
15. Let f : R −→ A be a ring homomorphism and let S be a multiplicative set in R. Show that T = f (S) is a multiplicative set in A. Let
M be an A-module. Show that S −1M and T −1M are isomorphic
as S −1R-modules.
Note that 1 = f (1) ∈ T .
Let x, y ∈ T . Then there exist a, b ∈ S such that f (a) = x and
f (b) = y.
Therefore, xy = f (a)f (b) = f (ab) ∈ T for all x, y ∈ T .
=⇒ T is a multiplicative set.
f (a)
a
=
.
s
f (s)
Verify that θ is well defined and is a ring homomorphism.
Let θ : S −1R −→ T −1A be the map given by θ
Therefore, by restriction of scalars, T −1M is also an S −1R-module.
Treat M as an R-module by restriction of scalars. Then we can
construct S −1M .
Define g : S −1M −→ T −1M by setting g
x
x
=
.
s
f (s)
We shall show that g is well defined.
x
y
Let = .
s
t
=⇒ There exists some u ∈ s such that (ut).x = (us).y.
=⇒ f (ut)x = f (us)y or f (u)f (t)x = f (u)f (s)y.
x
y
=⇒
=
as f (u) ∈ T .
f (s)
f (t)
a
x y
−1
Let ∈ S R and , ∈ S −1M .
s
s s
a x
y
a.x
y
a.x + s.y
=⇒ g
· +
=g
+
=g
s s
s
ss
s
ss
a.x + s.y
f (a)x + f (s)y
=
=
f (ss)
f (s)f (s)
=
f (a)x
f (s)y
+
f (s)f (s)
f (s)f (s)
f (a) x
y
+
=
f (s) f (s)
f (s)
a
x
y
=θ
g
+g
s
s
s
x
y
a
= · g
+g
.
s
s
s
=⇒ g is S −1R-linear.
Claim. g is one-one.
x
Let ∈ ker(g).
s
x
= 0.
=⇒
f (s)
=⇒ There exists some t ∈ S such that f (t)x = 0.
=⇒ t.x = 0.
x
=⇒ = 0.
s
=⇒ g is one-one.
Also, clearly, g is onto.
16. Let S be a multiplicative set in R and let M be a fg R-module.
Then S −1M = 0 iff there exists some s ∈ S such that sM = 0.
Assume that M is generated by S = {x1, . . . , xn}.
Assume that S −1M = 0.
xi
=⇒
= 0 for all i = 1, . . . , n.
1
=⇒ There exists si ∈ S such that sixi = 0, for all i = 1, . . . , n.
Put s = s1 . . . sn. Then sxi = 0 for all i = 1, . . . , n.
Let x ∈ M .
Then there exist a1, . . . , an ∈ R such that x =
Pn
i=1 ai xi .
=⇒ sx = s
Pn
Pn
a
x
=
i=1 i i
i=1 ai sxi = 0 for all x ∈ M .
=⇒ sM = 0.
Conversely, let sM = 0 for some s ∈ S.
Let X ∈ S −1M .
=⇒ There exist x ∈ M and t ∈ S such that X =
x
.
t
sx
0
=
= 0 for all X ∈ S −1M .
st
st
=⇒ S −1M = 0.
=⇒ X =
17. Let M be an R-module and let S be a multiplicative
set in R.
Let
x
T be a set of generators of M . Show that T 0 =
| x ∈ S is a
1
−1
−1
set of generators of S M as an S R-module. Hence if M is fg
as an R-module then S −1M as fg as an S −1R-module.
Let K be the S −1R-submodule of S −1M generated by T 0.
Let X ∈ S −1M .
x
.
s
=⇒ There exist x1, . . . , xn ∈ T and a1, . . . , an ∈ R such that x =
Then there exist x ∈ M and s ∈ S such that X =
Pn
i=1 ai xi .
Pn
n
X
a
x
ai xi
i i
=⇒ X = i=1
=
∈ K.
s
i=1 s 1
=⇒ T 0 generates S −1M as an S −1R-module.
If M is finitely generated then there exists a finite subset T of M
which generates M as an R-module.
Clearly, T 0 is finite in this case.
18. Let I be an ideal and let S = 1 + I = {1 + a | a ∈ I}. Show that
S −1I ⊆ J(S −1R). In particular, if M is fg and M = IM then there
exist a ∈ I such that (1 + a)M = 0.
a
−1
Let x ∈ S I. Then there exist a ∈ I and s ∈ S such that x = .
s
b
Let y = ∈ S −1R.
t
ba
st + ab
1
=⇒ 1 + xy = +
=
1
ts
st
Let s = 1 + α and t = 1 + β for some α, β ∈ I.
1 + α + β + αβ + ab
=⇒ 1 + xy =
st
Then u = 1 + α + β + αβ + ab ∈ S as α + β + αβ + ab ∈ I.
st
u
is a unit with multiplicative inverse .
Note that
st
u
=⇒ 1 + xy is a unit for all y ∈ S −1R.
=⇒ x ∈ J(S −1R) for all x ∈ S −1I.
=⇒ S −1I ⊆ J(S −1R).
Let M = IM .
Claim.
S −1M = (S −1I)(S −1M ).
Let X ∈ S −1(M ). Then there exist x ∈ M = IM and s ∈ S such
x
that X = .
s
=⇒ There exist a1, . . . , an ∈ I and x1, . . . , xn ∈ M such that x =
Pn
i=1 ai xi .
Pn
n
X
a
x
ai xi
i
i
i=1
=⇒ X =
=
∈ (S −1I)(S −1M ) for all X ∈ S −1M .
s
i=1 s 1
=⇒ S −1M ⊆ (S −1I)(S −1M ) and hence S −1M = (S −1I)(S −1M ).
(Note that the above claim is true for all multiplicative sets. )
As M is fg as an R-module, S −1M is fg as an S −1R-module. Thus
by Nakayama Lemma, S −1M = 0 as S −1I ⊆ J(S −1R).
Therefore, by a previous result, there exists some s ∈ S such that
sM = 0 (as M is fg).
=⇒ There exists an a ∈ I such that s = 1 + a.
=⇒ (1 + a)M = 0.
19. Let S be a multiplicative set in R. Let M be an R-module. Show
that every S −1R-submodule of S −1M is of the form S −1N for
some R-submodule N of M .
Let K be an S −1R-submodule of S −1M .
Let : M −→ S −1M be the natural map, that is, (x) =
x
for all
1
x ∈ M . Recall that is R-linear.
=⇒ N = −1(K) is an R-submodule of M .
Let X ∈ K. Then there exist some x ∈ M and s ∈ S such that
x
X= .
s
x
sx
=⇒ (x) = =
∈ K and hence x ∈ N .
1
1s
=⇒ X ∈ S −1N for all X ∈ K.
=⇒ K ⊆ S −1N .
Conversely, let X ∈ S −1N .
=⇒ There exist x ∈ N and s ∈ S such that X =
x
.
s
1x
x
∈ K as ∈ K.
s1
1
=⇒ S −1N ⊆ K and hence K = S −1N .
=⇒ X =
20. Let S be a multiplicative set in R. Let M be an R-module.
(a) Let N1, . . . , Nk be submodules of M .
∩ki=1S −1Ni.
−1
k
Then S
∩i=1Ni =
−1
k
Let X ∈ S
∩i=1Ni .
x
k
=⇒ There exists some x ∈ ∩i=1Ni and s ∈ S such that X = .
s
=⇒ X ∈ S −1Ni for all i = 1, . . . , k.
k
−1
−1
k
∩i=1Ni .
=⇒ X ∈ ∩i=1S Ni for all X ∈ S
=⇒ S −1 ∩ki=1Ni ⊆ ∩ki=1S −1Ni.
Conversely, let X ∈ ∩ki=1S −1Ni.
x
Then there exists some x ∈ M and s ∈ S such that X = .
s
xi
Also, there exists some xi ∈ Ni and ti ∈ S such that X = , for
ti
all i = 1, . . . , k.
=⇒ There exists some ui ∈ S such that uitix = uisxi for all
i = 1, . . . , k.
Put u = u1 . . . uk and t = t1 . . . tk . Then utx = ut1 . . . ti−1ti+1 . . . tk sxi ∈
Ni for all i = 1, . . . , k.
=⇒ utx ∈ ∩ki=1Ni.
utx
x
−1
k
∈S
=⇒ X = =
∩i=1Ni for all X ∈ ∩ki=1S −1Ni.
s
uts
k
−1
−1
k
=⇒ ∩i=1S Ni ⊆ S
∩i=1Ni .
∼
(b) Let {Mλ | λ ∈ Λ} be a family of R-modules. Then S −1 (⊕λ∈ΛMλ) =
⊕λ∈ΛS −1Mλ.
≺ xλ −1
−1
=≺
Define φ : S (⊕λ∈ΛMλ) −→ ⊕λ∈ΛS Mλ by setting φ
s
xλ
.
s
Note that xλ’s are 0 for all but finitely many values of λ. Therexλ
fore, so are
’s.
s
Claim. φ is well defined.
≺ yλ ≺ xλ =
.
Let
s
t
=⇒ There exists some u ∈ S such that ut ≺ xλ = us ≺ yλ or
≺ utxλ =≺ usyλ .
=⇒ utxλ = usyλ for all λ ∈ Λ.
xλ
yλ
=⇒
=
for all λ ∈ Λ.
s
t
x
y
=⇒ ≺ λ =≺ λ s
t
=⇒ φ is well defined.
Claim. φ is S −1R-linear.
≺ xλ ≺ yλ a
−1
,
∈ S −1 (⊕λ∈ΛMλ).
Let ∈ S R,
s
s
s
≺ yλ a ≺ xλ ≺ yλ a ≺ xλ +
=φ
+
=⇒ φ
s
s
s
ss
s
a ≺ xλ +s ≺ yλ =φ
ss
≺ axλ + syλ =φ
ss
axλ + syλ
=≺
ss
a xλ
y
+ λ
s s
s
a
x
y
= ≺ λ+≺ λ
s
s
s
≺ xλ ≺ yλ a
+φ
= φ
s
s
s
=≺
Claim. φ is one-one.
≺ xλ Let
∈ ker(φ).
s
x
=⇒ ≺ λ = 0.
s
x
=⇒ λ = 0 for all λ ∈ Λ.
s
=⇒ There exists tλ ∈ S such that tλxλ = 0 for all λ ∈ Λ.
Let A = {λ ∈ Λ | xλ 6= 0}. Then A is finite.
Put t =
Q
λ∈A tλ
=⇒ txλ = 0 for all λ ∈ A.
If λ 6∈ A then xλ = 0 and therefore, txλ = 0 in this case as well.
=⇒ txλ = 0 for λ ∈ Λ.
t ≺ xλ ≺ txλ ≺ xλ =
=
=0
=⇒
s
ts
ts
=⇒ φ is one-one.
Claim. φ is onto.
Let X ∈ ⊕λ∈ΛS −1Mλ.
=⇒ For every λ ∈ Λ, there exist xλ ∈ Mλ and sλ ∈ S such that
x
x
X =≺ λ and λ ’s are 0 for all but finitely many values of λ.
sλ
sλ
(
)
x
Put B = λ ∈ Λ | λ =
6 0 . Then B is finite.
sλ
Let B = {λ1, . . . , λn}.
Put s =
Qn
Qn
s
and
t
=
λj
i=1 λi
i=1,i6=j sλi .
Then for all j = 1, . . . , n,
xλj
sλj
=
tλj xλj
s
.
Put yλj = tλj xλj for all j = 1, . . . , n and yλ = 0 for all λ ∈ Λ \ B.
y
x
Then for all λ ∈ Λ, λ = λ .
s
sλ
≺ yλ y
x
=⇒ φ
=≺ λ =≺ λ = X
s
s
sλ
=⇒ φ is onto.
(c) Let I be an ideal of R. Then S −1
Let a ∈ S −1(
√
√ I). Then there exist b ∈
I =
√
S −1I.
I and s ∈ S such that
b
s
=⇒ There exists n ∈ N such that bn ∈ I.
bn
n
=⇒ a = n ∈ S −1I.
s
√
√
√
−1
−1
=⇒ a ∈ S I, that is, S ( I) ⊆ S −1I.
a=
√
√
Conversely, let a ∈
S −1I.
=⇒ There exists n ∈ N such that an ∈ S −1I.
b
Let b ∈ R and s ∈ S such that a = .
s
bn
=⇒ n ∈ S −1I.
s
c
bn
=⇒ There exist c ∈ I and t ∈ S such that n =
s
t
=⇒ There exist u ∈ S such that utbn = usnc ∈ I
√
n
=⇒ (utb) ∈ I or utb ∈ I.
√
b
utb
−1
=⇒ a = =
∈ S ( I).
s
uts
√
=⇒
S −1I ⊆ S −1(
√
I).
(d) S −1 (nil(R)) = nil(S −1R).
Take I = 0 in (c)
(e) If M is fg then S −1 (annR (M )) = annS −1R (S −1M ).
Let a ∈ S −1 (annR (M )). Then there exist b ∈ annR (M ) and s ∈ S
b
such that a = .
s
Let X ∈ S −1M . Then there exist x ∈ M and t ∈ S such that
x
X= .
t
0
bx
=
= 0 for all X ∈ S −1M .
st
st
=⇒ a ∈ annS −1R (S −1M ) for all a ∈ S −1 (annR (M )).
=⇒ aX =
=⇒ S −1 (annR (M )) ⊆ annS −1R (S −1M ).
Conversely, let a ∈ annS −1R (S −1M ). Then there exist b ∈ R and
b
s ∈ S such that a = .
s
Let T = {x1, . . . , xn} be a set of generators of M .
b xi
bxi
xi
=⇒ 0 = a
=
=
.
1
s1
s
=⇒ There exists some si ∈ S such that sibxi = 0, i = 1, . . . , n.
Put t = s1 . . . sn. Then tbxi = 0 for all i = 1, . . . , n.
Let x ∈ M .
Then there exist a1, . . . , an ∈ R such that x =
Pn
i=1 ai xi .
Pn
Pn
=⇒ tbx = tb i=1 aixi = i=1 aitbxi = 0 for all x ∈ M .
=⇒ tb ∈ annR (M ).
b
tb
=⇒ a = =
∈ S −1 (annR (M )) for all a ∈ annS −1R (S −1M ).
s
ts
=⇒ annS −1R (S −1M ) ⊆ S −1 (annR (M )).
Assignment – 3
MTH611A – 2020-21
1. Let I and J be ideal of R and let S be a multiplicative set in R.
If J is fg then show that (S −1I : S −1J) = S −1(I : J).
Assume that J is generated by {b1, . . . , bn}.
a
Let X = ∈ (S −1I : S −1J).
s
b
Note that i ∈ S −1J for all i = 1, . . . , n.
1
a bi
∈ S −1I for all i = 1, . . . , n.
=⇒
s 1
ci
ci
a bi
abi
=
= ;
=⇒ There exist ci ∈ I and ti ∈ S such that
or
s 1
ti
s
ti
i = 1, . . . , n.
=⇒ There exists ui ∈ S such that uitiab = uisci ∈ I; i = 1, . . . , n.
Put u = u1 . . . un and t = t1 . . . tn.
=⇒ utabi ∈ I for all i = 1, . . . , n.
Let b ∈ J.
Pn
=⇒ There exist a1, . . . , an ∈ R such that b = i=1 aibi.
P
Pn
=⇒ utab = uta n
a
b
=
i=1 i i
i=1 ai utabi ∈ I for all b ∈ J.
=⇒ uta ∈ (I : J).
a
uta
=⇒ X = =
∈ S −1(I : J) for all X ∈ (S −1I : S −1J).
s
uts
=⇒ (S −1I : S −1J) ⊆ S −1(I : J).
Conversely, let X ∈ S −1(I : J).
a
=⇒ There exist a ∈ (I : J) and s ∈ S such that X = .
s
b
−1
Let Y ∈ S J. Then there exist b ∈ J and t ∈ S such that Y = .
t
ab
=⇒ XY =
∈ S −1I (as ab ∈ I) for all Y ∈ S −1J.
st
=⇒ X ∈ (S −1I : S −1J) for all X ∈ S −1(I : J).
=⇒ S −1(I : J) ⊆ (S −1I : S −1J).
2. Let S be a multiplicative set in R such that 0 6∈ S. Then there is
a prime ideal P of R such that P ∩ S = ∅.
As 0 6∈ S, S −1R is a nonzero ring.
Therefore, S −1R contains prime ideals. Let Q be a prime ideal of
S −1R.
=⇒ There exists a prime ideal P of R such that S −1P = Q and
S ∩ P = ∅.
3. Let F = {S | S is a multiplicative set in R and 0 6∈ S}. Show that
F has maximal elements. Further show that S is maximal in F iff
S = R \ P for some minimal prime P of R.
We shall use Zorn’s Lemma to prove this result.
Note that {1} ∈ F and hence F 6= ∅.
We order F by inclusion.
Let T be a chain in S and let S 0 = ∪S∈T S.
Claim. S 0 is a multiplicative set and S 0 ∈ F .
Note that 1 ∈ S for all S ∈ T . Hence 1 ∈ S 0.
Let s, t ∈ S 0.
=⇒ There exist S1, S2 ∈ T such that s ∈ S1 and t ∈ S2.
As T is a chain, either S1 ⊆ S2 or S2 ⊆ S1. Assume that S1 ⊆ S2.
=⇒ s, t ∈ S2 and hence st ∈ S2 ⊆ S 0.
=⇒ S 0 is multiplicative.
Also, 0 6∈ S 0 as no member of T contains 0.
Therefore, S 0 ∈ F and S 0 is an upper bound of T .
By Zorn’s Lemma, F has maximal element.
Let S be a maximal element of F .
As 0 6∈ S, there exists a prime ideal P of R such that S ∩ P = ∅.
=⇒ S ⊆ R \ P .
As R \ P is a multiplicative set and 0 6∈ R \ P , R \ P ∈ F .
But S is maximal in F and hence S = R \ P .
Recall that the collection of all prime ideals of R has minimal
elements. Hence the exists a minimal prime P 0 of R such that
P0 ⊆ P.
=⇒ S = R \ P ⊆ R \ P 0 and again by the maximality of S in F ,
S = R \ P 0. Thus P = P 0, thas is, P is a minimal prime of R.
Conversely, let P be a minimal prime of R. Let T = R \ P .
As 0 6∈ T , we have T ∈ F .
=⇒ There exists a maximal element S of F such that T ⊆ S.
By first part, there exists a minimal prime Q of R such that S =
R \ Q.
=⇒ Q ⊆ P and hence P = Q (as P is minimal).
=⇒ T = R \ P = R \ Q = S, that is, T is a maximal element of F .
4. Let S be a multiplicative set. Then S is called saturated if the
following condition is satisfied: For x, y ∈ R, xy ∈ S iff x, y ∈ S.
Show that S is saturated iff R \ S is a union of prime ideals.
First Assume that S is saturated.
We need to show that R \ S is a union of prime ideals.
Case I. 0 ∈ S.
Let a ∈ R. Then a0 = 0 ∈ S.
=⇒ a ∈ S for all a ∈ R, that is, S = R.
=⇒ R \ S = ∅ = union of the empty collection of prime ideals.
Case II. 0 6∈ S.
Let F = {P ⊆ R | P is a prime ideal of R and P ∩ S = ∅ }.
Then F 6= ∅.
Let X = ∪P ∈F P .
=⇒ X ∩ S = (∪P ∈F P ) ∩ S = ∪P ∈F (P ∩ S) = ∅.
=⇒ S ⊆ R \ X.
Let a ∈ R \ X.
a
is unit of S −1R.
Claim.
1
a
Suppose, if possible, that
is a nonunit of S −1R.
1
a
∈ Q.
1
=⇒ There exists a prime ideal P 0 of R such that P 0 ∩ S = ∅ and
=⇒ There exists a prime ideal Q of S −1R such that
S −1P 0 = Q.
a
=⇒ ∈ S −1P 0 and hence a ∈ P 0.
1
As P 0 ∩ S = ∅, P 0 ∈ F . Hence P 0 ⊆ X.
=⇒ a ∈ X, a contradiction.
a
is unit of S −1R.
=⇒
1
ab
1
= .
1s
1
=⇒ There exists some u ∈ S such that uab = us ∈ S.
Thus there exist b ∈ R and s ∈ S such that
As S is saturated, a ∈ S (and ub ∈ S) for all a ∈ R \ X.
=⇒ R \ X ⊆ S, that is, S = R \ X.
=⇒ R \ S = X = a union of prime ideals.
Conversely, suppose that R \ S is a union of prime ideals.
Let G be a collection of prime ideals such that R \ S = ∪P ∈GP .
Let a, b ∈ R such that ab ∈ S.
=⇒ ab 6∈ R \ S = ∪P ∈GP .
=⇒ ab 6∈ P for every P ∈ G.
=⇒ a 6∈ P and b 6∈ P for every P ∈ G.
=⇒ a 6∈ ∪P ∈GP = R \ S and b 6∈ ∪P ∈GP = R \ S.
=⇒ a ∈ S and b ∈ S.
Therefore, S is saturated.
5. Let S be a multiplicative set. Show that there exists a smallest
saturated multiplicative set S containing S. Show that R\S equals
the union of all prime ideal which do not intersect with S. (This
S is called the saturation of S.)
Let F = {T ⊆ R | T is a saturated multiplicative set and S ⊆ T }.
Note that R is a saturated multiplicative set and hence R ∈ F .
Thus F is nonempty.
Put S = ∩T ∈F T .
Then S is a multiplicative set and S ⊆ S.
Let a, b ∈ R such that ab ∈ S.
=⇒ ab ∈ T for every T ∈ F .
=⇒ a, b ∈ T for every T ∈ F .
=⇒ a, b ∈ S, that is, S is saturated.
Therefore, S is a saturated multiplicative set containing S and it is
contained in every saturated multiplicative set which contains S.
This means it the smallest saturated multiplicative set containing
S.
Let X = {P ⊆ R | P is prime ideal of R such that P ∩ S = ∅}.
=⇒ S ⊆ R \ P for every P ∈ X.
=⇒ S ⊆ R \ ∪P ∈X P .
Put S 0 = R \ ∪P ∈X P .
By a previous problem, S 0 is a saturated multiplicative set.
=⇒ S 0 ∈ F (as S ⊆ S 0).
=⇒ S ⊆ S 0.
Let a ∈ S 0.
a
is a unit of S −1R.
Claim.
1
a
Suppose, if possible, that
is a nonunit of S −1R.
1
a
−1
=⇒ There exists a prime ideal Q of S R such that ∈ Q.
1
=⇒ There exists a prime ideal P 0 of R such that P 0 ∩ S = ∅ and
S −1P 0 = Q.
a
=⇒ ∈ S −1P 0 and hence a ∈ P 0.
1
Note that P 0 ∈ X (as P 0 ∩ S = ∅).
=⇒ a ∈ ∪P ∈X P (as P 0 ⊆ ∪P ∈X P ).
=⇒ a 6∈ S 0 = R \ ∪P ∈X P , a contradiction.
a
=⇒ is unit of S −1R. Thus there exist b ∈ R and s ∈ S such that
1
ab
1
= .
1s
1
=⇒ There exists some u ∈ S such that uab = us ∈ S ⊆ S.
As S is saturated, a ∈ S (and ub ∈ S) for all a ∈ S 0.
=⇒ S 0 ⊆ S, that is, S 0 = S.
=⇒ R \ S = ∪P ∈X P .
a
0
(As S = S , we note that if a ∈ S then
is a unit in S −1R.)
1
6. Show that S −1R and S
−1
R are isomorphic as rings, where S is
saturation of S.
If 0 ∈ S then 0 ∈ S and hence S −1R = 0 and S
−1
R = 0. In this
case they are isomorphic.
Assume, now that 0 6∈ S.
−1
Let φ : R −→ S R be the natural ring homomorphism, given by
a
φ(a) = .
1
s
−1
Let s ∈ S. Then φ(s) = is a unit of S R (as S ⊆ S).
1
a
−1
=
By universal property, the map ψ : S −1R −→ S R given by ψ
s
1 a
a
φ(s)−1φ(a) =
= is a ring homomorphism.
s 1
s
Claim. ψ is one-one.
a
Let ∈ ker(ψ).
s
a
−1
=⇒ = 0 in S R.
s
=⇒ There exists t ∈ S such that ta = 0.
t
As in the previous problem,
is a unit in S −1R.
1
−1
−1
−1
a
t
ta
t
ta
t
0
=⇒ =
=
=
= 0.
s
1
1s
1
s
1
s
=⇒ ψ is one-one.
Claim. ψ is onto.
Let X ∈ S
−1
a
R. Then there exist a ∈ R and t ∈ S such that X = .
t
t
is a unit in S −1R.
1
!
!
−1
−1
t
a
a
t
=⇒ ψ
ψ
=ψ
1
1
1
1
Again, as in the previous problem,
= ψ
=
−1
t
a
1
1
−1
a
t
1
1
1 a
a
−1
=
= = X for all X ∈ S R.
t 1
t
=⇒ ψ is onto.
7. Show that the set D of all zero-divisors of R is a union of prime
ideals. Show that every minimal prime of R is contained in D.
Let S = The set of all nonzero divisors of R.
Then S is multiplicative and saturated (check).
=⇒ D = R \ S is a union of prime ideals.
Let P be a minimal prime of R.
Consider, RP .
Claim. RP has exactly one prime ideal, namely, PP .
Let Q be a prime ideal of RP .
=⇒ There exists a prime ideal P 0 of R such that Q = T −1P 0 and
T ∩ P 0 = ∅, where T = R \ P .
=⇒ P 0 ⊆ R \ T = P , and hence P = P 0 as P is a minimal prime.
=⇒ Q = T −1P 0 = T −1P = PP .
=⇒ RP has exactly one prime ideal, namely, PP .
=⇒ nil(RP ) = PP .
We want to show that P ⊆ D.
Let a ∈ P .
a
As ∈ PP , it is nilpotent.
1
an
=⇒ There exists n ∈ N such that
=0=
.
1
1
n
=⇒ There exists t ∈ T such that ta = 0.
n
a
So we have t 6= 0, ta, ta2, . . . , tan = 0.
=⇒ There exists i ≥ 0 such that tai 6= 0 and tai+1 = 0 = (tai)a.
=⇒ a is a zero-divisor of R, that is, a ∈ D for all a ∈ P .
=⇒ P ⊆ D.
=⇒ D contains all minimal primes of R.
8. Let I1, . . . , In be ideals of R and let S be a multiplicative set of R.
Show that S −1(I1 . . . In) = S −1I1 . . . S −1In.
Let X = {a1 . . . an | aj ∈ Ij for all j = 1 . . . , n}.
=⇒ S −1(I1 . . . In) is generated by
a1 . . . an
Y =
| aj ∈ Ij for all j = 1 . . . , n .
1
Let x ∈ Y .
a1 . . . an
=⇒ There exist a1 ∈ I1, . . . , an ∈ In such that x =
=
1
a1
an
... .
1
1
=⇒ x ∈ S −1I1 . . . S −1In for all x ∈ Y .
=⇒ Y ⊆ S −1I1 . . . S −1In.
=⇒ S −1(I1 . . . In) ⊆ S −1I1 . . . S −1In.
Note that S −1I1 . . . S −1In is generated by Z = {x1 . . . xn | xj ∈
S −1Ij for all j = 1 . . . , n}.
Let x ∈ Z.
=⇒ There exist x1 ∈ S −1I1, . . . , xn ∈ S −1In such that x = x1 . . . xn.
aj
=⇒ There exist aj ∈ Ij and sj ∈ S such that xj = ; j = 1, . . . , n.
sj
a
an
a . . . an
=⇒ x = 1 . . .
= 1
∈ S −1(I1 . . . In) (as a1 . . . an ∈ I1 . . . In).
s1
sn
s1 . . . s n
=⇒ Z ⊆ S −1(I1 . . . In).
=⇒ S −1I1 . . . S −1In ⊆ S −1(I1 . . . In).
9. Show that Q ⊗Z Z/nZ = 0 for all n ∈ N.
a
∈ Q and y ∈ Z/nZ.
b
a
na
=⇒ x ⊗ y =
⊗y =
⊗ y.
b
nb
a
a
= n
⊗y =
⊗ ny
nb
nb
a
=
⊗ 0 = 0.
nb
=⇒ x ⊗ y = 0 for all x ∈ Q and y ∈ Z/nZ.
Let x =
=⇒ Q ⊗Z Z/nZ = 0 for all n ∈ N.
10. Show that Q is flat as a Z-module.
Q ⊗Z
Q
n∈N Z/nZ
Also,
show that
6= 0.
Let K be the quotient field of Z.
Note that K = S −1Z, where S = Z \ {0}.
We shall show that K and Q are isomorphic as Z-modules.
We have the inclusion map φ : Z −→ Q.
=⇒ The map ψ : K −→ Q given by ψ
x
s
= φ(s)−1φ(x) =
ring homomorphism.
Then ψ is one-one (as φ is one-one) and clearly, onto.
=⇒ ψ is an isomorphism.
In particular, ψ is Z-linear (as ψ respects addition).
x
is a
s
=⇒ K and Q are isomorphic as Z-modules.
Therefore, Q is flat over Z as K is flat over Z.
Let θ : Z −→
Q
n∈N Z/nZ be given by θ(x) =≺ x + nZ for all x ∈ Z.
Claim. θ is Z-linear and one-one.
Let x, y ∈ Z.
=⇒ θ(x + y) =≺ (x + y) + nZ =≺ (x + nZ) + (y + nZ) =≺ (x + nZ) + ≺ (y + nZ) = θ(x) + θ(y) for all x, y ∈ Z.
=⇒ θ is Z-linear.
Now let, x ∈ ker(θ).
=⇒ θ(x) =≺ (x + nZ) = 0.
=⇒ x + nZ = 0 = nZ in Z/nZ for all n ∈ N.
=⇒ x ∈ nZ or n | x for all n ∈ N.
=⇒ x = 0 and hence θ is one-one.
=⇒ IQ ⊗ θ : Q ⊗Z Z −→ Q ⊗Z
map.
∼ Q, Q ⊗ Z 6= 0.
As Q ⊗Z Z =
Z
Q
=⇒ Q ⊗Z
n∈N Z/nZ 6= 0.
However,
Q
Q
n∈N Z/nZ
is a one-one Z-linear
n∈N Q ⊗Z (Z/nZ) = 0 as Q⊗Z (Z/nZ) = 0 for all n ∈ N.
(We have shown that if {Mλ | λ ∈ Λ} is family of R-modules and N
∼⊕
is an R-module then (⊕λ∈ΛMλ) ⊗N =
λ∈Λ (Mλ ⊗ N ). However, in
general, ( λ∈Λ Mλ)⊗N need not be isomorphic to
Q
Q
λ∈Λ (Mλ ⊗ N ).)
11. Let R be an integral domain and let S be a multiplicative set in
R such that 0 6∈ S. Show that
(a) The natural map : R −→ S −1R is one-one.
(b) S −1R is an integral domain.
(a) Let a ∈ ker().
a
=⇒ = 0 in S −1R.
1
=⇒ There exists some t ∈ S such that ta = 0.
As R is an integral domain and t 6= 0 (as 0 6∈ S), a = 0.
=⇒ is one-one.
(b) X, Y ∈ S −1R such that XY = 0.
Let a, b ∈ R and s, t ∈ S such that X =
a
b
and Y = .
s
t
ab
Then
= 0.
st
=⇒ There exists some u ∈ S such that uab = 0.
As u 6= 0, either a = 0 or b = 0.
=⇒ Either X = 0 or Y = 0.
12. Let R be an integral domain with the quotient field K. Let S be
a multiplicative set in R such that 0 6∈ S. Show that there exists a
one-one ring homomorphism from S −1R to K. Further show that
the quotient field of S −1R is isomorphic to K.
Let φ : R −→ K be the natural ring homomorphism. Then φ is
one-one.
s
6= 0 for all s ∈ S.
1
=⇒ φ(s) is a unit of K for all s ∈ S.
=⇒ φ(s) =
By
universal
property,
there
exists a ring homomorphism
1 a
a
a
−1
−1
= φ(s) φ(a) =
ψ : S R −→ K given by ψ
=
for
s
s 1
s
all a ∈ R and s ∈ S.
Claim. ψ is one-one.
a
Let ∈ ker(ψ), where a ∈ R and s ∈ S.
s
a
=⇒ = 0 in K.
s
a
=⇒ φ(a) = = 0 in K.
1
=⇒ a = 0 and hence
a
= 0 in S −1R.
s
=⇒ ψ is one-one.
Let L be the quotient field of S −1R (S −1R is an integral domain).
Then there
exists a one-one ring homomorphism θ : L −→ K given
!
x
= ψ(y)−1ψ(x) for all x, y ∈ S −1R, y 6= 0.
by θ
y
Claim. θ is onto.
a
Let X = ∈ K, where a, b ∈ R and b 6= 0.
b
b
Then 6= 0 in S −1R.
1
b
a
Let x = and y = be elements of S −1R.
1
1
x
=⇒ θ
y
!
= ψ(y)−1ψ(x) =
−1
a
b
1
1
=
1 a
a
= = X.
b 1
b
=⇒ θ is onto and hence an isomorphism.
13. Let R be an integral domain and let a ∈ R \ 0 be a non-unit such
n ) = 0 (for example, 2 in Z). Put I = (an ). Let S
that ∩∞
(a
n
n=1
be a multiplicative set in R such that a ∈ S and 0 6∈ S. Show that
∞
−1
−1
−1
∞
∩n=1S In = S R but S
∩n=1In = 0.
As a ∈ S,
a
is a unit in S −1R.
1
an
is a unit in S −1R for all n ∈ N.
=⇒
1
=⇒ S −1In = S −1R for all n ∈ N (as an ∈ In).
−1 I = S −1 R.
=⇒ ∩∞
n
n=1 S
∞
∞
−1
∩n=1In = 0.
As ∩n=1In = 0, we have S
(This means if {Nλ | λ ∈ Λ} is a family of submodules of an
R-module M , then in general, S −1 (∩λ∈ΛNλ) 6= ∩λ∈ΛS −1Nλ. However, if Λ is finite then the equality is guaranteed.)
f
g
14. Let 0 −→ K −→ M −→ L −→ 0 be an exact sequence of Rmodules. Let g1 : L −→ M be an R-linear map such that g og1 = IL.
Show that there exists an R-linear map f1 : M −→ K such that
f1of = IK . Furthermore, if M is fg then so are K and L.
Let x ∈ M .
=⇒ (g1og)(x) ∈ M and
g(x − (g1og)(x)) = g(x) − (g og1og)(x)
= g(x) − (g og1)(g(x))
= g(x) − (IL)(g(x))
= g(x) − g(x) = 0.
=⇒ x − (g1og)(x) ∈ ker(g) = Im(f ).
=⇒ There exists some yx ∈ K such that f (yx) = x − (g1og)(x).
As f is one-one this yx is unique (for a given x).
Define f1 : M −→ K by setting f1(x) = yx for all x ∈ M .
Claim. f1 is R-linear.
Let x, x0 ∈ M and a ∈ R.
=⇒ f (ayx + yx0 ) = af (yx) + f (yx0 )
= a(x − (g1og)(x)) + x0 − (g1og)(x0)
= (ax + x0) − (g1og)(ax + x0)
= f (yax+x0 ).
As f is one-one, yax+x0 = ayx + yx0 .
=⇒ f1(ax + x0) = af1(x) + f1(x0) for all x, x0 ∈ M and a ∈ R.
=⇒ f1 is R-linear.
Let y ∈ K. Put x = f (y).
=⇒ f (yx) = x − (g1og)(x)
= f (y) − (g1og)(f (y))
= f (y) − g1(g(f (y)) = f (y).
=⇒ yx = y as f is one-one.
=⇒ (f1of )(y) = f1(f (y)) = f1(x) = yx = y for all y ∈ K.
=⇒ f1of = IK .
Assume now that M is fg.
As g is onto, L is fg.
Let y ∈ K. Then f1(f (y)) = y and hence y ∈ Im(f1) for all y ∈ K.
=⇒ f1 is onto.
=⇒ K is fg.
15. Let R be a local ring and P be a fg projective R-module. Show
that P is free.
Let m be the maximal ideal of R. Put F = R/m.
=⇒ P/mP is fg as an R-module.
=⇒ P/mP is fg as an F = R/m-module.
=⇒ P/mP is fg as a vector-space over the field F .
=⇒ P/mP has a finite basis as a vector space over the field F .
Let B = {X1, . . . , Xn} be a basis of P/mP as a vector space over
F.
=⇒ There exist xi’s in R such that Xi = xi +mP for all i = 1, . . . , n.
By a result proved in the lectures, P is generated by S = {x1, . . . , xn}.
Let M be a free R-module with basis {e1, . . . , en}.
=⇒ There exists an R-linear map φ : M −→ P such that φ(ei) = xi
for all i = 1, . . . , n.
The map φ is onto as S generates P and S ⊆ Im(φ).
Let K = ker(φ) and let : K −→ M be the inclusion map.
φ
Then 0 −→ K −→ M −→ P −→ 0 is exact.
As P is projective and φ is onto, there exists an R-linear map
ψ : P −→ M such that φoψ = IP .
Then by the previous result K is fg (as M is fg).
e
e
φ
Claim. 0 −→ F ⊗R K −→ F ⊗R M −→ F ⊗R P −→ 0 is exact.
We only need to show that e is one-one.
By the previous problem, there exists an R-linear map θ : M −→ K
such that θo = IK .
=⇒ θeoe = θg
o = I F ⊗ R K .
=⇒ e is one-one.
Let α : F ⊗R K −→ K/mK
α((a + m) ⊗ x) = ax + mK.
Let β : F ⊗R M
be the isomorphism given by
−→ M/mM be the isomorphism given by
β((a + m) ⊗ x) = ax + mM .
Let γ : F ⊗R P −→ P/mP
γ((a + m) ⊗ x) = ax + mP .
be the isomorphism given by
Let ¯
: K/mK −→ M/mM be the R-linear map induced by . Then
¯
(x + mK) = (x) + mM .
Let φ̄ : M/mM −→ P/mP be the R-linear map induced by φ. Then
φ̄(x + mM ) = φ(x) + mP .
Now consider the digram
e
φ
e
0
−→ F ⊗R K −→ F ⊗R M −→ F ⊗R P −→ 0



0y

αy
0 −→ K/mK
γ
y

βy
¯
φ̄
−→ M/mM −→
P/mP

0y
−→ 0.
The squares in above digram are commutative (check).
As the upper sequence is exact, so is the lower sequence.
Claim. φ̄ is an isomorphism.
Note that (1) M/mM and P/mP are F -modules, that is, vector
spaces over F .
(2) As φ̄ is R-linear, it is also F -linear.
(3) M is generated by n elements as R-module and therefore,
M/mM is generated by n elements as an R-module.
Hence,
M/mM is generated by n elements as an F -module. This means,
dimF (M/mM ) ≤ n.
(4) As φ̄ is onto, dimF (M/mM ) ≥ dimF (P/mP ) = n. Therefore,
dimF (M/mM ) = dimF (P/mP ) = n.
=⇒ φ̄ is an isomorphism of vector spaces over F .
=⇒ ker(φ̄) = 0 = Im(¯
)
=⇒ K/mK = 0 (as ¯
is one-one).
=⇒ K = mK.
By Nakayama Lemma, K = 0 (as K is fg and J(R) = m).
=⇒ φ : M −→ K is an isomorphism.
As M is free, so is P .
16. Let I be an ideal of R and let M be an R-module. Let X =
{m ⊆ R | m is a maximal ideal of R and I ⊆ m}. If Mm = 0 for all
m ∈ X then show that M = IM .
Let m be a maximal ideal of R.
We shall show that Mm = (IM )m.
. Case I. m 6∈ X.
=⇒ I 6⊆ m.
=⇒ There exists some s ∈ I such that s 6∈ m, that is, s ∈ R \ m .
s
=⇒ ∈ Im.
1
s
As
is a unit in Rm, Im = Rm.
1
=⇒ (IM )m = ImMm = Mm (as S −1(IM ) = (S −1I)(S −1M )).
Case II. m ∈ X.
Then Mm = 0 = (IM )m.
Therefore, Mm = (IM )m for all maximal ideals of R.
Consider, the exact sequence
π
0 −→ IM −→ M −→ M/IM −→ 0,
where is inclusion and π is the natural map.
Let m be a maximal ideal of R.
m
πm
=⇒ 0 −→ (IM )m −→
Mm −→
(M/IM )m −→ 0 is exact.
As (IM )m = Mm, we have ker(πm) = Im(m) = Mm.
=⇒ πm is the 0 map.
As πm is also onto, (M/IM )m = 0, for all maximal ideal m of R.
=⇒ M/IM = 0 or M = IM .
Lecture – 31
42. Integral Extensions
Let B be a ring and let A be a subring of B.
Then we say that A ⊆ B is a ring extension of extension of rings.
Let A[x] be the polynomial ring over A in variable x.
Pn
Let f (x) = i=0 aixi ∈ A[x].
Then n is called the degree of f (x) if an 6= 0.
Note that the degree of the polynomial 0 is not defined.
Then f (x) is called monic if n = deg(f (x)) and an = 1.
Let b ∈ B. Then
Pn
i
i=0 ai b ∈ B. We denote this element by f (b).
We say that f (b) has been obtained by substituting x = b.
We also say that f (b) is a polynomial in b with coefficients in A.
Let C = {f (b) | f (x) ∈ A[x]}.
Then C is a subring of B containing A ∪ {b}.
We denote this subring by A[b].
If f (b) = 0 for some monic polynomial f (x) ∈ A[x], then we say that
b is integral over A.
Let R = {b ∈ B | b is integral over A}.
Then R is called the integral closure of A in B.
Note that if b ∈ A then it satisfies the monic polynomial x − b ∈ A[x].
=⇒ A ⊆ R ⊆ B.
We shall show that R is a subring of B.
We shall need the some lemmas.
Lemma 57 Let A be a subring of R and R be a subring of R0. If R
and R0 are fg as modules over A and R respectively, then R0 is fg as
a module over A.
Proof. Suppose that R is generated by {x1, . . . , xn} as a module over
A and R0 is generated by {y1, . . . , ym} as a module over R.
Let x ∈ R0.
Pm
=⇒ There exist b1, . . . , bm ∈ R such that x = j=1 bj yj .
As bj ∈ R, there exist aj1, . . . , ajn ∈ A such that bj =
j = 1, . . . , m.
Pn
i=1 ajixi ,
Pm Pn
=⇒ x = j=1 i=1 ajixiyj for all x ∈ R0.
Let N be the A-submodule of R0 generated by S = {xiyj | i =
1, . . . , n; j = 1, . . . , m}.
=⇒ x ∈ N for all x ∈ R0.
=⇒ R0 = N , that is, R0 is generated by the finite set S as A-module. Lemma 58 Let A ⊆ B be a ring extension and let b ∈ B. Then the
following statements are equivalent:
(a) b is integral over A.
(b) A[b] is finitely generated as an A-module.
(c) There exists a subring C of B containing A such that b ∈ C and
C is fg as an A-module.
Proof.
(a) =⇒ (b) Note that A[b] is generated by S = {1, b, b2, . . .}
as A-module.
Pn
Let f (x) = i=0 aixi ∈ A[x] be a monic polynomial (of degree n) such
that f (b) = 0. Put T = {1, b, . . . , bn−1}.
Claim. A[b] is generated by the set T as an A-module.
Let N be the A-submodule of A[b] generated by T .
By induction, we shall show that bk ∈ N for all k ≥ n.
Pn
i = 0. Thus bn = − Pn−1 a bi ∈ N .
a
b
i=0 i
i=0 i
Now let k ≥ n + 1 and bn, . . . , bk−1 ∈ N .
P
i+k−n ∈ N .
=⇒ bk = − n−1
a
b
i
i=0
Note that an = 1 and
=⇒ S ⊆ N .
As A[b] is generated by S as an A-module, A[b] ⊆ N , that is, A[b] = N .
=⇒ A[b] is finitely generated as an A-module.
(b) =⇒ (c) As A[b] is finitely generated as an A-moudle, A ⊆ A[b] ⊆ B
and b ∈ A[b], we can take C = A[b].
(c) =⇒ (a) Assume that C is generated by S = {x1, . . . , xn} as an
A-module.
Let φ : C −→ C given by φ(x) = bx for all x ∈ C.
=⇒ φ is C-linear and therefore, A-linear.
By Proposition 2.4 of Atiyah Macdonald, there exist a0, · · · , an−1 ∈ A
such that φn + an−1φn−1 + · · · + a0IC = 0.
=⇒ φn(1) + an−1φn−1(1) + · · · + a0IC (1) = 0.
=⇒ bn + an−1bn−1 + · · · + a0 = 0.
=⇒ f (b) = 0, where f (x) = xn + an−1xn−1 + · · · + a0 ∈ A[x] (and f (x)
is monic).
=⇒ b is integral over A.
Lemma 59 Let A ⊆ B be a ring extension and let C be the integral
closure of A in B. Then C is a subring of B and A ⊆ C.
Proof. We have already shown that A ⊆ C.
Let α, β ∈ C.
Then A1 = A[α] is fg as an A-module as α is integral over A.
Note that β is also integral over A1 as A ⊆ A1.
=⇒ A2 = A1[β] is fg as an A1-module.
=⇒ A2 is fg as an A-module.
As α − β, αβ ∈ A2, they are integral over A.
=⇒ α − β, αβ ∈ C for all α, β ∈ C.
As A ⊆ C, unity of B= unity of A ∈ C.
=⇒ C is a subring of B.
Let A ⊆ B be a ring extension.
Let C be the integral closure of A in B.
We say that B is integral over A if C = B, that is, all elements of B
are integral over A. In this case, we also say that A ⊆ B is integral.
If C = A, then we say that A is integrally closed in B.
Lemma 60 Let A ⊆ B be a ring extension and let b1, . . . , bn ∈ B be
integral over A. Then there exists a subring R of B such that A ⊆ R,
b1, . . . , bn ∈ R and R is fg as an A-module.
Proof. Induction on n.
If n = 1, then by Lemma 58, we are done.
Assume that n ≥ 2.
Then, by induction, there exists a subring R0 of B such that A ⊆ R0,
b1, . . . , bn−1 ∈ R0 and R0 is fg as an A-module.
Note that bn is integral over R0 as A ⊆ R0.
=⇒ R0[bn] is fg as an R0-module.
=⇒ R0[bn] is fg as an A-module (as R0 is fg as an A-module).
Take R = R0[bn]. Clearly, b1, . . . , bn ∈ R, and A ⊆ R.
Lemma 61 Let A ⊆ B and B ⊆ C be ring extensions, both integral.
Then A ⊆ C is also integral.
Proof. Let c ∈ C.
=⇒ There exist b1, . . . , bn ∈ B such that cn + b1cn−1 + · · · + bn = 0.
As b1, . . . , bn ∈ B, they are integral over A.
=⇒ There exists a subring R of B such that b1, . . . , bn ∈ R, A ⊆ R and
R is fg as an A-module.
As b1, . . . , bn ∈ R, c is also integral over R.
=⇒ R[c] is fg as an R-module.
=⇒ R[c] is fg as an A-module (as R is fg as an A-moduel).
By Lemma 58, c is integral over A, for all c ∈ C.
=⇒ A ⊆ C is integral.
Corollary. Let A ⊆ B be a ring extension and let C be the integral
closure of C in B. Then C is integrally closed in B.
Proof. Note that A ⊆ C is integral.
Let R be the integral closure of C in B.
=⇒ C ⊆ R is integral.
By Lemma 61, A ⊆ R is integral.
=⇒ R ⊆ C ⊆ R and hence C = R.
=⇒ C is integrally closed in B
Lemma 62
Let A ⊆ B be an integral ring extension. Let S be a
multiplicative set in A. Then S −1A ⊆ S −1B is integral.
Proof. Let x ∈ S −1B.
b
Let b ∈ B and s ∈ S such that x = .
s
As b ∈ B, it is integral over A.
=⇒ There exist a1, . . . , an ∈ A such that bn + a1bn−1 + · · · + an = 0.
bn + a1bn−1 + · · · + an
0
= n = 0.
=⇒
n
s
s
n
b
a1 b n−1
a2 b n−2
an
=⇒
+
+ 2
+ · · · + n = 0.
s
s s
s
s
s
b
=⇒ x = is integral over S −1A.
s
As x ∈ S −1B is arbitrary, S −1A ⊆ S −1B is integral.
Definition. Let A and B be rings and φ : A −→ B be a ring homomorphism. If B is integral over φ(A) then we shall say that B is integral
over A. (This may cause a bit of ambiguity, but is manageable.)
Lemma 63 Let A ⊆ B be an integral extension and let J be an ideal
of B. Let I = A ∩ J. Then B/J is integral over A/I.
Proof.
Note that we have a ring homomorphism φ : A/I −→ B/J
given by φ(a + I) = a + J for all a ∈ A. Furthermore, φ is one-one.
Let x ∈ B/J.
=⇒ There exists some b ∈ B such that x = b + J.
As b is integral over A, there exist a1, . . . , an ∈ A such that bn +
a1bn−1 + · · · + an = 0.
=⇒ (b + J)n + (a1 + J)(b + J)n−1 + · · · + (an + J) = 0.
=⇒ (b + J)n + φ(a1 + I)(b + J)n−1 + · · · + φ(an + I) = 0.
=⇒ x = b + J is integral over Im(φ), for all x ∈ B/J.
=⇒ Im(φ) ⊆ B/J is integral.
=⇒ B/J is integral over A/I.
Comments. Let A ⊆ B be a ring extension and b ∈ B be integral
over A. Assume that B is an integral domain. If b 6= 0 then there
exist a1, . . . , an ∈ A such that an 6= 0 and bn + a1bn−1 + · · · + an = 0.
Let f (x) = xm + a1xm−1 + · · · + am ∈ A[x] be a monic polynomial such
that f (b) = 0.
Claim. At least one of the ai’s is nonzero.
Assume, the contrary.
=⇒ a1 = a2 = · · · = am = 0, that is, bm = 0.
=⇒ b = 0 (as B is an integral domain), a contradiction.
Let n = max{i | ai 6= 0}.
=⇒ an 6= 0 and an+1 = an+2 = · · · = am = 0.
=⇒ bm + a1bm−1 + · · · + anbm−n = 0 = bm−n(bn + a1bn−1 + · · · + an).
As b 6= 0, we must have bn + a1bn−1 + · · · + an = 0.
As an 6= 0, we are done.
Lecture – 32
43. Going-up theorem
Lemma 64 Let A ⊆ B be an integral ring extension such that B is
an integral domain. Then A is a field iff B is a field.
Proof. Assume first that B is a field.
Let a ∈ A be nonzero.
=⇒ There exists b ∈ B such that ab = 1.
As A ⊆ B is integral, there exist a1, . . . , an ∈ A such that bn + a1bn−1 +
· · · + an = 0.
=⇒ 0 = an−1(bn + a1bn−1 + · · · + an) = b + a1 + a2a + · · · + anan−1.
=⇒ b = −(a1 + a2a + · · · + anan−1) ∈ A.
=⇒ a has inverse in A.
As a ∈ A (a 6= 0) is arbitrary, A is a field.
Conversely, let A be a field.
Let b ∈ B, b 6= 0.
=⇒ As A ⊆ B is integral and B is an integral domain, there exist
a1, . . . , an ∈ A such that an 6= 0 and bn + a1bn−1 + · · · + an = 0.
As A is a field and an ∈ A is nonzero, there exists a ∈ A such that
aan = 1.
=⇒ (−a)(bn + a1bn−1 + · · · + an−1b) = (−a)(−an) = aan = 1.
=⇒ b(−a)(bn−1 + a1bn−2 + · · · + an−1) = 1.
=⇒ b has a multiplicative inverse, namely, −a(bn−1 + a1bn−2 + · · · +
an−1) in B.
As b ∈ B, b 6= 0 is arbitrary, B is a field.
Corollary. Let A ⊆ B be an integral extension. Let Q be a prime
ideal of B and let P = A ∩ Q. Then P is a maximal ideal of A iff Q is
a maximal ideal of B.
Proof.
Note that we have a ring homomorphism φ : A/P −→ B/Q
given by φ(a + P ) = a + Q for all a ∈ A. Furthermore, φ is one-one
B/Q is integral over Im(φ).
Note that B/Q is an integral domain.
Assume that Q is maximal.
=⇒ B/Q is a field.
=⇒ Im(φ) is a field.
∼ Im(φ) as rings).
=⇒ A/P is a field (as φ is one-one, A/P =
=⇒ P is a maximal ideal of A.
Conversely, let P be a maximal ideal of A.
=⇒ A/P is a field.
∼ Im(φ) as rings).
=⇒ Im(φ) is a field (as φ is one-one, A/P =
=⇒ B/Q is a field (as B/Q is an integral domain and Im(φ) ⊆ B/Q is
integral).
=⇒ Q is a maximal ideal of B.
Theorem. Let A ⊆ B be an integral extension and let P be prime
ideal of A. Then there exists a prime ideal Q of B such that Q∩A = P .
Proof. Let S = A \ P .
=⇒ S −1A ⊆ S −1B is integral.
b
Let m be a maximal ideal of S −1B. Then Q = b ∈ B | ∈ m
1
−1
prime ideal of B such that Q ∩ S = ∅ and S Q = m.
is a
Note that m ∩ S −1A is a maximal ideal of S −1A.
=⇒ m ∩ S −1A = S −1P (as S −1A is local with maximal ideal S −1P ).
Claim. Q ∩ A = P .
Let a ∈ P .
a
=⇒ ∈ S −1P ⊆ m = S −1Q.
1
=⇒ a ∈ Q for all a ∈ P or P ⊆ Q ∩ A.
Conversely, let a ∈ Q ∩ A.
a
=⇒ a ∈ Q and hence ∈ m.
1
a
a
As ∈ S −1A as well, ∈ S −1A ∩ m = S −1P .
1
1
=⇒ a ∈ P for all a ∈ Q ∩ A.
=⇒ Q ∩ A ⊆ P or Q ∩ A = P .
Theorem. Let A ⊆ B be an integral extension and let P ⊆ P 0 be prime
ideals of A. Let Q be a prime ideal of B such that Q ∩ A = P . Then
there exists a prime ideal Q0 of B such that Q ⊆ Q0 and Q0 ∩ A = P 0.
Proof. Let φ : A/P −→ B/Q be the map given by φ(a + P ) = a + Q.
Then φ is a one-one ring homomorphism and B/Q is integral over
A/P .
Let α : A −→ A/P and β : B −→ B/Q be the natural maps.
=⇒ α(P 0) is prime ideal of A/P (as P ⊆ P 0).
=⇒ φ(α(P 0)) is prime ideal of Im(φ).
As B/Q is integral over Im(φ), there exists a prime ideal Q0 of B/Q
such that Q0 ∩ Im(φ) = φ(α(P 0)).
Put Q0 = β −1(Q0).
Then Q0 is a prime ideal of B and Q ⊆ Q0.
Claim. Q0 ∩ A = P 0.
Let a ∈ P 0.
=⇒ φ(α(a)) ∈ φ(α(P 0)) ⊆ Q0.
=⇒ φ(a + P ) ∈ Q0.
=⇒ β(a) = a + Q = φ(a + P ) ∈ Q0.
=⇒ a ∈ Q0 for all a ∈ P 0.
=⇒ P 0 ⊆ Q0 ∩ A.
Conversely, let a ∈ Q0 ∩ A.
=⇒ β(a) = a + Q ∈ Q0.
=⇒ φ(α(a)) = φ(a + P ) = a + Q ∈ Q0 ∩ Im(φ) = φ(α(P 0)).
=⇒ There exists a0 ∈ P 0 such that φ(α(a0)) = φ(α(a)).
=⇒ α(a0) = α(a) as φ is one-one.
=⇒ a − a0 ∈ ker(α) = P ⊆ P 0.
As a0 ∈ P 0, a ∈ P 0 for all a ∈ Q0 ∩ A.
=⇒ Q0 ∩ A ⊆ P 0 or Q0 ∩ A = P 0.
Corollary. [Going-up Theorem] Let A ⊆ B be an integral extension
and let P1, . . . , Pn be prime ideals of A and Q1, . . . , Qr (r < n) be a
prime ideal of B such that P1 ⊆ . . . ⊆ Pn, Q1 ⊆ . . . ⊆ Qr and Qi ∩A = Pi
for all i = 1, . . . , r. Then there exist prime ideal Qr+1, . . . , Qn of B such
that Qr+1 ⊆ . . . ⊆ Qn and Qi ∩ A = Pi for all i = 1, . . . , n.
Proof. Let P = Pr , P 0 = Pr+1 and Q = Qr .
Then by the Theorem above, there exists a prime ideal Q0 of B such
that Q0 ∩ A = P 0.
Put Qr+1 = Q0.
Continuing this way, we get the result.
Lemma 65 Let A ⊆ B be an integral extension. Let Q and Q0 be
primes ideals of B such that Q ⊆ Q0 and Q ∩ A = Q0 ∩ A. Then Q = Q0.
Proof. Assume, if possible, that Q0 6= Q.
=⇒ There exists some b ∈ Q0 such that b 6∈ Q.
Let φ : A/P −→ B/Q be the natural map.
Then B/Q is integral over Im(φ).
Put c = b + Q. Then c is a nonzero element of B/Q.
=⇒ There exist a1, . . . , an ∈ A such that cn + φ(a1 + P )cn−1 + · · · +
φ(an + P ) = 0 and φ(an + P ) 6= 0.
=⇒ (b + Q)n + (a1 + Q)(b + Q)n−1 + · · · + (an + Q) = 0 = Q and
an + Q 6= Q.
=⇒ bn + a1bn−1 + · · · + an ∈ Q and an 6∈ Q.
=⇒ an ∈ Q0 as Q ⊆ Q0 and b ∈ Q0.
=⇒ an ∈ Q0 ∩ A = P = Q ∩ A ⊆ Q, a contradiction.
Therefore, Q = Q0.
Assignment – 4
MTH612A – 2020-21
1. Let R be a ring and let S ⊆ R. Let F = {A ⊆ R | A is a subring of
R and S ⊆ A}. Show that F has a unique minimal element. (This
element is called the subring generated by S and denoted by RS .)
Note that R ∈ F .
Therefore, F 6= ∅.
Let B = ∩A∈F A.
Then B is subring of R and S ⊆ B.
=⇒ B ∈ F and B is contained in every member of F .
=⇒ B is the unique minimal element of F .
2. Let R be a ring and let A be a subring of R. Let S ⊆ R. Then the
subring generated by A ∪ S is also called the subring generated by
S over A and it is denoted by A[S]. Show that A[S ∪T ] = (A[S])[T ]
for all S, T ⊆ R. If S = {b1, . . . , bn} then find the structure of A[S].
(We denote A[S] by A[b1, . . . , bn] in this case.)
Since A ∪ S ⊆ A ∪ (S ∪ T ) ⊆ A[S ∪ T ] hence A[S] ⊆ A[S ∪ T ].
Also, as T ⊆ A[S ∪ T ], A[S] ∪ T ⊆ A[S ∪ T ].
=⇒ (A[S])[T ] ⊆ A[S ∪ T ].
Conversely, A ∪ S ⊆ A[S] ⊆ (A[S])[T ] and T ⊆ (A[S])[T ].
=⇒ A ∪ (S ∪ T ) ⊆ (A[S])[T ].
=⇒ A[S ∪ T ] ⊆ (A[S])[T ].
=⇒ A[S ∪ T ] = (A[S])[T ].
We now prove the second part. Put X = (Z+)n.
α
n
For α = (α1, . . . , αn) ∈ X, put bα = b1 1 . . . bα
n .
Note that X = (Z+)n is closed under componentwise addition and
(0, . . . , 0) is the identity for this binary composition in X.
Let α, β ∈ X, α = (α1, . . . , αn) and β = (β1, . . . , βn).
Then α + β = (α1 + β1, . . . , αn + βn).
α +β
α +β
=⇒ bα+β = b1 1 1 . . . bn n n
α β
n b βn
= b1 1 b11 . . . bα
n n
α
β
β
n
n 1
= b1 1 . . . b α
n b1 . . . b n
= bαbβ for all α, β ∈ X.
Note that R is an A-module as A is a subring of R.
Let R0 be the A-submodule of R generated by the set Y = {bα |
α ∈ X}.
Claim. A ⊆ R0, b1, . . . , bn ∈ R0 and R0 is a subring of R.
0
Note that b(0,...,0) = b0
1 . . . bn = 1 . . . 1 = 1 ∈ Y .
=⇒ a = a1 ∈ R0 for all a ∈ A, that is, A ⊆ R0.
Also, bi = b(0,...,0,1,0,...,0) (1 at ith place) and hence bi ∈ Y ⊆ R0 for
all i = 1, . . . , n.
Now, let x, y ∈ R0.
P
=⇒ There exist xα’s and yβ ’s in A such that x = α∈X xαbα and
y=
P
β
β∈X yβ b such that xα ’s (resp. yβ ’s) are 0 for all but finitely
many values of α (resp. β).
Let P = {α ∈ X | xα 6= 0} and Q = {β ∈ X | yβ 6= 0}.
P
P
α
=⇒ x = α∈P xαb and y = β∈Q yβ bβ .
P
P
P
α bβ = P
α+β .
x
y
b
α∈P β∈Q α β
α∈P β∈Q xα yβ b
As α + β ∈ X, we have bα+β ∈ Y for all α ∈ P and β ∈ Q.
=⇒ xy =
=⇒ bα+β ∈ R0 for all α ∈ P and β ∈ Q.
P
P
=⇒ xy = α∈P β∈Q xαyβ bα+β ∈ R0 for all x, y ∈ R0.
Note that R0 is an abelian group and hence x − y ∈ R0 for all
x, y ∈ R0.
Also 1 ∈ R0 and hence R0 is a subring of R containing A ∪ S.
=⇒ A[S] ⊆ R0.
Conversely, note that b1, . . . , bn ∈ A[S].
α
n
=⇒ bα = b1 1 . . . bα
n ∈ A[S] (as A[S] is a subring) for all α =
(α1, . . . , αn) ∈ X.
=⇒ Y ⊆ A[S].
Also, as A ⊆ A[S], A[S] is an A-submodule of R.
=⇒ R0 ⊆ A[S], that is, R0 ⊆ A[S].
Hence elements of A[S] are precisely all polynomials in b1, . . . , bn
with coefficients in A.
3. Let A ⊆ B be a ring extension. Let b1, . . . , bn ∈ B be integral over
A. Show that A[b1, . . . , bn] is finitely generated as an A-module.
Induction on n.
For n = 1, proof is given in the lectures.
Let n ≥ 2 and let S = {b1, . . . , bn−1} and T = {bn}.
As bn is integral over A and A ⊆ A[S], bn is integral over A[S].
=⇒ (A[S])[bn] is fg as an A[S]-module.
By induction, A[S] is fg as an A-module.
=⇒ (A[S])[bn] is fg as an A-module.
Note that (A[S])[bn] = (A[S])[T ] = A[S ∪ T ] = A[b1, . . . , bn].
4. Let A be a ring and let B, C and D be A-algebras. Let f : B −→ C
be a ring homomorphism which is also A-linear. If C is integral
over B then show that so is C ⊗A D over B ⊗A D.
We have the A-linear map fe: B ⊗A D −→ C ⊗A D.
We shall show that fe is a ring homomorphism.
Note that fe respects addition.
Let x, y ∈ B ⊗A D.
Then there exist x1, . . . , xn, y1, . . . , ym ∈ B
Pn
and z1, . . . , zn, t1, . . . , tm ∈ D such that x = i=1 xi ⊗ zi and y =
Pm
j=1 yj ⊗ tj .
P
Pm
=⇒ fe(xy) = fe(( n
x
⊗
z
)(
i
i=1 i
j=1 yj ⊗ tj ))
Pn Pm
e
= f ( i=1 j=1 xiyj ⊗ zitj )
=
=
=
=
=
=
Pn Pm
e
i=1 j=1 f (xi yj ⊗ zi tj )
Pn Pm
i=1 j=1 f (xi yj ) ⊗ zi tj
Pn Pm
i=1 j=1 f (xi )f (yj ) ⊗ zi tj
Pn
Pm
( i=1 f (xi) ⊗ zi)( j=1 f (yj ) ⊗ tj )
P
e(Pm y ⊗ t )
fe( n
x
⊗
z
)
f
i
j
i=1 i
j=1 j
fe(x)fe(y) for all x, y ∈ B ⊗A D.
Also, fe(1 ⊗ 1) = f (1) ⊗ 1 = 1 ⊗ 1 = unity of C ⊗A D.
Therefore, fe is a ring homomorphism.
We now need to show that Im(fe) ⊆ C ⊗A D is integral.
Let R be the integral closure of Im(fe) in C ⊗A D.
Let c ∈ C and d ∈ D.
Then c is integral over f (B).
=⇒ There exist b1, . . . , bn ∈ B such that cn+f (b1)cn−1+f (b2)cn−2+
· · · + f (bn) = 0.
=⇒ (c ⊗ 1)n + fe(b1 ⊗ 1)(c ⊗ 1)n−1 + · · · + fe(bn ⊗ 1)
= cn ⊗ 1 + (f (b1) ⊗ 1)(cn−1 ⊗ 1) + · · · + (f (bn) ⊗ 1)
= cn ⊗ 1 + (f (b1)cn−1) ⊗ 1 + · · · + f (bn) ⊗ 1
= (cn + f (b1)cn−1 + · · · + f (bn)) ⊗ 1 = 0 ⊗ 1 = 0
=⇒ c ⊗ 1 ∈ R.
Also, 1 ⊗ d = f (1) ⊗ d = fe(1 ⊗ d) ∈ Im(fe) and hence, 1 ⊗ d ∈ R.
=⇒ c ⊗ d = (c ⊗ 1)(1 ⊗ d) ∈ R.
As every element of C ⊗A D is a finite sum of elements of the
form c ⊗ d and R is closed under addition, we have C ⊗A D ⊆ R.
=⇒ R = C ⊗A D, that is, C ⊗A D is integral over Im(fe).
5. Let A be a ring and let I = nil(A). Show that the integral closure
of A in A[x] is A + I[x].
Let R be the integral closure of A in A[x].
=⇒ A ⊆ R.
Let a ∈ I.
=⇒ There exists n ∈ N such that an = 0
=⇒ axi satisfies the monic polynomial tn ∈ A[t] for all i ≥ 0.
=⇒ axi ∈ R.
Pn
Let f (x) = i=0 aixi ∈ I[x]. Then ai ∈ I for all i = 0, . . . , n.
=⇒ aixi ∈ R for all i = 0, . . . , n.
=⇒ f (x) ∈ R for all f (x) ∈ I[x].
=⇒ I[x] ⊆ R and as A ⊆ R, we have A + I[x] ⊆ R.
Pn
Conversely, let f (x) = i=0 aixi ∈ R.
If f (x) = 0, then obviously, f (x) ∈ A + I[x].
Assume that f (x) 6= 0 and n = deg(f (x)).
By induction on n, we shall show that f (x) ∈ A + I[x].
If n = 0, then f (x) ∈ A ⊆ A + I[x].
Let n ≥ 1 and let f (x)m + b1f (x)m−1 + · · · + bm = 0, where
b1, . . . , bn ∈ A.
=⇒ 0 = coefficient of xmn in f (x)m + b1f (x)m−1 + · · · + bm = am
n.
=⇒ an ∈ I.
=⇒ anxn ∈ R.
Put g(x) =
Pn−1
i.
a
x
i
i=0
=⇒ g(x) = f (x) − anxn ∈ R.
Note that either g(x) = 0 or deg(g(x)) < n.
If g(x) = 0 then g(x) ∈ A + I[x] and if deg(g(x)) < n then, by
induction, g(x) ∈ A + I[x].
=⇒ f (x) = g(x) + anxn ∈ A + I[x] (as anxn ∈ I[x] ⊆ A + I[x]).
As f (x) ∈ R is arbitrary, we have R ⊆ A + I[x].
=⇒ R = A + I[x].
6. Let f ∈ A[x] be monic. Show that A[x] is integral over A[f ].
Let R be the integral closure of A[f ] in A[x].
Then A[f ] ⊆ R and therefore, A ⊆ R (as A ⊆ A[f ]).
Let f = xn + a1xn−1 + · · · + an ∈ A[x].
Let g(t) = tn + a1tn−1 + · · · + (an − f ) ∈ (A[f ])[t].
=⇒ g(x) = xn + a1xn−1 + · · · + (an − f ) = f − f = 0.
=⇒ x is integral over A[f ], that is, x ∈ R.
=⇒ xk ∈ R for all k ≥ 0.
=⇒ axk ∈ R for all k ≥ 0 and all a ∈ A (as A ⊆ R).
=⇒ A[x] ⊆ R, that is, R = A[x].
7. Let A ⊆ B be integral. Let a ∈ A be a unit in B. Show that a is
also a unit in A. Hence, show that J(A) = J(B) ∩ A.
As a is unit in B, there exists b ∈ B such that ab = 1.
As A ⊆ B is integral, there exist a1, . . . , an ∈ A such that bn +
a1bn−1 + · · · + an = 0.
=⇒ 0 = an−1(bn + a1bn−1 + · · · + an) = b + a1 + a2a + · · · + anan−1.
=⇒ b = −(a1 + a2a + · · · + anan−1) ∈ A.
=⇒ a is also a unit in A.
We now show that J(A) = J(B) ∩ A.
Let m be a maximal ideal of B.
=⇒ m ∩ A is a maximal ideal of A.
=⇒ J(A) ⊆ m ∩ A ⊆ m for all maximal ideals m of B.
=⇒ J(A) ⊆ J(B) ∩ A.
Conversely, let a ∈ J(B) ∩ A. Let x ∈ A.
1 + xa ∈ A and 1 + xa is a unit in B (as a ∈ J(B)).
=⇒ 1 + xa is a unit in A for all x ∈ A.
=⇒ a ∈ J(A) for all a ∈ J(B) ∩ A, that is, J(A) = J(B) ∩ A.
8. Let M1, . . . , Mn be finitely generated R-modules. Show that
is a finitely generated R-module.
We shall use induction on n.
If n = 1, then
Qn
i=1 Mi = M1 and hence we are done.
Assume that n ≥ 2.
Qn
i=1 Mi
Define f :
Qn−1
Qn
M
−→
i
i=1 Mi by f (x1 , . . . , xn−1 ) = (x1 , . . . , xn−1 , 0).
i=1
Claim. f is R-linear.
Qn−1
Let x, y ∈ i=1 Mi and a ∈ R.
Let x = (x1, . . . , xn−1) and y = (y1, . . . , yn−1).
=⇒ f (ax + y) = f (ax1 + y1, . . . , axn−1 + yn−1)
= (ax1 + y1, . . . , axn−1 + yn−1, 0)
= a(x1, . . . , xn−1, 0) + (y1, . . . , yn−1, 0)
Qn−1
= af (x) + f (y) for all x, y ∈ i=1 Mi and a ∈ R.
=⇒ f is R-linear.
Qn
Let g : i=1 Mi −→ Mn be the projection map.
Note that g(x1, . . . , xn) = xn and that g is R-linear.
Qn
=⇒ ker(g) = {(x1, . . . , xn) ∈ i=1 Mi | xn = 0} = Im(f ) and g is
onto.
Qn−1
f Qn
g
Therefore, i=1 Mi −→ i=1 Mi −→ Mn −→ 0 is exact.
Q
As Mn is fg and by induction, n−1
i=1 Mi is also fg, by a previous
Q
problem, n
i=1 Mi is fg as an R-module.
9. An A algebra B is called an integral A-algebra if B is integral over
A. Let B1, . . . , Bn be integral A-algebras. Show that
Qn
i=1 Bi is
also an integral A-algebra.
Qn
Put B = i=1 Bi.
Let φi : A −→ Bi be the ring homomorphism; i = 1, . . . , n.
Then φ : A −→ B be given by φ(a) = (φ1(a), . . . , φn(a)) for all a ∈ A
is a ring homomorphism, which makes B an A-algebra.
We need to show that Im(φ) ⊆ B is integral.
Let b = (b1, . . . , bn) ∈ B.
=⇒ bj ∈ Bj is integral over φj (A) for all j = 1 . . . , n.
=⇒ There exists a subring Rj of Bj such that bj ∈ Rj and Rj is
fg as a φj (A)-module.
Put R = {(x1, . . . , xn) ∈ B | xj ∈ Rj for all j = 1, . . . , n}.
Note that b ∈ R.
Claim. R is a subring of B and φ(A) ⊆ R.
Let x = (x1, . . . , xn) ∈ R and y = (y1, . . . , yn) ∈ R.
=⇒ xj , yj ∈ Rj for all j = 1, . . . , n.
=⇒ xj − yj , xj yj ∈ Rj for all j = 1, . . . , n.
=⇒ x − y = (x1 − y1, . . . , xn − yn) ∈ R and
xy = (x1y1, . . . , xnyn) ∈ R for all x, y ∈ R.
Let ej denote the unity of Bj .
Then ej ∈ Rj for all j = 1, . . . , n and therefore, (e1, . . . , en) ∈ R.
(Note that unity of B is (e1, . . . , en)).
=⇒ R is a subring of B.
Now let a ∈ A.
Then φj (a) ∈ Rj for all j = 1, . . . , n.
=⇒ φ(a) = (φ1(a), . . . , φn(a)) ∈ R.
=⇒ φ(A) ⊆ R.
Claim. R is fg as a φ(A)-module.
Define θ :
Qn
i=1 Ri −→ R by setting θ(x1 , . . . , xn ) = (x1 , . . . , xn ).
Note that each Ri is an A-module via φi : A −→ Bi and R is an
Q
A-module via φ : A −→ n
i=1 Bi .
Clearly, θ respects addition and it is a bijection.
Let a ∈ A and x = (x1, . . . , xn) ∈
Qn
i=1 Ri .
=⇒ θ(ax) = θ(ax1, . . . , axn)
= θ(φ1(a)x1, . . . , φn(a)xn)
= (φ1(a)x1, . . . , φn(a)xn)
= (φ1(a), . . . , φn(a))(x1, . . . , xn)
= φ(a)θ(x) = aθ(x) for all a ∈ A and x ∈
Qn
i=1 Ri .
Therefore, θ is an isomorphism of A-modules.
By previous problem,
Qn
i=1 Ri is fg as an A-module and therefore,
so is R. This means R is fg as a φ(A)-module.
As b ∈ R, b is integral over φ(A).
As b ∈ B is arbitrary, B is integral over φ(A).
This also means that B is integral over A.
10. Let A ⊆ B be an extension of rings such that B \ A is closed under
multiplication. Show that A is integrally closed in B.
Let b ∈ B be integral over A.
Suppose, if possible, b 6∈ A. Then b ∈ B \ A.
Let f (x) ∈ A[x] be a monic polynomial such that f (b) = 0.
Let F = {g(x) ∈ A[x] | g(x) is monic and g(b) = 0}.
Then f (x) ∈ F and therefore, F 6= ∅.
Let n = min{deg(g(x)) | g(x) ∈ F }.
Therefore, if g(x) ∈ F then deg(g(x)) ≥ n.
Let h(x) ∈ F be such that deg(h(x)) = n.
=⇒ Let h(x) = xn + a1xn−1 + · · · + an, where a1, . . . , an ∈ A.
We have h(b) = 0 = bn + a1bn−1 + · · · + an.
Note that if n = 1 then h(x) = x + a1 and hence, b = −a1 ∈ A.
A contradiction, hence n ≥ 2.
Also, b(bn−1 + a1bn−2 + · · · + an−1) = −an ∈ A.
=⇒ b(bn−1 + a1bn−2 + · · · + an−1) 6∈ B \ A.
=⇒ bn−1 + a1bn−2 + · · · + an−1 6∈ B \ A as b ∈ B \ A and B \ A is
closed under multiplication.
=⇒ bn−1 + a1bn−2 + · · · + an−1 ∈ A.
Let a = bn−1 + a1bn−2 + · · · + an−1.
Then a ∈ A.
Put h1(x) = xn−1 + a1xn−2 + · · · + an−1 − a.
Then h1(x) ∈ A[x] and h1(b) = 0.
=⇒ h1(x) ∈ F (as h1(x) is monic).
A contradiction, as F can not contain a monic polynomial of
degree less than n and deg(h1(x)) = n − 1.
=⇒ b ∈ A. Therefore, A is integrally closed in B.
Lecture – 34
11. Let f ∈ A[x] be a monic polynomial of degree ≥ 1 and let B =
A[x]/(f ). Show that the natural map from A to B is a one-one
ring homomorphism and that B is integral over A.
Let I = (f ).
Let : A −→ A[x] be the inclusion and let π : A[x] −→ A[x]/I be
the natural map. Then both are ring homomorphisms.
=⇒ φ = π o : A −→ A[x]/I is a ring homomorphism.
Let a ∈ ker(φ).
We want to show that a = 0.
Suppose, if possible, that a 6= 0.
=⇒ φ(a) = a + I = 0 = I.
=⇒ a ∈ I or a = gf for some g ∈ A[x].
Let deg(f ) = n and deg(g) = m.
Let f =
Pn
i and g = Pm b xi .
a
x
i=0 i
i=0 i
Then an = 1 and bm 6= 0.
=⇒ Coefficient of xm+n in f g = anbm = bm 6= 0.
=⇒ deg(gf ) = n + m = deg(a) = 0.
=⇒ n = 0 (and m = 0), a contradiction as n ≥ 1.
Therefore, φ is one-one.
Let R be the integral closure of φ(A) in A[x]/I.
Note that f + I = I = 0.
P
i ) + I = Pn (a + I)(x + I)i
=⇒ 0 = ( n
a
x
i=0 i
i=0 i
=
Pn
i
i=0 φ(ai )(x + I) .
As an = 1, φ(an) = 1 + I is the unity of A[x]/I.
=⇒ x + I is integral over φ(A), that is, x + I ∈ R.
Let y ∈ R.
Then y = h + I for some h ∈ A[x].
Pk
Let h = i=0 hixi.
P
P
=⇒ y = ( ki=0 hixi) + I = ki=0(hi + I)(x + I)i
P
= ki=0 φ(hi)(x + I)i ∈ R as φ(A) ⊆ R and x + I ∈ R.
=⇒ All elements of A[x]/I are integral over φ(A).
=⇒ A[x]/I is integral over A.
12. Let A be a subring of an integral domain B. Let C be the integral
closure of A in B. Let f, g ∈ B[x] be monic polynomials such that
f g ∈ C[x]. Show that f, g ∈ C[x].
Proof of this uses some field theory.
Let f (x) = xn +a1xn−1 +· · ·+an and g(x) = xm +b1xm−1 +· · ·+bm.
Let h(x) = f (x)g(x) = xn+m + c1xn+m−1 + · · · + cn+m.
Then a1, . . . , an, b1, . . . , bm ∈ B and c1, . . . , cn+m ∈ C.
Let K be the quotient field of B and let : B −→ K be the natural
map.
Put f 0(x) = xn + (a1)xn−1 + · · · + (an),
g 0(x) = xm + (b1)xm−1 + · · · + (bm) and
h0(x) = xn+m + (c1)xn+m−1 + · · · + (cn+m).
=⇒ h0(x) = f 0(x)g 0(x).
Let L be the algebraic closure of K.
Then all roots of f 0(x) and g 0(x) are in L.
Let α1 . . . , αn be the roots of f 0(x) in L and
β1 . . . , βm be the roots of g(x) in L.
=⇒ f 0(x) = (x − α1) . . . (x − αn) and g 0(x) = (x − β1) . . . (x − βm).
=⇒ h0(x) = f 0(x)g 0(x) = (x − α1) . . . (x − αn)(x − β1) . . . (x − βm).
=⇒ Roots of h0(x) are α1 . . . , αn, β1 . . . , βm.
Note that f 0(x), g 0(x) ∈ (B)[x] and h0(x) ∈ (C)[x].
Let R be the integral closure of (C) in L.
Note that h0(x) ∈ (C)[x] is monic and h0(αi) = 0 = h0(βj ) for all
i = 1, . . . , n and j = 1, . . . , m.
=⇒ α1 . . . , αn, β1 . . . , βm ∈ R.
Note that (a1), . . . , (an) are polynomial in α1 . . . , αn and hence
(a1), . . . , (an) ∈ R.
Claim. ai ∈ C for all i = 1, . . . , n.
Fix an i ∈ {1, . . . , n}.
As (ai) ∈ R, it is integral over (C), there exist t1, . . . , tk ∈ C such
that (ai)k + (t1)(ai)k−1 + · · · + (tk ) = 0.
=⇒ (aki + t1aik−1 + · · · + tk ) = 0.
As is one-one, aki + t1ak−1
+ · · · + tk = 0.
i
=⇒ As t1, . . . , tk ∈ C, ai is integral over C.
However, C is integrally closed in B and hence ai ∈ C, for all
i = 1, . . . , n.
=⇒ f (x) ∈ C[x] and similarly, g(x) ∈ C[x].
13. Prove the previous results without assuming that B is an integral
domain.
Let f (x) = xn +a1xn−1 +· · ·+an and g(x) = xm +b1xm−1 +· · ·+bm.
Let h(x) = f (x)g(x) = xn+m + c1xn+m−1 + · · · + cn+m.
Then a1, . . . , an, b1, . . . , bm ∈ B and c1, . . . , cn+m ∈ C.
Let B 0 = Z[y1, . . . , yn, z1, . . . , zm] be the polynomial ring in variables
y1, . . . , yn, z1, . . . , zm over Z.
Then as Z is an integral domain so is B 0.
Let f 0(x), g 0(x) ∈ B 0[x], where f 0(x) = xn + y1xn−1 + · · · + yn and
g 0(x) = xm + z1xm−1 + · · · + zm.
Put h0(x) = f 0(x)g 0(x) and let ti denote the coefficient of xm+n−i
in h0(x); i = 1, . . . , m + n.
=⇒ h0(x) = xn+m + t1xn+m−1 + · · · + tn+m.
Put S = {t1, . . . , tm+n} and let A0 = Z[S].
=⇒ h0(x) ∈ A0[x].
Let C 0 be the integral closure of A0 in B 0.
Then h0(x) ∈ C 0[x].
By previous problem, f 0(x), g 0(x) ∈ C 0[x].
=⇒ y1, . . . , yn, z1, . . . , zm ∈ C 0.
As Z ⊆ C 0, we must have B 0 = C 0.
By theory of polynomial rings, there exists a ring homomorphism
φ : B 0 −→ B such that φ(yi) = ai and φ(zj ) = bj for all i = 1, . . . , n
and j = 1, . . . , m.
Claim. φ(A0) ⊆ C.
Put y0 = 1 = z0 = t0 and a0 = 1 = b0 = c0.
=⇒ φ(yi) = ai and φ(zj ) = bj for all i = 0, . . . , n and j = 0, 1, . . . , m.
Note that tk = coefficient of xm+n−k in h0(x).
P
= i+j=m+n−k yn−izm−j
P
=⇒ φ(tk ) = i+j=m+n−k φ(yn−i)φ(zm−j )
P
= i+j=m+n−k an−ibm−j
= ck ∈ C for all k = 0, . . . , n + m.
Let t ∈ A0.
=⇒ t has a polynomial expression in t1, . . . , tn+m with coefficients
in Z.
As φ respects addition and multiplication, φ(t) has (an identical)
polynomial expression in φ(t1) = c1, . . . , φ(tn+m) = cn+m with coefficients in Z.
=⇒ φ(t) ∈ C for all t ∈ A0.
=⇒ φ(A0) ⊆ C.
Let u ∈ B 0.
Claim. φ(u) ∈ C.
As A0 ⊆ B 0 is integral there exist s1, . . . , sk ∈ A0 such that
uk + s1uk−1 + · · · + sk = 0.
=⇒ φ(uk + s1uk−1 + · · · + sk ) = 0.
=⇒ φ(u)k + φ(s1)φ(u)k−1 + · · · + φ(sk ) = 0.
=⇒ φ(u) is integral over C (as φ(si) ∈ C for all i = 1, . . . , k).
=⇒ φ(u) ∈ C (as C is integrally closed in B).
In particular, ai = φ(yi) ∈ C and bj = φ(zj ) ∈ C, i = 1, . . . , n and
j = 1, . . . , m.
14. Let A ⊆ B be an extension of rings and let C be the integral
closure of A in B. Show that the integral closure of A[x] in B[x]
is C[x].
Let R be the integral closure of A[x] in B[x].
We need to show that R = C[x].
Note that x ∈ R as x ∈ A[x] ⊆ R.
Let c ∈ C.
=⇒ c is integral over A and hence over A[x] (as A ⊆ A[x]).
=⇒ c ∈ R for all c ∈ C.
=⇒ cxk ∈ R for all c ∈ C k ≥ 0.
As every element of C[x] is a finite sum of elements of the form
cxk for c ∈ C and k ≥ 0 and R is closed under addition, C[x] ⊆ R.
Conversely, let f (x) ∈ R. Then f (x) ∈ B[x] is integral over A[x].
If f (x) = 0 then f (x) ∈ C[x].
Assume that f (x) 6= 0.
=⇒ There exist g1(x), . . . , gn(x) ∈ A[x] such that
f (x)n + g1(x)f (x)n−1 + · · · + gn(x) = 0.
Let m0 = deg(f (x)) and mi = deg(gi(x)) if gi(x) 6∈ 0 and mi = 0
if gi(x) = 0.
Let m = max{m0, . . . , mn}.
Choose k > m.
Put f1(x) = f (x) − xk . Then f (x) = f1(x) + xk .
=⇒ (xk + f1(x))n + g1(x)(xk + f1(x))n−1 + · · · + gn(x) = 0.
Put A0 = A[x]. Consider polynomial ring A0[t] in variable t over
A0 .
Then xk + t ∈ A0[t], and g1(x), . . . , gn(x) ∈ A0 ⊆ A0[t].
=⇒ (xk + t)n + g1(x)(xk + t)n−1 + · · · + gn(x) ∈ A0[t].
Then (xk + t)n + g1(x)(xk + t)n−1 + · · · + gn(x)
= tn + h1(x)tn−1 + · · · + hn−1(x)t + hn(x), . . . . . . . . . . . . . . . . . . (1)
where h1(x), . . . , hn(x) ∈ A0.
Note that hn(x) = xkn + g1(x)xk(n−1) + · · · + gn(x).
Claim.
gi(x)xk(n−i) = 0 or deg(gi(x)xk(n−i)) < kn for all i =
1, . . . , n.
Assume that gi(x)xk(n−i) 6= 0.
=⇒ deg(gi(x)xk(n−i)) = deg(gi(x)) + deg(xk(n−i))
= mi + k(n − i)
< k + k(n − i) = k(n + 1 − i) ≤ kn as i ≥ 1.
Therefore, deg(hn(x)) = kn and it is monic.
In (1), we take t = f1(x). This gives us
f1(x)n + h1(x)f1(x)n−1 + · · · + hn−1(x)f1(x) + hn(x)
= (xk + f1(x))n + g1(x)(xk + f1(x))n−1 + · · · + gn(x)
= f (x)n + g1(x)f (x)n−1 + · · · + gn(x) = 0.
=⇒ (−f1(x))(f1(x)n−1 + h1(x)f1(x)n−2 + · · · + hn−1(x)) = hn(x).
As −f1(x) and hn(x) are both monic,
f1(x)n−1 + h1(x)f1(x)n−2 + · · · + hn−1(x) is also monic.
Furthermore, hn(x) ∈ A[x] ⊆ C[x].
Therefore, by previous problem, −f1(x) ∈ C[x].
=⇒ f (x) = f1(x) − xk ∈ C[x] for all f (x) ∈ R.
=⇒ R ⊆ C[x], that is, R = C[x].
Lecture – 35
44. A result
Let M be a non-zero free R-module with a basis B and let S be
set which is bijective to B. Then there exists a free R-module N
∼ M.
containing S such that S is a basis of N and N =
Let σ : S −→ B be a bijection.
Put N = (M \ B) ∪ S.
Define φ : N −→ M by
φ(x) =


x if x ∈ M \ B

σ(x) if x ∈ S.
Claim. φ is a bijection.
We first show that φ is one-one.
Let x, y ∈ N such that φ(x) = φ(y).
Case I. x, y ∈ M \ B.
In this case φ(x) = x and φ(y) = y and hence x = y.
Case II. x, y ∈ S.
In this case φ(x) = σ(x) and φ(y) = σ(y) and hence x = y (as σ is
one-one).
Case III. x ∈ S and y ∈ M \ B.
In this case φ(x) = σ(x) ∈ B and φ(y) = y.
=⇒ y ∈ (M \ B) ∩ B = ∅, a contradiction.
We now show that φ is onto. Let x ∈ M .
If x ∈ M \ B, then x ∈ N and φ(x) = x.
If x ∈ B, then there exists s ∈ S such that x = σ(s) = φ(s).
=⇒ φ is also onto and hence a bijection.
Put ψ = φ−1.
Define a binary ⊕ in N by x ⊕ y = ψ(φ(x) + φ(y)), for all x, y ∈ N .
=⇒ φ(x ⊕ y) = φ(x) + φ(y) for all x, y ∈ N .
We now show that (N, ⊕) is an abelian group.
Let x, y, z ∈ N .
(1) φ(x ⊕ y) = φ(x) + φ(y) = φ(y) + φ(x) = φ(y ⊕ x).
=⇒ x ⊕ y = y ⊕ x (as φ is one-one) for all x, y ∈ N .
(2) φ((x ⊕ y) ⊕ z) = φ(x ⊕ y) + φ(z)
= (φ(x) + φ(y)) + φ(z)
= φ(x) + (φ(y) + φ(z))
= φ(x) + φ(y ⊕ z)
= φ(x ⊕ (y ⊕ x)).
=⇒ (x ⊕ y) ⊕ z = x ⊕ (y ⊕ x) (as φ is one-one) for all x, y, z ∈ N .
(3) Let x0 = ψ(0).
Then φ(x0) = φ(ψ(0)) = φoψ(0) = 0.
=⇒ φ(x ⊕ x0) = φ(x) + φ(x0) = φ(x) + 0 = φ(x)
=⇒ x ⊕ x0 = x = x0 ⊕ x (as φ is one-one) for all x ∈ N .
=⇒ x0 is the identity for ⊕.
(4) Let x ∈ N and let x0 = ψ(−φ(x)).
=⇒ φ(x ⊕ x0) = φ(x) + φ(x0)
= φ(x) + φ(ψ(−φ(x)))
= φ(x) + φoψ(−φ(x))
= φ(x) + (−φ(x))
= 0 = φ(x0).
=⇒ x ⊕ x0 = x0 = x0 ⊕ x (as φ is one-one).
Therefore, every x ∈ N has an inverse wrt ⊕.
Hence, (N, ⊕) is an abelian group.
Henceforth, we shall use + instead for ⊕ and use standard notations
in N .
Note that φ and ψ are group homomorphism.
We define . in N by setting a.x = ψ(aφ(x)).
Note that φ(a.x) = aφ(x) for all a ∈ R and x ∈ N .
Let a, b ∈ R and let x, y ∈ N .
(1) 1.x = ψ(1φ(x)) = ψ(φ(x)) = x for all x ∈ N .
(2) (a + b).x = ψ((a + b)φ(x))
= ψ(aφ(x) + bφ(x))
= ψ(aφ(x)) + ψ(bφ(x))
= a.x + b.x for all a, b ∈ R and x ∈ N .
(3) a.(x + y) = ψ(aφ(x + y))
= ψ(a(φ(x) + φ(y)))
= ψ(aφ(x) + aφ(y))
= ψ(aφ(x)) + ψ(aφ(y))
= a.x + a.y for all a ∈ R and x, y ∈ N .
(4) (ab).x = ψ(abφ(x))
= ψ(a(bφ(x)))
= ψ(a(bφ(x)))
= ψ(a(φ(b.x)))
= a.(b.x) for all a, b ∈ R and x ∈ N .
Therefore, . defines an R-module structure on N .
As φ(a.x) = aφ(x) for all a ∈ R and x ∈ N , φ is R-linear.
Note that φ is an isomorphism of R-modules with inverse ψ.
Claim. S is a basis of N .
Note that if s ∈ S then φ(s) = σ(s) and hence ψ(σ(s)) = s.
Let t ∈ B.
=⇒ σ −1(t) ∈ S.
=⇒ ψ(σ(σ −1(t))) = σ −1(t).
=⇒ ψ(t) = σ −1(t), for all t ∈ B.
We shall now show that N is generated by S, that is, Se = N .
Let x ∈ N . Then φ(x) ∈ M .
=⇒ There exist a1, . . . , an ∈ R and x1, . . . , xn ∈ B such that
P
φ(x) = n
i=1 ai xi .
P
=⇒ x = ψ( n
i=1 ai xi )
Pn
i=1 ai ψ(xi )
Pn
= i=1 aiσ −1(xi) ∈ Se for all x ∈ N .
e that is, N is generated by S.
Therefore, N = S,
=
We shall now show that S is LI.
Let s1, . . . , sn be distinct elements of S.
Pn
i=1 ai si = 0.
Pn
Pn
Pn
=⇒ 0 = φ( i=1 aisi) = i=1 aiφ(si) = i=1 aiσ(si).
Let a1, . . . , an ∈ R such that
As, σ is one-one, σ(s1), . . . , σ(sn) are distinct elements of B.
=⇒ σ(s1), . . . , σ(sn) are LI.
=⇒ a1 = · · · = an = 0.
Therefore, every finite subset of S is LI.
=⇒ S is a basis of N .
In other worlds, N is a free R-module which contains S and S is a
∼ M as an R-module.
basis of N . Furthermore, N =
45. Another result
We learn about matrices in linear algebra, where these matrices are
considered over a field. However, as matrices are only arrangements
of elements, they can be constructed over any set S. However, if S
has no binary operations, the addition or multiplication of matrices
over S can not be defined. In particular, we have matrices over any
commutative ring with unity. The matrices over a commutative ring
with unity fail to exhibit many properties of matrices over a field but
some of these properties still hold good. For example, addition and
multiplication of matrices over a commutative ring with unity R can
be defined. Also, the notion of determinant of such square matrices
exists. In fact, if A = [aij ] is an n × n matrix then its determinant is
defined to be equal to
P
σ∈Sn sgn(σ)a1σ(1) . . . anσ(n) , where Sn is the
symmetric group of {1, 2, . . . , n} and sgn(σ) is the signature of σ ∈ Sn.
We denote determinant of A by det(A).
The notion of transpose and adjoint are also defined and A adj(A) =
det(A)In = adj(A)A, is valid for all n × n square matrices A over R,
where adj(A) denotes the adjoint of square matrix A and In denote
the n × n identity matrix.
Let M be R-module and let x1, . . . , xn be elements of M .
Put X = the transpose of 1 × n matrix [x1 x2 . . . xn] or
X = [x1 x2
...
xn]t.
Let A = [aij ] be an n × n matrix over R.
We define A ∗ X = Y , where Y = [y1 y2 . . . yn]t and yi =
for all j = 1, . . . , n.
Pn
j=1 aij xj
It is easy to prove that (as in case of matrices over a field) A∗(B∗X) =
(AB) ∗ X, where A and B are n × n square matrices over R.
We say that an R-module M is faithful if annR (M ) = 0.
We now prove the result.
Theorem. Let M be a faithful R-module generated by a set S =
{x1, . . . , xn}. Let φ : M −→ M be an R-linear map. Then there exist
a1, . . . , an ∈ R such that φn + a1φn−1 + · · · + anIM = 0.
Proof. Let R0 = R[t] be the polynomial ring in variable t over R.
We have shown in Lecture 9 (second theorem) that M becomes an
R0-module via φ.
More precisely the R0-module structure on M is given by
Pk
i (x) for all x ∈ M
a
φ
i
i=0
Pk
(if f (t) = i=0 aiti ∈ R0, where φ0 = IM and φi = φo · · · oφ (i-times)).
f (t).x =
As txi = φ(xi) ∈ M , there exist ai1, . . . , ain ∈ R such that txi =
Pn
j=1 aij xj for all i = 1, . . . , n.
Let δij denote the Kronecker delta symbol.
P
=⇒ n
j=1 (aij − δij t)xj = 0 for all i = 1, . . . , n.
Put bij = aij − δij t for all i, j = 1, . . . , n.
Let A denote the n × n matrix [bij ] over R0.
Then A ∗ X = Y , where X = [x1
...
xn]t and Y = [0
=⇒ adj(A) ∗ (A ∗ X) = adj(A) ∗ Y = Y .
=⇒ (adj(A)A) ∗ X = Y .
=⇒ (det(A)In) ∗ X = Y .
=⇒ det(A)xi = 0 for all i = 1, . . . , n.
Let x ∈ M .
Pn
Then there exist a1, . . . , an ∈ R such that x = i=1 aixi.
P
=⇒ det(A)x = n
i=1 ai det(A)xi = 0 for all x ∈ M .
P
Note that det(A) = σ∈Sn sgn(σ)b1σ(1) . . . bnσ(n).
We shall look at b1σ(1) . . . bnσ(n) for various σ ∈ Sn.
Let e be the identity of Sn. Put T = {1, . . . , n}.
...
0]t.
Case I. σ = e.
In this case, b1σ(1) . . . bnσ(n) = b11b22 . . . bnn
= (a11 − t)(a22 − t) . . . (ann − t) = (−1)ng(t),
where g(t) ∈ R0 is a monic polynomial in t of degree n.
Case II. σ 6= e.
In this case, there is at least one i ∈ T such that σ(i) 6= i.
Let X = {i ∈ T | σ(i) 6= i} and Y = {i ∈ T | σ(i) = i}.
Note that |Y | < n.
b1σ(1) . . . bnσ(n) =
=
Q
Q
i∈X biσ(i)
i∈X (aiσ(i) − δiσ(i) t)
Q
Q
i∈Y biσ(i)
i∈Y (aiσ(i) − δiσ(i) t)
=
Q
i∈X aiσ(i)
Q
i∈Y (aii − t)
=a polynomial in t of degree less than n.
Therefore, det(A) is polynomial of degree n in t such that the coefficient of tn is (−1)n.
=⇒ There exist a1, . . . , an ∈ R such that det(A) = (−1)n(tn +a1tn−1 +
· · · + an).
=⇒ (tn + a1tn−1 + · · · + an)x = (−1)ndet(A)x = 0 for all x ∈ M .
=⇒ φn(x) + a1φn−1(x) + · · · + anIM (x) = 0 for all x ∈ M .
=⇒ (φn + a1φn−1 + · · · + anIM )(x) = 0 for all x ∈ M .
=⇒ φn + a1φn−1 + · · · + anIM = 0, (the map 0).
15. Show that Z is integrally closed in Q.
Let b ∈ Q be integral over Z.
=⇒ There exist a1, . . . , an ∈ Z such that bn + a1bn−1 + · · · + an = 0.
c
Let b = , where c, s ∈ Z, s 6= 0 and gcd(c, s) = 1.
s
n
n−1
c
c
Then
+ a1
+ · · · + an = 0.
s
s
=⇒ cn + a1scn−1 + · · · + ansn = 0.
=⇒ cn = −(a1scn−1 + · · · + ansn) = −s(a1cn−1 + · · · + ansn−1).
=⇒ s divides cn.
As gcd(c, s) = 1, gcd(cn, s) = 1 and therefore, s = ±1.
=⇒ b = ±c ∈ Z.
16. Let (M, +) be an abelian group (and therefore, a Z-module). Let
R be a ring and let φ : R −→ HomZ(M, M ) be a group homomorphism such that φ(1) = IM and φ(ab) = φ(a)oφ(b) for all a, b ∈ R.
Show that M becomes an R-module via φ.
Define a.x = φ(a)(x) for all a ∈ R and x ∈ M .
Let a, b ∈ R and x, y ∈ M . Then
(1) 1.x = φ(1)(x) = IM (x) = x for all x ∈ M .
(2) (a + b).x = φ(a + b)(x) = (φ(a) + φ(b))(x).
= φ(a)(x) + φ(b)(x) = a.x + b.x
for all a, b ∈ R and x ∈ M .
(3) a.(x + y) = φ(a)(x + y) = φ(a)(x) + φ(a)(y) = a.x + a.y
for all a ∈ R and x, y ∈ M .
(4) (ab).x = φ(ab)(x) = (φ(a)oφ(b))(x)
= φ(a)(φ(b)(x)) = φ(a)(b.x)
= a.(b.x) for all a, b ∈ R and x, y ∈ M
for all a, b ∈ R and x ∈ M .
=⇒ M is an R-module via φ.
17. Let (R, +) be an abelian group (and therefore, a Z-module). Let
φ : R −→ HomZ(R, R) be a group homomorphism. Put φa = φ(a)
for all a ∈ R. The map φ satisfies the following conditions
(1) There exists an e ∈ R such that φe = IR
(2) φaoφb = φφa(b) for all a, b ∈ R.
(3) φa(b) = φb(a) for all a, b ∈ R.
Show that R becomes a commutative ring with unity via φ.
Define a.b = φa(b) for all a, b ∈ R. Then
(1) a.b = φa(b) = φb(a) = b.a for all a, b ∈ R.
(2) (a.b).c = φa.b(c) = φφa(b)(c) = (φaoφb)(c) = φa(φb(c)) = a.(b.c)
for all a, b, c ∈ R.
(3) e.a = φe(a) = IR (a) = a = a.e for all a ∈ R.
(4) a.(b + c) = φa(b + c) = φa(b) + φa(c) = a.b + a.c for all a, b, c ∈ R.
(5) (a + b).c = c.(a + b) = c.a + c.b = a.c + b.c, for a, b, c ∈ R.
Therefore, R is a commutative ring with unity via φ.