(WINTER 2016) 110BH - RING/MODULE THEORY TOPICS
AIDEN GIM
Contents
1. Examples of rings
2. Rngs satisfying x n = x for all x ∈ R
3. Maximal ideals of a ring of continuous functions
4. Sums of two squares
5. An example of a PID which is not a Euclidean domain
6. Classes of rings
7. Polynomial functions
8. Exact
sequences
M
9.
Z and ∏ Z
10. Composition series
11. Take-home midterm
References
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19
1. Examples of rings
Example 1.1. Consider the ring R = Z/6Z and S = {0, 3} ⊆ R. Even though S is a subset of R and is a
ring itself with 0S = 0 and 1S = 3, S is not a subring of R because 1R = 1 6= 3 = 1S . It is a subrng of R.
Note that 3 ∈ S× but 3 ∈
/ R× .
Example 1.2. Let X be a set and P ( X ) be the power set of X. We define +, · on P ( X ) by A + B =
( A \ B) ∪ ( B \ A) = ( A ∪ B) \ ( A ∩ B) and A · B = A ∩ B. Then we can check that (P ( X ), +, ·) is a ring.
(Exercise : Check all the details! For example, you need to check that the distributive law holds by comparing
( A + B) · C = (( A \ B) ∪ ( B \ A)) ∩ C and ( A · C ) + ( B · C ) = (( A ∩ C ) \ ( B ∩ C )) ∪ (( B ∩ C ) \ ( A ∩ C )).
You can use Venn diagram to check these are equal.) We have 0P (X ) = ∅, 1P (X ) = X, P ( X )× = { X } and
− A = A for all A ∈ P ( X ).
Example 1.3. Let R be a nontrivial ring (0R 6= 1R ), then Mn ( R) is a ring with usual matrix addition/multiplication for n ≥ 1. Note that Mm,n ( R) is not a ring if m 6= n because multiplication is
not well-defined. For A ∈ Mn ( R), we denote the (i, j)-th component of A by ( A)ij . We define Eij ∈ Mn ( R)
to be the matrix satisfying
(
1R if (i, j) = (u, v)
( Eij )uv =
0R otherwise
for 1 ≤ u, v ≤ n. Then we can write A =
∑
( A)ij Eij . Given Eij , Ek` ∈ Mn ( R), we have
1≤i,j≤n
(
Eij Ek` =
Ei`
O
Proposition 1.4. Let n ≥ 2.
(1) Mn ( R) is not commutative.
Date: March 30, 2016.
1
if j = k
otherwise.
(2) Mn ( R) is not a division ring.
(3) Suppose R is commutative. If B ⊆ Mn ( R) is an ideal, then there exists an ideal A of R such that
B = Mn ( A ) .
(4) If R is a commutative simple ring, then Mn ( R) is simple.
Proof. (1) We have E11 E12 = E12 6= O = E12 E11 .
(2) Suppose Mn ( R) is a division ring. Then there is A ∈ Mn ( R) such that AE11 = In . Since
!
∑
AE11 =
Aij Eij
E11 =
1≤i,j≤n
∑
Ai1 Ei1 ,
1≤ i ≤ n
we have ( In )22 = 1 6= 0 = ( AE11 )22 , which is a contradiction.
(3) Let B ⊆ Mn ( R) be an ideal. Define A to be the ideal of R generated by all entries of elements in B.
Then clearly we have B ⊆ Mn (A).
Conversely, suppose A ∈ Mn (A). Let A = ∑ aij Eij with aij ∈ A, then we can write
i,j
t
aij =
∑ r k bk
k =1
for t ∈ Z≥1 , rk ∈ R and bk = ( Bk ) xk yk for some Bk ∈ B and 1 ≤ xk , yk ≤ n. Note that
!
∑
Eixk Bk Eyk j = Eixk
( Bk )uv Euv
1≤u,v≤n
Eyk j =
∑
1≤u,v≤n
( Bk )uv Eixk Euv Eyk j = bk Eij
for all 1 ≤ i, j ≤ n. This shows
aij Eij =
thus A =
t
t
k =1
k =1
∑ rk bk Eij = ∑ (rk Eixk ) Bk (Eyk j ) ∈ B,
∑ aij Eij ∈ B. Hence, B = Mn (A).
i,j
(4) Suppose R is a commutative simple ring. If B ⊆ Mn ( R) is an ideal, then by (3) there exists an ideal
A ⊆ R such that B = Mn (A). Since R is simple, A = 0 or R. Thus B = Mn (0) = 0 or B = Mn ( R).
2. Rngs satisfying x n = x for all x ∈ R
Let R be a rng (R may not have a multiplicative identity) and Z ( R) = { a ∈ R | ra = ar for all r ∈ R} be
the center of R. Note that Z ( R) is a subrng of R.
Lemma 2.1. Let n ∈ Z≥2 and suppose x n = x for all x ∈ R. Then the followings hold.
(1) If ab = 0 for a, b ∈ R, then ba = 0.
(2) If a2 = a for a ∈ R, then a ∈ Z ( R).
(3) an−1 ∈ Z ( R) for all a ∈ R.
Proof. (1) If ab = 0, then ba = (ba)n = b( ab)n−1 a = 0.
(2) Let x, a ∈ R and suppose a2 = a. Since 0 = ( a2 − a) x = a( ax − x ), we have axa − xa = ( ax − x ) a = 0 by
(1). Similarly, 0 = x ( a2 − a) = ( xa − x ) a gives axa − ax = a( xa − x ) = 0. Hence, ax = xa for all x ∈ R.
(3) For a ∈ R, we have ( an−1 )2 = an an−2 = aan−2 = an−1 . By (2), an−1 ∈ Z ( R).
Theorem 2.2. If x2 = x for all x ∈ R, then R is commutative.
Proof. By Lemma (3), R = Z ( R).
Theorem 2.3. If x3 = x for all x ∈ R, then R is commutative.
2
Proof. By Lemma (3), we have a2 ∈ Z ( R) for all a ∈ R. Thus any square in R commutes with all elements
in R. For given x, y ∈ R,
xy = ( xy)3 = ( x (yx )2 )y = ((yx )2 x )y = yx (yx2 )y = yx ( x2 y)y = y( x3 y2 ) = y(y2 x3 ) = yx.
Theorem 2.4. If x4 = x for all x ∈ R, then R is commutative.
Proof. Note that for any a ∈ R, we have − a = (− a)4 = a4 = a. Since
( a2 + a)2 = a4 + 2a3 + a2 = a + a2 ,
we have a2 + a ∈ Z ( R) for all a ∈ R by Lemma (2). Also we have
ab + ba = (( a + b)2 + ( a + b)) − ( a2 + a) − (b2 + b) ∈ Z ( R)
for all a, b ∈ R because Z ( R) is a subrng. For given x, y ∈ R,
xy = x4 y = x2 ( x2 y + yx2 ) − x2 yx2 = ( x2 y + yx2 ) x2 − x2 yx2 = yx4 = yx
since ( x2 )y + y( x2 ) ∈ Z ( R).
Theorem 2.5. If x6 = x for all x ∈ R, then R is commutative.
Proof. For any a ∈ R, we have − a = (− a)6 = a6 = a. Note that
a2 + a = ( a2 + a)6 = a6 ( a6 + 6a5 + 15a4 + 20a3 + 15a2 + 6a + 1) = a2 + a5 + a3 + a
gives a5 = − a3 = a3 for all a ∈ R. For given x ∈ R, we have x = x6 = xx5 = xx3 = x4 . By the previous
theorem, R is commutative.
At this point, we need to question if this statement is true for all positive integers n ≥ 2. In fact, the
following stronger statement holds.
Theorem 2.6 (Jacobson). Let R be a ring and suppose for each x ∈ R there is n = n( x ) ∈ Z≥2 such that x n( x) = x.
(n( x ) is not fixed and depends on x.) Then, R is commutative.
Proof. See [Jac45].
3. Maximal ideals of a ring of continuous functions
Let R = C ([0, 1], R) = { f : [0, 1] → R | f is continuous}, then R is a commutative ring with
( f + g)( a) = f ( a) + g( a),
( f g)( a) = f ( a) g( a)
for f , g ∈ R. Let Spm( R) be the set of maximal ideals of R.
Theorem 3.1. There is a bijection α : [0, 1] → Spm( R) given by
α ( a ) = m a : = { f ∈ R | f ( a ) = 0}.
Proof. We first prove that ma is maximal for all a ∈ [0, 1]. Define φa : R → R by φa ( f ) = f ( a), then φa is a
ring homomorphism. It is surjective since constant maps are continuous. By first isomorphism theorem, we
have
R/ma = R/ ker φa ∼
= R,
which is a field. Thus, ma is a maximal ideal.
Now let I be an ideal such that I 6⊆ ma for any a ∈ [0, 1]. Then we can choose f a ∈ I \ ma for all a ∈ [0, 1]
satisfying f ( a) 6= 0. By continuity of f a , there exist x a , y a ∈ [0, 1] such that x a < a < y a and f ( a) 6= 0 on
( x a , y a ). (In the case of endpoints, we regard ( x0 , y0 ) = [0, y0 ) and ( x1 , y1 ) = ( x1 , 1].) Then we have
[0, 1] =
[
a∈[0,1]
3
( x a , y a ).
By compactness of the closed interval [0, 1], there exist a1 , . . . , an ∈ [0, 1] such that
[0, 1] =
n
[
( x a k , y a k ).
k =1
n
Define f =
∑
k =1
f a2k , then f ≥ f a2k > 0 on ( x ak , y ak ). Since the intervals ( x ak , y ak ) cover [0, 1], we have f > 0 on
[0, 1], i.e., f ∈ R× . On the other hand, since
f = f a1 f a1 + · · · + f an f an ∈ ( f a1 , . . . , f an ) ⊆ I,
we have I = R. This proves that any proper ideal is contained in ma for some a ∈ [0, 1].
Now we consider S = C ((0, 1), R). We will show that in this case there is a maximal ideal which is not
of the form ma for some a ∈ (0, 1). Define
1
I = f ∈ S f
= 0 for all but finitely many n ∈ Z≥2 ,
n
π
then we can easily see that I is an ideal of S. Let f ( x ) = sin on (0, 1), then f ∈ I. Define
x

π
1

sin
if 0 < x ≤
x
k
f k (x) =
1
1

x −
if ≤ x < 1,
k
k
then f k ∈ I for all k ≥ Z≥2 .
Proposition 3.2.
(1) I 6⊆ ma for any a ∈ (0, 1).
(2) I is not prime (thus not maximal).
(3) There exists a maximal ideal m of S containing I with m 6= ma for any a ∈ (0, 1).
Proof. (1) Let a ∈ (0, 1) and choose k ∈ Z≥2 such that 0 <
f k ∈ I \ ma .
1
1
< a. Since f k ( a) = a − > 0, we have
k
k
π
π 1
(2) Consider g1 ( x ) = sin
and g2 ( x ) = sin
− 1 . Then we have
2x
2 x
1
1
1
1
g1
= 0 = g2
, g1
6= 0, g2
6= 0
2n
2n + 1
2n + 1
2n
for all n ≥ 1. This shows g1 g2 ∈ I but g1 , g2 ∈
/ I.
(3) By (2), I is not maximal. Thus by Zorn’s lemma, there exists a maximal ideal m of S properly containing
I. By (1), m 6= ma for any a ∈ (0, 1).
Question 3.3. Can you describe the ideal m and the field S/m?
4. Sums of two squares
Some integers can be written as sums of two squares in Z (e.g. 5 = 12 + 22 , 1 = 12 + 02 ), while there are
integers that can’t be (e.g. 3). In this section, we classify all integers that can be written as sums of two
squares.
Let Z[i ] = { x + yi | x, y ∈ Z} ⊆ C, then Z[i ] is a subring of C generated by Z ∪ {i }. We define the norm
map N : C → R≥0 by N (z) = zz, then N is multiplicative (i.e., N (z1 z2 ) = N (z1 ) N (z2 ) for z1 , z2 ∈ C) and
N (Z[i ]) ⊆ Z≥0 . Note that N (Z[i ]) is the set of sums of two squares in Z.
Theorem 4.1. Z[i ] is a Euclidean domain. (Thus, it is a PID and a UFD.)
4
Proof. We will show that the norm map N is a Euclidean function on Z[i ]. For given x, y ∈ Z[i ] with y 6= 0,
x
write x = a + bi, y = c + di ∈ Z[i ] for a, b, c, d ∈ Z. Let = r + si for r, s ∈ Q, then we can choose m, n ∈ Z
y
1
1
such that |r − m| ≤ , |s − n| ≤ . Let q = m + ni and r = x − qy, then we have x = qy + r. We have either
2
2
r = 0 or
x
N (r ) = N ( x − qy) = N (y) N
− q = N (y) N ((r − m) + (s − n)i )
y
1
= N ( y ) (r − m )2 + ( s − n )2 ≤ N ( y ) < N ( y ).
2
Lemma 4.2.
(1) For r ∈ Z[i ], r is a unit if and only if N (r ) = 1 if and only if r ∈ {±1, ±i }.
(2) Let p be a prime satisfying p ≡ 1 (mod 4), then there exists an integer x such that p | x2 + 1 in Z.
(3) Let p be a prime satisfying p ≡ 1 (mod 4), then p is a sum of two squares in Z.
(4) If a, b ∈ Z are sums of two squares in Z, then so is ab.
Proof. (1) Firstly, suppose r ∈ Z[i ]× , then there exists s ∈ Z[i ] such that rs = 1. Taking the norms gives
N (r ) N (s) = N (1) = 1.
Since N (r ), N (s) are nonnegative integers, we have N (r ) = 1. Now suppose N (r ) = 1. Write r = u + vi
for u, v ∈ Z, then we have u2 + v2 = 1. Thus (u, v) = (±1, 0), (0, ±1), i.e., r ∈ {±1, ±i }. Lastly, suppose
r ∈ {±1, ±i }, then it is easy to see that r is a unit in Z[i ].
(2) By Wilson’s theorem, we have ( p − 1)! ≡ −1 (mod p) for all primes p. Thus,
p−1 p+1
·
· · · ( p − 2)( p − 1)
2 2
p−1
p−1
≡ 1·2···
−
· · · (−2)(−1)
2
2
p −1
p−1 2
2
≡ (−1)
1·2···
2
2
p−1
≡ 1·2···
(mod p)
2
−1 ≡ ( p − 1) ! ≡ 1 · 2 · · ·
p−1
if p ≡ 1 (mod 4). Let x =
!, then p | x2 + 1.
2
(3) (Note: This is the part where we apply ring theory to number theory!) By (2), we have p | x2 + 1 in
Z, or p | ( x + i )( x − i ) in Z[i ] for some x ∈ Z. We will show that p ∈ Z[i ] is not irreducible. Suppose
p is irreducible in Z[i ], then it is a prime because Z[i ] is a UFD. Thus p | ( x + i )( x − i ) implies p | x + i
or p | x − i in Z[i ]. If p | x + i in Z[i ], then there exists a, b ∈ Z such that ( x + i ) = p( a + bi ). This gives
pb = 1 in Z, which is a contradiction. Similarly, p - x − i. So p is not irreducible in Z[i ]. Now we can write
p = αβ for some nonzero, nonunit α, β ∈ Z[i ]. If we take the norms on both sides, we get
p2 = N ( p ) = N ( α ) N ( β ).
Since α, β are nonunits, we have N (α), N ( β) ∈ Z>1 by (1). Hence, N (α) = N ( β) = p. If α = u + vi for
u, v ∈ Z, then we have p = N (α) = u2 + v2 , which is a sum of two squares.
(4) If a = x2 + y2 , b = z2 + w2 for x, y, z, w ∈ Z, then
ab = ( x2 + y2 )(z2 + w2 ) = ( xz + yw)2 + ( xw − yz)2 .
Alternatively you can argue that if a, b are sums of two squares, then there are z, w ∈ C such that
a = N (z), b = N (w). Hence, ab = N (z) N (w) = N (zw) is also a sum of two squares.
e
f
f
Theorem 4.3. Let n ∈ Z+ and n = 2e0 p11 · · · prer q11 · · · qs s be the prime factorization of n where e0 , r, s ≥ 0,
ei , f j ≥ 1 and pi ≡ 1 (mod 4), q j ≡ 3 (mod 4) for 1 ≤ i ≤ r, 1 ≤ j ≤ s. Then, n is a sum of two squares if and
only if f j ’s are even for all 1 ≤ j ≤ s.
5
Proof. (⇐) Suppose f j ’s are even for all j. We have 2 = 12 + 12 , q2j = q2j + 02 , and by Lemma (3), pi ’s are
also sums of two squares. Since n is a product of these sums of two squares, so is n by Lemma (4).
(⇒) Suppose n = x2 + y2 for some x, y ∈ Z. Let d = ( x, y) and x = dx0 , y = dy0 with ( x0 , y0 ) = 1. Let
q be a prime such that q | n and q ≡ 3 (mod 4). We will show that the exact power of q dividing n is
even. Suppose first that q | x02 + y20 . If q | x0 , then q | y20 = ( x02 + y20 ) − x02 , thus q | y0 . This contradicts the
fact that ( x0 , y0 ) = 1. Thus q - x0 , y0 . Consider x0 , y0 ∈ (Z/qZ) \ {0}. We have x02 + y20 ≡ 0 (mod q), i.e.,
x0 2 + y0 2 = 0 in Z/qZ. By Fermat’s little theorem,
1 = ( x 0 · y 0 −1 ) q −1 = ( x 0 2 · y 0 −2 )
q −1
2
= (−1)
q −1
2
q−1
is even, i.e., q ≡ 1 (mod 4) which is a contradiction. Hence, q - x02 + y20
2
and q | d. If q f kd (i.e., q f | d but q f +1 - d, or f is the exact power of q dividing d), then by the previous
argument we have q2 f kd2 ( x02 + y20 ) = n.
in Z/qZ. This implies that
We can also consider the set of integers which can be written as sums of three (or more) squares. The
followings are generalizations of the previous theorem.
Theorem 4.4. Let n ∈ Z+ , then n is a sum of three squares if and only if n is not of the form 4m (8k + 7) for some
m, k ∈ Z≥0 .
Theorem 4.5 (Four-square theorem). Every positive integer can be written as a sum of four (thus, or more) squares.
Theorem 4.6 (Waring’s problem). For each k ∈ Z+ , there is an integer N such that every positive integer can be
written as a sum of N k-th powers.
Note 4.7. Let g(k ) denote the minimum number of k-th powers needed to represent all positive integers.
Four-square theorem above says g(2) ≤ 4. Since 7 can’t be written as a sum of three squares, we have
g(2) = 4. Several values of g(k)’s are known, for$example,
we have g(3) = 9, g(4) = 19. For general k, let’s
%
k
3
− 1. Here b x c denotes the largest integer less than or
consider possibly the worst scenario. Let n = 2k
2k
$ %
3k
equal to x. Note that 2k
is the largest multiple of 2k less than or equal to 3k . Now let’s try to write n
2k
as a sum of k-th powers. Since
$ %
!
k
3k
k 3
k
−1 ≤ 2
− 1 < 3k ,
n=2
2k
2k
we can only use 2k and 1k to write n. To use the least number of k-th powers to represent n, we can write
$ %
!
3k
n=
− 1 2k + (2k − 1 )1k
2k
$ %
3k
k
which uses 2 + k − 2 k-th powers. This example shows
2
$ %
3k
k
g(k) ≥ 2 + k − 2
2
for all k. It is conjectured that the above is an equality for all k.
6
√
5. An example of a PID which is not a Euclidean domain
−19
and R = Z[θ ] = { a + bθ | a, b ∈ Z} ⊆ C be a subring. In this section, we will show
2
that R is a PID, but not a Euclidean domain. We consider the same norm map N : C → R≥0 defined by
N (z) = zz from the last section. Note that the norm of an element a + bθ ∈ R is given by
b
b√
(2a − b)2 + 19b2
N ( a + bθ ) = N a − −
−19 =
= a2 − ab + 5b2 ∈ Z≥0 .
2 2
4
Let θ =
−1 −
Lemma 5.1.
(1) For r ∈ R, r is a unit if and only if N (r ) = 1 if and only if r = ±1.
(2) There is no element r ∈ R such that N (r ) = 2 or 3.
(3) 2, 3 ∈ R are irreducible.
Proof. (1) Firstly, suppose r ∈ R× , then there exists s ∈ R such that rs = 1. Taking norms gives
N (r ) N (s) = N (1) = 1.
Since N (r ), N (s) are nonnegative integers, we have N (r ) = 1. Now suppose N (r ) = 1. Write r = a + bθ
for a, b ∈ Z, then we have N (r ) = a2 − ab + 5b2 = 1, thus (2a − b)2 + 19b2 = 4. This gives ( a, b) = (±1, 0),
and r = ±1. Lastly, if r = ±1, then r is a unit.
(2) Suppose N ( a + bθ ) = 2, then we have (2a − b)2 + 19b2 = 8. In this case, we have b = 0 because otherwise
we will have 19b2 > 8 and N ( a + bθ ) > 2. But there is no element a ∈ Z satisfying N ( a) = a2 = 2. Hence,
there is no element whose norm is 2. The same argument shows there is no element of norm 3.
(3) Suppose 2 = αβ for α, β ∈ R. Then, we have 4 = N (2) = N (α) N ( β). By (2), we have either N (α) = 1 or
N ( β) = 1. By (1), α ∈ R× or β ∈ R× , thus 2 is irreducible. The same argument works for 3.
Theorem 5.2. R is not a Euclidean domain.
Proof. Suppose R is a Euclidean domain. Then, there exists a Euclidean function δ : R \ {0} → Z≥0 such
that for a, b ∈ R with b 6= 0, there exist q, r ∈ R such that
a = bq + r
r = 0 or δr < δb.
We choose m ∈ R \ R× so that δ(m) is minimal. There exist q, r ∈ R such that 2 = qm + r and we have
r = 0 or δ(r ) < δ(m). By the choice of m, we have either r = 0 or r ∈ R× = {±1}. Since m | qm = 2 − r, we
have m | 1, 2 or 3. Since m ∈
/ R× , we have m | 2 or 3, thus N (m) | 4 or 9.
On the other hand, there exist q0 , r 0 ∈ R such that θ = q0 m + r 0 with r 0 = 0, ±1 by the same argument. Thus,
m | θ or θ ± 1, which implies N (m) | 5 = N (θ ) = N (θ + 1) or N (m) | 7 = N (θ − 1).
From two results above, we conclude N (m) = 1, but it contradicts the fact that m is not a unit.
Theorem 5.3. R is a PID.
Proof. Let I ⊆ R be an ideal and a ∈ I \ {0} be an element with √
minimal N ( a) > 0. We will show that
b
19
I = ( a). Choose b ∈ I and consider
∈ Q[θ ]. Since Im θ = −
, we can choose m ∈ Z such that
a
2
√
Im b + mθ ≤ 19 .
a
4
√
b
3
Case 1 Im
+ mθ <
.
a
2
b
1
Choose n ∈ Z such that Re
+ mθ + n ≤ , then we have
a
2
2 2
b
b
b
N
+ mθ + n = Re
+ mθ + n + Im
+ mθ + n < 1
a
a
a
by assumption. This shows N (b + a(mθ + n)) < N ( a). Since b + a(mθ + n) ∈ ( a, b) ⊆ I, it is zero by
minimality of N ( a). Thus, b = − a(mθ + n) ∈ ( a).
7
√
b
19
+ mθ ≤
.
Case 2
a
4
√
√
b
19
In this case, we have 3 ≤ Im 2
+ mθ
≤
and
a
2
√
√
√
√
√
3
3 3
19
b
−
= 3−
< 3−
≤ Im 2
+ mθ + θ ≤ 0.
2
2
2
a
b
1
Choose n ∈ Z such that Re 2
+ mθ + θ + n ≤ , then we have
a
2
b
+ mθ + θ + n < 1,
N 2
a
√
3
≤ Im
2
thus N (2(b + amθ ) + a(θ + n)) < N ( a). By similar argument in the previous case, we have 2(b + amθ ) +
a(θ + n)
aθ
a(θ + n)
n
a(θ + n) = 0. This shows
= −b − amθ ∈ I. If n is even, then
=
− a ∈ I and
2
2
2
a2
a
aθ
< N ( a). If n is odd, then
we have =
θ − 2a ∈ I. But this is a contradiction since 0 < N
2
2
2
a ( θ + 1)
a ( θ + 1)
a(θ + n) n − 1
a
θ + 1 − 2a ∈ I. This is also a contradiction.
=
−
a ∈ I and we have =
2
2
2
2
2
Hence, this case
√ doesn’t happen. √
19
b
3
Case 3 −
≤ Im
+ mθ ≤ −
.
4
a
2
b
You can apply the same argument in Case 2 for −
+ mθ .
a
6. Classes of rings
We have the following inclusion among classes of rings.
(1)
(2)
(3)
(4a)
(5)
(6)
(7)
6⊆
6⊆
{Fields} ( {ED} ( {PID} ( {UFD} ( {Integral Domains} ( {Commutative Rings} ( {Rings}
(
(
{ND}
(4b)
where ED (resp. ND) stands for Euclidean domain (resp. Noetherian domain.)
(1) ⇒ (2) ⇒ (3) ⇒ (4a) ⇒ (5) ⇒ (6) ⇒ (7) We already proved all of these implications.
(3) ⇒ (4b) ⇒ (5) Any ideal in a PID is generated by one element, in particular, finitely generated.
(2) 6⇒ (1) Z is a ED, but not a field.
√
(3) 6⇒ (2) We proved in last section that R = Z[θ ] for θ = −1− 2 −19 is a PID, but not a ED.
(4a), (4b) 6⇒ (3) Z[t] is a UFD and a Noetherian domain because Z is both (See Theorem 32.8 and
Theorem 33.3.) But the ideal (2, t) ⊆ Z[t] is not principal.
(4a) 6⇒ (4b) Let Z[t1 , t2 , . . . ] be the polynomial ring over Z with countably many variables {ti }in=1 . It
is a UFD by the following argument. For any f ∈ Z[t1 , t2 , . . . ], we can find i1 , i2 , . . . , ik ∈ Z+ such that
f ∈ Z[ti1 , ti2 , . . . , tik ] ⊆ Z[t1 , t2 , . . . ]. Since Z[ti1 , ti2 , . . . , tik ] is a UFD, f can be uniquely written as a product
of irreducible polynomials in Z[ti1 , ti2 , . . . , tik ]. Note that the existence and uniqueness of the factorization
still holds in a larger domain Z[t1 , t2 , . . . ]. On the other hand, Z[t1 , t2 , . . . ] is not noetherian because
(t1 ) ( (t1 , t2 ) ( · · ·√( (t1 , t2 , . . . , tn ) ( · · · is a strictly increasing chain of ideals in Z[t1 , t2 , . . . ].
(4b) 6⇒ (4a) Z[ −5] is noetherian √
because it is the homomorphic
image of a noetherian domain
√
2
∼
Z[t] under Z[t] Z[t]/(t + 5) = Z[ −5]. Since 2 ∈ Z[ −5] is irreducible but not prime (consider
8
√
√
√
√
2 · 3 = (1 + −5)(1 − −5) in Z[ −5]), Z[ −5] is not a UFD.
(6) 6⇒ (5) Z/4Z is commutative, but is not an integral domain.
(7) 6⇒ (6) M2 (Z) is a non-commutative ring.
7. Polynomial functions
Let R be a commutative ring and define Func( R, R) = { f : R → R} as the set of all (set-theoretic)
functions from R to R. Then Func( R, R) is a commutative ring with the following operations.
( f + g)( a) = f ( a) + g( a)
( f g)( a) = f ( a) g( a).
Let R[t] be the polynomial ring over R, and let’s consider the following map
Φ R : R[t] → Func( R, R)
defined by (Φ R ( f ))( a) = f ( a) for a ∈ R. Note that Φ R is a ring homomorphism.
Definition 7.1. A function h : R → R is called a polynomial function if h ∈ im Φ R .
Note 7.2. If R is not commutative, then Φ R is not a ring homomorphism. Choose a, b ∈ R such that ab 6= ba.
Let f (t) = t, g(t) = a ∈ R[t], then we have
(Φ R ( f )Φ R ( g))(b) = (Φ R ( f ))(b)(Φ R ( g))(b) = ba 6= ab = Φ R ( f g)(b).
This is basically because the variable t commutes with all elements in R[t].
Note 7.3. In general, Φ R is not injective nor surjective. Let R = Z/pZ, then we have a p = a for all
a ∈ Z/pZ. This shows ΦZ/pZ (t p ) = ΦZ/pZ (t). Let R = R, then the exponential function g : R → R
defined by g( x ) = e x is not in the image of ΦR .
Theorem 7.4. If R is an infinite domain, then Φ R is injective.
Proof. Let f , g ∈ R[t] and suppose Φ R ( f ) = Φ R ( g), i.e., f (r ) = g(r ) for all r ∈ R. Let h(t) = f (t) − g(t),
then h has infinitely many roots in R. Suppose h 6= 0. Since R is a domain, a non-zero polynomial h can
have at most deg h roots. This is a contradiction. Hence we have f = g and Φ R is injective.
Theorem 7.5. If R = Fq is a finite field of order q, then Φ R is surjective and ker Φ R = (tq − t). Therefore, we have
Fq [ t ] / ( t q − t ) ∼
= Func(Fq , Fq ).
In particular, any function h : Fq → Fq is a polynomial function.
Proof. We will first show that ker Φ R = (tq − t). Since |F×
q | is a (cyclic) multiplicative group of order
q
q
q − 1, we have aq−1 = a for all a ∈ F×
q . This shows b − b = 0 for all b ∈ Fq , thus Φ R ( t − t ) = 0
q
and (t − t) ⊆ ker Φ R . Suppose f (t) ∈ ker Φ R . By division algorithm, there exist q, r ∈ Fq [t] such that
f (t) = (tq − t)q(t) + r (t) and r = 0 or deg r < deg(tq − t) = q. Since f (t), tq − t ∈ ker Φ R , so is r (t). If
r 6= 0, then r can have at most deg r roots in Fq . Since Φ R (r ) = 0, r has q roots in R, thus r should be zero.
This shows tq − t | f and ker Φ R ⊆ (tq − t).
By the first isomorphism theorem, we have
Fq [ t ] / ( t q − t ) ∼
= im Φ R ⊆ Func(Fq , Fq ).
Note that for any f ∈ Fq [t], we can find r ∈ Fq [t] with deg r < q such that f = r in Fq [t]/(tq − t) by
division algorithm. On the other hand, if r1 , r2 ∈ Fq [t] are distinct polynomials of degree less than q,
then deg(r1 − r2 ) < q and r1 6= r2 in Fq [t]/(tq − t) because tq − t - r1 − r2 in Fq [t]. This shows that we
can choose the set of polynomials in Fq [t] with degree at most q − 1 as the complete representative set of
Fq [t]/(tq − t). Thus, we have
qq = |Fq [t]/(tq − t)| = |im Φ R | ≤ |Func(Fq , Fq )| = qq .
This shows im Φ R = Func(Fq , Fq ).
9
In general, we have
q
q
q
Fq [ t 1 , t 2 , . . . , t n ] / ( t 1 − t 1 , t 2 − t 2 , . . . , t n − t n ) ∼
= Func(Fnq , Fq ),
and you can also use counting argument to prove this.
The following theorem gives a way to find the polynomial which defines a given polynomial function.
Theorem 7.6 (Lagrange interpolation). Let F be a field and consider xi , yi ∈ F for i = 0, 1, . . . , k with distinct
xi ’s. In this case, we can find a polynomial P(t) ∈ F [t] of degree at most k satisfying P( xi ) = yi for all i.
Proof. We first define
pi ( t ) =
∏
j 6 =i
0≤ j ≤ k
t − xj
∈ F [t]
xi − x j
for each i. Since xi ’s are distinct, it is well-defined. Then we have
(
1 if j = i
pi ( x j ) = δij =
0 if j 6= i.
Now we let P(t) =
∑
yi pi (t) ∈ F [t], then we have P( xi ) = yi for all i.
0≤ i ≤ k
Note 7.7. Consider k + 1 points on the real xy-plane R2 with distinct x-coordinates. The theorem above
tells you that there exists a polynomial in R[t] with degree at most k whose graph on the xy-plane passes
through all k + 1 points.
Example 7.8. Note that any function f : Z/5Z → Z/5Z is a polynomial function by surjectivity of ΦZ/5Z .
Consider f : Z/5Z → Z/5Z defined by f (0) = 3, f (1) = 0, f (2) = 1, f (3) = 1, f (4) = 0. We can find a
polynomial P[t] ∈ (Z/5Z)[t] defining f . We have
p0 ( t ) =
∏
j 6 =i
0≤ j ≤4
t−j
= −t4 + 1 ∈ (Z/5Z)[t],
i−j
and similarly we can compute p2 (t) = −(t3 − t)(t + 2), p3 (t) = −(t3 − t)(t − 2). Hence we get
P(t) = 3p0 (t) + p2 (t) + p3 (t) = 2t2 + 3.
We can actually check that P( a) = f ( a) for all a ∈ Z/5Z. Note that this P(t) is not the only polynomial
which defines f . For any h(t) ∈ (Z/5Z)[t], we have ΦZ/5Z ( P(t) + (t5 − t)h(t)) = f .
8. Exact sequences
Let R be a ring. All the objects in this section are R-modules and all arrows are R-homomorphisms.
f
g
Definition 8.1. We say that A −
→ B −
→ C is exact at B if im f = ker g. A sequence of R-modules and
R-homomorphisms is called exact if it is exact at all places.
f
Note that for any R-homomorphism A −
→ B, we have the following exact sequence.
f
0 → ker f → A −
→ B → coker f → 0.
f
g
Theorem 8.2. Suppose 0 → A −
→B−
→ C → 0 is a short exact sequence, i.e., we have ker f = 0, im f = ker g and
im g = C. Let N be an R-module, then the following holds.
10
(1) HomR ( N, −) : R-Mod → R-Mod is left exact, i.e.,
f∗
g∗
0 → HomR ( N, A) −
→ HomR ( N, B) −→ HomR ( N, C )
is exact. N is called projective if g∗ is surjective. In other words, HomR ( N, −) preserves short exact
sequences.
(2) HomR (−, N ) : R-Mod → R-Mod is left exact, i.e.,
g∗
f∗
0 → HomR (C, N ) −→ HomR ( B, N ) −→ HomR ( A, N )
is exact. N is called injective if f ∗ is surjective. In other words, HomR (−, N ) preserves short exact sequences.
(3) − ⊗ R N : R-Mod → R-Mod is right exact, i.e.,
f ⊗id N
g⊗id N
A ⊗ R N −−−→ B ⊗ R N −−−→ C ⊗ R N → 0
is exact. N is called flat if f ⊗ id N is injective. In other words, − ⊗ R N preserves short exact sequences.
Example 8.3.
(1) Z/2Z is not a projective Z-module. Consider the projection π : Z Z/2Z and
π∗ : HomZ (Z/2Z, Z) → HomZ (Z/2Z, Z/2Z).
If f : Z/2Z → Z is a Z-module homomorphism (i.e., an abelian group homomorphism), then f is
determined by f (1). Since 0 = f (0) = f (2 · 1) = 2 f (1) in Z, we have f (1) = 0, thus f = 0. This
shows HomZ (Z/2Z, Z) = 0. Since HomZ (Z/2Z, Z/2Z) ∼
= Z/2Z 6= 0, π∗ is not surjective.
(2) Z is not an injective Z-module. Consider the inclusion i : Z ,→ Q and
i∗ : HomZ (Q, Z) → HomZ (Z, Z).
1
= f (1) for all n ∈ Z+ . This
n
means f (1) = 0, thus f = 0. This shows HomZ (Q, Z) = 0 and i∗ is not surjective.
(3) Z/2Z is not a flat Z-module. Consider the injection f : Z ,→ Z defined by f (n) = 2n and
If f : Q → Z is a Z-module homomorphism, then we have n | n f
f ⊗ idZ/2Z : Z ⊗Z Z/2Z → Z ⊗Z Z/2Z.
For any a ⊗ b ∈ Z ⊗Z Z/2Z, we have f ⊗ idZ/2Z ( a ⊗ b) = (2a) ⊗ b = 2( a ⊗ b) = a ⊗ (2b) = a ⊗ 0 =
0 by bilinearity. Since f ⊗ idZ/2Z maps all generators of Z ⊗Z Z/2Z to 0, we have f ⊗ idZ/2Z = 0.
However, we have Z ⊗Z Z/2Z ∼
= Z/2Z, thus f ⊗ idZ/2Z is not injective.
Note 8.4. Any free R-module is projective (this was one of the homework problems.) Any projective
R-module is flat. You can prove that R-modules Mi are projective (resp. flat) for all i if and only if ⊕ Mi is
projective (resp. flat). Since R is R-flat (for any R-module N, we have R ⊗ R N ∼
= N), any free module is
flat. Since a projective module is a direct summand of a free module, it is flat as well. Z/2Z is a projective
Z/6Z-module, but not a free Z/6Z-module. Q is a flat Z-module, but not a projective Z-module.
Note 8.5. The properties above depend on the base ring. For example, Z/2Z is Z/2Z-free, but is not
Z/6Z-free. Q is Q-projective since it is Q-free, but is not Z-projective. Z/2Z is Z/2Z-flat since it is
Z/2Z-free, but is not Z-flat.
Theorem 8.6 (Snake lemma). Suppose we have the following commutative (i.e., f 0 α = β f and g0 β = γg) diagram
of R-modules with exact rows.
f
A
B
f0
A0
B0
C
0
γ
β
α
0
g
g0
C0
Then, we have the following exact sequence of R-modules.
i
fe
ge
f0
g0
idC
0 → ker f −
→ ker α −
→ ker β −
→ ker γ −
→ coker α −
→ coker β −
→ coker γ −→ coker g0 → 0.
δ
The homomorphisms are defined as follows. i is an inclusion and fe = f |ker α , ge = g|ker β are restrictions of f , g.
f 0 , g0 , idC are the quotient maps defined as f 0 ( a0 ) = f 0 ( a0 ) for all a0 ∈ A0 and similar for g0 and idC . To define
11
δ : ker γ → coker α, given c ∈ ker γ, choose b ∈ B such that g(b) = c. By commutativity of the diagram, we have
0 = γ(c) = γg(b) = g0 β(b). Thus β(b) ∈ ker g0 = im f 0 since the rows are exact. Choose a0 ∈ A0 such that
f 0 ( a0 ) = β(b) and define δ(c) = a0 ∈ A0 / im α = coker α.
Proof. We need to check that all the maps are well-defined and the sequence is exact at all places. Let’s
prove that δ is well-defined. Note that the choice of a0 in the definition is unique because f 0 is injective.
Suppose we choose another element b0 ∈ B such that g(b0 ) = c. Let a00 ∈ A be the element satisfying
f 0 ( a00 ) = β(b0 ). Since b − b0 ∈ ker g = im f , there is a ∈ A such that f ( a) = b − b0 . Thus, f 0 ( a0 − a00 ) =
β(b − b0 ) = β f ( a) = f 0 α( a). We know that f 0 is injective, hence a0 − a00 = α( a) ∈ im α, i.e., a0 = a00 in
A0 / im α = coker α. Proving that the other maps are well-defined is easier, and left as an exercise.
Let’s prove the exactness at ker γ. Suppose first c ∈ im ge. Choose b ∈ ker β such that ge(b) = c, then
δ(c) = a0 for some a0 ∈ A0 satisfying f 0 ( a0 ) = β(b) = 0. Since f 0 is injective, we have a0 = 0 thus δ(c) = 0
and c ∈ ker δ. On the other hand, suppose c ∈ ker δ. Then we have a0 ∈ im α and b ∈ B such that
f 0 ( a0 ) = β(b) and g(b) = c by definition of δ. Choose a ∈ A with α( a) = a0 , then we have β(b − f ( a)) =
β(b) − f 0 α( a) = β(b) − f 0 ( a0 ) = 0. Hence, b − f ( a) ∈ ker b and c = g(b) = ge(b − f ( a)) ∈ im ge. Other
exactness can be proved similarly, and left as an exercise. You need to check all the details at least once (or
more if you teach this in the future) in your life!
The name ”Snake” came from the following picture.
0
ker f
f
A
0
γ
B0
f0
C0
g0
coker β
coker α
C
β
δ
A0
g
B
α
0
ker γ
ker β
ker α
coker g0
coker γ
0
Note that the vertical lines are all exact.
f
g
Corollary 8.7 (Kernel-cokernel lemma). Let A −
→B−
→ C be a sequence of R-homomorphisms (not necessarily
exact). Then we have the following exact sequence.
0 → ker f → ker g f → ker g → coker f → coker g f → coker g → 0.
Proof. We apply Snake lemma to the following commutative diagram.
A
f
C
coker f
idC
0
g
gf
0
B
C
0
12
9.
M
Z and
∏Z
Definition 9.1. We define the following Z-modules.
∞
∏ Z = {(ai )i∞=0 = (a0 , a1 , . . . ) | ai ∈ Z}
i =0
∞
M
Z = {( ai )i∞=0 | ai ∈ Z, ai = 0 for all but finitely many i’s} ⊆
∞
∏ Z.
i =0
i =0
∼ Z[[t]]
Z∼
= Z[t] and ∏ Z =
as Z-modules via ( ai ) 7→ ∑ ai t . (They are not isomorphic as rings if you consider component-wise
multiplication on ∏ Z.)
We will use the notations
M
Z,
i
∏ Z for simplicity.
Note that we have
Lemma 9.2. Let e j = (δij )i∞=0 = (0, 0, . . . , 0, 1, 0, . . . ) ∈
component and 0 elsewhere.
M
Z⊆
M
∏ Z be the element which has 1 on the j-th
(1) Let f : ∏ Z → Z be a Z-module homomorphism. Then f (e j ) = 0 for all but finitely many j’s.
(2) If f (e j ) = 0 for all j, then f = 0.
Proof. (1) Given f , we choose integers nk for each k ∈ Z+ satisfying 0 < n1 < n2 < · · · and
k −1
ni
∑ 2 f ( e i ) < 2n k −1
i =0
for all k. Consider x = (2ni )i∞=0 ∈ ∏ Z, then we have
!
k −1
∞
ni
ni
| f ( x )| = f ∑ 2 ei + ∑ 2 ei i =0
i =k
!
k −1
∞
ni
nk
ni − n k
= ∑ 2 f ( ei ) + 2 f ∑ 2
ei i =0
i =k
! k −1
∞
nk ni − n k
ni
nk ≥ 2 f ∑ 2
ei − ∑ 2 f ( ei ) > 2 f
i =0
i=k
!
∞
for all k ≥ 1. Let bk = f
∑ 2n i − n k e i
∞
∑2
ni − n k
i =k
!
e i − 2n k −1
. If bk 6= 0 for infinitely many k, then | f ( x )| > 2nk |bk | − 2nk −1 ≥ 2nk −1
i =k
for such k’s. This is a contradiction since f ( x ) is an integer and lim nk = ∞. Hence, bk = 0 for sufficiently
k→∞
large k. Note that we have
nk
∞
2 bk = f
∑2
∞
!
ni
ei
i =k
= f
∑
nk
2 ek +
!
ni
2 ei
= 2 n k f ( e k ) + 2 n k +1 bk + 1
i = k +1
for all k. This shows f (ek ) = 0 for sufficiently large k as well.
(2) Suppose f (e j ) = 0 for all j. Given x = ( xi )i∞=0 ∈ ∏ Z, choose integers ai , bi ∈ Z satisfying ai 2i + bi 3i = xi
for all i. This is possible because 2i and 3i are relatively prime. Now we have
!
!
f (( ai 2i )i∞=0 ) =
k −1
∞
i =0
i =k
∑ a i 2i f ( e i ) + f ∑ a i 2i e i
= 2k f
∞
∑ a i 2i − k e i
i=k
for all k ≥ 1. Since f (( ai 2i )i∞=0 ) is an integer and 2k | f (( ai 2i )i∞=0 ) for all k ≥ 1, we have f (( ai 2i )i∞=0 ) = 0.
Similarly, we have f ((bi 3i )i∞=0 ) = 0 and f ( x ) = f (( ai 2i )i∞=0 ) + f ((bi 3i )i∞=0 ) = 0.
Note 9.3. Note that we can’t use f (∑ ai ei ) = ∑ ai f (ei ) to prove (2) when infinitely many ai ’s are nonzero.
We only have the identity for finite sums by using the fact that f is a Z-module homomorphism.
13
(1)
Theorem 9.4.
(2)
M
Z is Z-free with basis E = {ei }i∞=0 .
∏ Z is not Z-free. (This is an example of a torsion-free but not free module over a PID.)
(3) We have
M
Z
∗
= HomZ
M
∼
Z, Z =
∏ Z and ∏ Z
∗
∼
=
M
Z.
Z. We call E the standard basis.
M
(2) Suppose ∏ Z is Z-free with basis B. Suppose B is countable, then we have span B ∼
Z. We
=
Proof. (1) It is easy to see that E is linearly independent and spans
can think of
M
Z=
[
Zn , thus
M
M
Z is countable as being a countable union of countable sets. Since
n
∏ Z is uncountable, this is a contradiction. We can find a countable subset B0 ⊆ B which spans all e j ’s
because each e j is a finite linear combination of elements in B. Choose b ∈ B \ B0 and define a projection
πb : ∏ Z → Z as follows. For each x ∈ ∏ Z, we can write it as a unique linear combination x = ∑ xb b
b∈ B
where xb = 0 for all but finitely many b’s. We define πb ( x ) = xb . It is easy to see that πb is a Z-module
homomorphism. Also we have e j ∈ span B0 ⊆ ker πb for all j. By Lemma (2) above, πb = 0. This is a
contradiction because we have πb (b) = 1.
(3) We first define
M
Φ
HomZ
Z, Z
∏Z
( f (ei )i∞=0 ).
f
Clearly, Φ
is a well-defined Z-module homomorphism. If Φ( f ) = 0, then f (ei ) = 0 for all i. Since {ei } is a
M
basis of
Z, we have f = 0. This shows Φ is injective.
M
For each y = (yi )i∞=0 ∈ ∏ Z, define f y :
Z → Z by f y ∑ ai ei = ∑ ai yi . Then, f y is a well-defined
(only finitely many ai ’s are nonzero) Z-module homomorphism and Φ( f y ) = y. This shows Φ is surjective.
We also define
M
Ψ
HomZ ∏ Z, Z
Z
( g(ei )i∞=0 ).
g
The above map Ψ is a well-defined Z-module homomorphism by Lemma (1) above. If Ψ( g) = 0, then we
have g(ei ) = 0 for all i. By Lemma
(2), we have g = 0, thus Ψ is injective.
M
∞
For each z = (zi )i=0 ∈
Z, define gz : ∏ Z → Z by gz (∑ bi ei ) = ∑ bi zi . Then gz is a welldefined (only finitely many zi ’s are nonzero) Z-module homomorphism and Ψ( gz ) = z. This shows Ψ is
surjective.
Note 9.5. The first part of (3) in the above theorem holds in general. Let R be a ring and Mi , N be R-modules
for i ∈ I. Then the same proof shows
!
M
HomR
Mi , N ∼
= ∏ HomR ( Mi , N ),
i∈ I
i∈ I
M
∗
∼ ∏( M∗ ) when N = R above. The second part of the theorem doesn’t hold
thus in particular
Mi =
i
in general, and it is basically from the Lemma which uses some properties of Z.
Theorem 9.6. Let R = EndZ ∏ Z , then we have Rm ∼
= Rn as R-modules for all m, n ∈ Z+ . In other words, the
rank of a free R-module is not well-defined.
Proof. Consider f 1 , f 2 , g1 , g2 ∈ R defined by
f 1 ( a0 , a1 , a2 , . . . ) = ( a0 , a2 , a4 , a6 , . . . )
f 2 ( a0 , a1 , a2 , . . . ) = ( a1 , a3 , a5 , a7 , . . . )
g1 ( a0 , a1 , a2 , . . . ) = ( a0 , 0, a1 , 0, a2 , 0, . . . )
g2 ( a0 , a1 , a2 , . . . ) = (0, a0 , 0, a1 , 0, a2 , . . . ).
14
It’s easy to see that f 1 g1 = id R = f 2 g2 , f 1 g2 = 0 = f 2 g1 , and g1 f 1 + g2 f 2 = id R . We will show that { f 1 , f 2 }
is an R-basis of R.
For any h ∈ R, we have
h = h ◦ id R = h( g1 f 1 + g2 f 2 ) = (hg1 ) f 1 + (hg2 ) f 2 ∈ span( f 1 , f 2 ).
If we have h1 f 1 + h2 f 2 = 0 for some h1 , h2 ∈ R, then
0 = 0 ◦ g1 = ( h 1 f 1 + h 2 f 2 ) g1 = h 1 ( f 1 g1 ) + h 2 ( f 2 g1 ) = h 1
shows h1 = 0, and similarly h2 = 0. Hence, { f 1 , f 2 } is an R-basis of R and we have
R = R f1 ⊕ R f2 ∼
= R2 .
By induction, we have Rn ∼
= Rm for all n, m ∈ Z+ .
10. Composition series
Definition 10.1. Let R be a ring and M be a nonzero R-module. Recall that M is called simple if it has no
nonzero proper submodules. Consider the following series of submodules of M.
0 = M0 ⊆ M1 ⊆ · · · ⊆ Mn = M
(∗)
(1) (∗) is called proper if Mi−1 ( Mi for all i = 1, 2, . . . , n. In this case, we say (∗) has length n.
(2) (∗) is called a composition series if it is proper and Mi /Mi−1 is simple for all i = 1, 2, . . . , n.
Consider another series of submodules of M.
0 = M00 ⊆ M10 ⊆ · · · ⊆ Mn0 0 = M
(∗∗)
(3) (∗∗) is called a refinement of (∗) if there is an increasing function f : {0, 1, . . . , n} → {0, 1, . . . , n0 }
such that f (0) = 0, f (n) = n0 and Mi = M0f (i) . It is called a proper refinement if n < n0 .
(4) (∗) and (∗∗) are called equivalent if there is a bijection between factors (up to isomorphism)
{ Mi /Mi−1 | i = 1, 2, . . . , n} ←→ { M0j /M0j−1 | j = 1, 2, . . . , n0 = n}.
Example 10.2. Consider R = Z and M = Z.
(1) 0 ⊆ 6Z ⊆ 2Z ⊆ Z is a proper refinement of 0 ⊆ 6Z ⊆ Z.
(2) 0 ⊆ 6Z ⊆ 2Z ⊆ Z and 0 ⊆ 6Z ⊆ 3Z ⊆ Z are equivalent.
(3) Z does not have a composition series because it has no nonzero simple submodule. (Any nonzero
submodule of Z is of the form nZ for some n ≥ 1.)
Theorem 10.3. Let R be a ring and M be a nonzero R-module.
(1) A composition series of M doesn’t have a proper refinement.
(2) (Schreier) Any two proper series of submodules of M have equivalent refinements.
(3) (Jordan-Hölder) If M has a composition series, then all composition series of M are equivalent.
Proof. (1) Suppose M1 , M2 are submodules of M satisfying M1 ⊆ M2 ⊆ M. By Correspondence Principle,
there is a bijection between submodules M0 with M1 ⊆ M0 ⊆ M2 and submodules of M2 /M1 defined by
M0 7→ M0 /M1 . If M1 /M2 is simple, there’s no submodule M0 satisfying M1 ( M0 ( M2 .
(2) See [Elm, Theorem 16.10.] in the case of groups. Same argument works for modules.
(3) By (1) and (2).
Theorem 10.4. Let M be an R-module. Then, M has a composition series if and only if M satisfies ACC and DCC
on submodules (i.e., M is both a Noetherian R-module and an Artinian R-module.)
Proof. (⇒) Suppose
0 = M0 ⊆ M1 ⊆ · · · ⊆ Mn = M (∗)
is a composition series of M. If M does not satisfy ACC (ascending chain condition), then we can construct
another proper series of submodules of length n + 1
0 = M00 ⊆ M10 ⊆ · · · ⊆ Mn0 +1 = M
15
(∗∗)
(you can take a strictly increasing sequence of submodules and add 0 and M to it.) By Schreier’s theorem,
(∗) and (∗∗) have equivalent refinements which have length at least n + 1. This is a contradiction because
(∗) has length n, and it does not have a proper refinement. This shows M satisfies ACC on submodules.
The same proof shows M satisfies DCC on submodules as well.
(⇐) Suppose M satisfies ACC and DCC on submodules. For any submodule 0 6= N ⊆ M, the set of proper
submodules of N is nonempty. By ACC (ACC on M implies ACC on N), we can choose a maximal proper
submodule NM ( N. Let C be the set of all submodules of M and we define f : C → C by
(
NM if N 6= 0
f (N) =
0
if N = 0
for all N ⊆ M. Note that N/ f ( N ) is simple if N 6= 0. By DCC,
M ⊇ f ( M) ⊇ f ( f ( M)) ⊇ · · · ⊇ f n ( M) ⊇ . . .
has to stabilize. This means there is N ≥ 1 such that f N ( M ) = 0. Let Mi = f i ( M), then
M ⊇ M1 ⊇ · · · ⊇ M N = 0
is a composition series of M.
Example 10.5. Let R = Mn ( D ) for a division ring D. Note that REkk is a simple left R-module for 1 ≤ k ≤ n.
(Rkk is the set of matrices whose j-th columns are zero for all j 6= k.) Let Mi be the left R-module defined
by Mi = RE11 + · · · + REii , then we have
∼ REii
Mi /Mi−1 =
for all 1 ≤ i ≤ n. Thus,
0 =: M0 ⊆ M1 ⊆ · · · ⊆ Mn = R
is a composition series of R with left R-modules. By the theorem above, this shows that R is a left
Noetherian, left Artinian ring. The same argument using the right R-modules Ni = E11 R + · · · + Eii R
shows R is also a right Noetherian, right Artinian ring. Thus, R = Mn ( D ) is a Noetherian, Artinian ring.
11. Take-home midterm
Problem 3. Let R be a noetherian commutative ring and I be an ideal of R. Show that there are only a
finite number of minimal prime ideals containing I.
Proof. (Solution 1) We will prove that every proper radical ideal is a finite intersection of prime ideals. Let
√
S = { I ( R | I is an ideal and I is not a finite intersection of prime ideals}
p
and suppose S 6= ∅. Since R p
is noetherian,p
there is a maximal
∈ S. Since J ispnot prime,
p element J p
p there
are a, b ∈ R such that a, b ∈
/ J but ab
∈
J.
Let
J
=
J
+
(
a
)
,
J
=
J
+
(
b
)
,
then
J
and
J2 are
1
p
p
p1
p 2p
finite intersection of primes. We have J ⊆ J1 ∩ J2 . If x ∈ J1 ∩ J2 , then we have
x n2 = j2 + r2 b
x n1 = j1 + r1 a,
for some n1 , n2 ∈ Z, j1 , j2 ∈
p
J and r1 , r2 ∈ R. This means
p
x n1 +n2 = ( j1 + r1 a)( j2 + r2 b) ∈ J,
p
p
p
p
thus x ∈ J. Hence, J = J1 ∩ J2 is also a finite intersection of primes, which is a contradiction.
Suppose
J is a minimal prime containing I. By above argument, there are primes P1 , . . . , Pr such that
√
I = P1 ∩ · · · ∩ Pr . Since
√
\
P1 · · · Pr ⊆ P1 ∩ · · · ∩ Pr = I =
P ⊆ J,
I ⊆ P⊆ R
P : prime
16
we have Pi ⊆ J for some i. By minimality, we have Pi = J. Thus, there are only finitely many minimal
primes containing I.
(Solution 2) Let
S = { I ⊆ R | I is an ideal and there are infinitely many minimal primes containing I }
and suppose S 6= ∅. Since R is noetherian, there is a maximal element J ∈ S. Since J is not prime (if it is, J
would be the only minimal prime containing itself), there are ideals J1 , J2 such that J ( J1 , J2 and J1 J2 ⊆ J.
Let P be a minimal prime containing J, then we have J1 J2 ⊆ J ⊆ P, thus J1 ⊆ P or J2 ⊆ P. So P is either a
minimal prime containing J1 or a minimal prime containing J2 . This shows at least one of J1 and J2 has
infinitely many minimal primes containing it, which contradicts the maximality of J.
Let R be a noetherian integral domain. If f , g ∈ R, we define
Problem 4.
( f ) : ( g) = {h ∈ R | hg ∈ ( f )}.
(c) Show that if ( f ) : ( g) is principal for all f , g ∈ R, then R is a UFD.
Proof. (Existence) We will use noetherian induction. Let
S = {( a) | 0 ( ( a) ( R, a ∈ R is not a product of finitely many irreducibles in R}
and suppose S 6= ∅. Since R is noetherian, there is a maximal element (m) ∈ S. Since m is not irreducible,
there are nonunits a, b ∈ R such that m = ab. This means (m) ( ( a), (m) ( (b) and a, b are products of
finitely many irreducibles in R, thus so is m. This is a contradiction.
(Uniqueness) It is enough to show that every irreducible is prime. Let p ∈ R be irreducible and suppose
p | ab for some a, b ∈ R. Let pz = ab and
( a) : (z) = {h ∈ R | hz ∈ ( a)} = (c)
for some c ∈ R. Since p ∈ (( a) : (z)) = (c), there is y ∈ R such that p = cy. By assumption, one of c and y is
a unit. If c ∈ R× , then 1 ∈ ( a) : (z). This means a | z, thus p | b. If y ∈ R× , then a ∈ (( a) : (z)) = (c) = ( p),
thus p | a.
Problem 6.
Let R be a commutative ring and M be an R-module. Suppose
R = ( f1 , f2 , . . . , fk )
for some f 1 , f 2 , . . . , f k ∈ R.
(a) For any positive integer N, show that there are ri ∈ R such that
k
∑ ri fiN = 1.
i =1
(b) Define U ( f ) = { f , f , f , . . . } for f ∈ R. Let Mi = U ( f i )−1 M and Mij = U ( f i f j )−1 M for i, j = 1, 2, . . . , k.
Find a natural sequence
M
M
0→M→
Mi →
Mij
2
3
i
i,j
and show this sequence is exact.
Proof. (a) Since 1 ∈ R = ( f 1 , . . . , f k ), we can write 1 =
∑ ci fi for some ci ∈ R. For N ≥ 1, we have
i
1 = 1 Nk =
∑ ci f i
! Nk
∈ ( f 1N , . . . , f kN ).
i
(b) We define
f
0→M−
→
M
g M
Mi −
→
i
i,j
17
Mij
by f (m) =
fi m
fi
mi
n
fi i
and g
i
n
!!
f j i mi
=
( f i f j ) ni
i
nj
f mj
− i nj
( fi f j )
!
. We can easily check g is well-defined.
i,j
fi m
= 0 in Mi for all i, thus there is ti ∈ Z≥1 such that
fi
ti
f i m = 0 for all i. Let T = max{ti }, then f iT m = 0 for all i. By (a), this means m = 1 · m = 0.
ker f = 0
Suppose f (m) = 0. This means
im f ⊆ ker g
ker g ⊆ im f
i
Clear.
mi
n
fi i
Suppose g
!!
= 0. For simplicity, let N = max{ni } and si = mi f iN −ni , then
i
!!
!
N
N
f j si − f i s j
si
=
= 0, there is tij ∈ Z≥1 such that
( fi f j )N
f iN
i
i,j
i
mi
si
in Mi for all i. Since g
ni =
f iN
fi
( f i f j )tij ( f jN si − f iN s j ) = 0 for all i, j. Let T = max{tij }, then we have
i,j
f iT f jT + N si = f jT f iT + N s j
for all i, j. By (a), we can choose a j ∈ R satisfying 1 =
∑ a j f jT+ N . We have
j
∑ a j f jT+ N
f iT si =
!
j
for all i. Define a =
f iT si = f iT + N
∑ a j f jT s j
!
j
∑ a j f jT s j , then we have
j
mi
n
fi i
!
=
i
si
f iN
!
=
i
f iT si
f iT + N
!
=
i
f iT + N a
!
f iT + N
=
i
fi a
fi
= f ( a ).
i
Problem 8. Let R be a noetherian ring and I be an ideal of R. Suppose that I is maximal with respect to
the property that R/I has no composition series. Show that I is prime.
Proof. Suppose I is not prime. We have a, b ∈ R such that a, b ∈
/ I and ab ∈ I. We have
I ( I + ( a)
and
I ( I + (b) ⊆ ( I : ( a)).
Consider the exact sequence
0→
Define f : R →
I + ( a)
R
R
→ →
→ 0.
I
I
I + ( a)
( a)
by f (r ) = ra. Then clearly f is surjective, and ker f = ( I : ( a)). This shows
I ∩ ( a)
R ∼ ( a) ∼ I + ( a)
=
=
I : ( a)
I ∩ ( a)
I
by first and second isomorphism theorem. By maximality of I, both R/( I : ( a)) and R/( I + ( a)) have
composition series. Thus, both satisfy ACC and DCC, and so does R/I. But this implies R/I has a
composition series, which is a contradiction. (See the previous section.)
18
References
[Con]
[Elm]
[H]
[Jac45]
[L]
K. Conrad, Expository Papers, http://www.math.uconn.edu/~kconrad/blurbs/.
R. Elman, Lectures on Abstract Algebra, http://www.math.ucla.edu/~rse/.
T. Hungerford, Algebra, Springer, 1974.
N. Jacobson, Structure Theory for Algebraic Algebras of Bounded Degree, Annals of Mathematics 46
(1945), 695-707.
S. Lang, Algebra (Revised Third Edition), Springer, 2002.
19