IMMERSE 2008: Assignment 1
1.1)
Let R be a ring. Prove
(a) 0a = a0 = 0 for all a ∈ R.
Proof:
Let a ∈ R. Then
0a = (0 + 0)a = 0a + 0a
From the additive inverse law we get 0a + (−(0a)) = 0a + 0a + (−(0a)) which
becomes 0 = 0a. Similarly, if we start with a0 and follow the same type of
argument we get a0 = 0. Thus 0a = a0 = 0.
(b) (−a)b = a(−b) = −(ab) for all a, b ∈ R.
Proof:
Let a, b ∈ R.
(−a)b = (−a)b + 0 = (−a)b + (ab + (−(ab))) = ((−a)b + (ab)) + (−(ab)) =
b((−a) + a) + (−(ab)) = b0 + (−(ab)) = −(ab)
Similarly, if we follow the same proof but add (a(−b) + (−(a(−b))) then we get
(−a)b = a(−b). Thus (−a)b = a(−b) = −(ab).
(c) (−a)(−b) = ab for all a, b ∈ R.
Proof: Let a, b ∈ R.
(−a)(−b) = (−a)(−b) + 0 = (−a)(−b) + (ab + (−(ab)))
By 1.b, −(ab) = (−a)b which means we have
(−a)(−b) + ab + (−a)b = (−a)(−b) + (−a)b + ab = (−a)((−b) + b) + ab =
(−a)0 + ab = ab.
Thus (−a)(−b) = ab.
(d) If R has an identity 1, then the identity is unique and −a = (−1)a.
Proof: Let x, y ∈ R. Suppose xa = ya where a 6= 0 and x and y are two
identities.
xa + (−ya) = 0
(x + (−y))a = 0
x + −y = 0
x=y
Thus the identity is unique and by hypothesis it is equal to 1.
Then -a=(-a)1 and by 1b and the commutative property (-a)1=a(-1)=(-1)a. 1.2)
Problems involving zerodivisors:
(a) Prove that a unit element of a ring cannot be a zerodivisor.
Solution. Let a be a unit element and a zero divisor. Then by definition, a 6= 0
and for b ∈ R and b 6= 0, ab = 0. Since a is a unit, a−1 exists. So a−1 (ab) =
(a−1 a)b = 0 · b, so b = 0. But this contradicts our assumption that b is non-zero.
Therefore, a unit element cannot be a zero divisor.
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(b) Let a and b be elements of a ring whose product ab is a zerodivisor. Show that
either a or b is a zerodivisor.
Solution. Let a,b ∈ R and suppose ab is a zero divisor. Then ab 6= 0 and for
c ∈ R and c 6= 0, (ab)c = 0, and so a(bc) = 0. Note that a 6= 0 and b 6= 0
because ab 6= 0. So consider the cases when bc = 0 or when bc 6= 0. If bc 6= 0,
then since a 6= 0, a is a zero divisor. On the other hand, if bc = 0, then since
b 6= 0 and c 6= 0, this implies b is a zero-divisor. Therefore, either a or b is a
zero-divisor.
(c) Is the sum of two zerodivisors necessarily a zerodivisor? If so, give a prove. If
not, give a counterexample.
Solution. There are cases where the sum of two zerodivisors is a zerodivisor. In
Z/8, for instance, 4 and 6 are zerodivisors since 4 · 6 = 0 (mod 8) and 4 + 6 = 10
and 4 · 10 = 0 (mod 8). The claim is not true in general, however. Consider
Z/6 where 2 and 3 are zero-divisors since 2 · 3 = 0 (mod 6), but 2 + 3 = 5 is not
a zero-divisor in Z/6.
1.3)
Let R be an integral domain. Determine the units of R[x].
Claim: The units of R[x] are the units of R.
Proof: Let R be an integral domain. We know that units of R[x] are elements u
for which there exists an r such that u · r = 1. Furthermore, we know u and r are of
the form
u = a0 x0 + a1 x1 + ... + an xn
0
1
m
r = b0 x + b1 x + ...bm x .
(1)
(2)
It is apparent, because of the form of our elements, that an and bm can both be
defined as nonzero, because if the last term of u or r were zero, we could simply
decide our last term were the term before it. Also, if u or r were simply the zero
polynomial, the product u · r 6= 1.
The last term, or the term of highest power in u·r is (an xn )(bm xm ) = an bm xm+n .
Because of how we constructed u and r, either m + n = 0 or m + n > 0. If m + n > r,
then xm+n 6= 1, so for u · r = 1, an bm = 0. But we know neither an or bm is equal
to zero, and because R is an integral domain, there are no zero divisors, and their
product cannot be zero. Thus m + n = 0, so that m = 0 and x = 0, and we are left
with u = a0 x0 and r = b0 x0 . The product u · r = a0 b0 = 1, and since a0 and b0 are
both elements in R and their product is 1, they must be units in R. Hence units in
R[x] are units in R.
1.4)
Let R be an integral domain. Determine the units of R[[x]].
1.5)
Let A be the ring of all functions from [0, 1] to R.
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(a) What are the units of A?
Claim: The units of A are all functions from [0, 1] to R which have entirely
nonzero ranges.
Proof: Since these functions are real-valued, it follows that functions whose
ranges are all units in the real numbers are units in A. The units of R are all
nonzero real numbers, thus the units of A are those with nonzero ranges.
(b) Prove that if f is not a unit and not zero, then f is a zero divisor.
Proof: Assume f is not a unit and is not zero. then define g to be the function
such that where f is not zero, g is zero and where f is zero, g is not zero. Then
g is also not zero, and the multiplication f · g = 0, so f is a zero divisor.
1.6)
Let A be the ring of all continuous functions from [0, 1] to R.
(a) What are the units of A?
Proof: A∗ = {f (x) ∈ A|f (x) 6= 0 for all x ∈ [0, 1]}.
Let f (x) ∈ A∗ , with f (x) 6= 0. Thus there exists some g(x) ∈ A such that
f (x)g(x) = 1, for all x ∈ [0, 1]
That is, f (x) 6= 0 for all x ∈ [0, 1]. Thus
A∗ ⊆ {f (x) ∈ A|f (x) 6= 0 for all x ∈ [0, 1]}.
Next let f (x) ∈ A such that f (x) 6= 0 for all x ∈ [0, 1]. Since f is continuous,
g := f1 is also continuous. Thus since
f (x)
1
= 1,
f (x)
1
such that
for all non-zero f (x) ∈ A there exists g(x) defined by g(x) := f (x)
f (x)g(x) = 1. Therefore {f (x) ∈ A|f (x) 6= 0 for all x ∈ [0, 1]} ⊆ A∗ , so
A∗ = {f (x) ∈ A|f (x) 6= 0 for all x ∈ [0, 1]}.
(b) Give an example of an element which is neither a unit nor a zero divisor.
Proof: The function f (x) = x is neither a zero divisor nor a unit in A. Since
f (0) = 0, f (x) ∈
/ A∗ , and g(x) = 0 is the only continuous function such that
f (x)g(x) = 0.
Thus f (x) is not a zero divisor.
(c) Give an example of a zero divisor in A.
Proof:
The following two functions are zero divisors in the ring:
(
0
for all x ∈ [0, 12 ]
h(x) =
x − 21 for all x ∈ ( 12 , 1]
and
g(x) =
(
x−
0
1
2
for all x ∈ [0, 21 ]
for all x ∈ ( 21 , 1]
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1.7)
Determine whether the following polynomials are irreducible in the rings indicated.
For those that are reducible, determine their factorization into irreducibles. The
notation Fp denotes the finite field Z/pZ, where p is a prime.
(a) x2 + x + 1 in F2 [x].
(b) x3 + x + 1 in F3 [x].
(c) x4 + 1 in F5 [x].
(d) x4 + 10x2 + 1 in Z[x].
1.8)
Let R be a non-zero ring. Prove that the following are equivalent:
(a) R is a field.
(b) The only ideals in R are (0) and (1).
(c) Every homomorphism of R into a non-zero ring B is injective.
Proof: (a)⇒(b). Assume R is a field. Suppose that there exists an ideal, I, of R
such that I 6= (0) and I 6= (1). Since I 6= (0), there exists an a ∈ I such that a 6= 0.
Now, a−1 ∈ F since F is a field. By the definition of ideal, a−1 a ∈ I. But a−1 a = 1
which implies that I = R, a contradiction.
(b)⇒(c). Now assume that the only ideals in R are (0) and (1). Say φ is a
homomorphism such that φ : R → B. We much show that φ is injective. To do so,
say φ(x) = 0 for some x ∈ R. Then x ∈ Ker φ, which is an ideal of R. Since the only
ideals of R are (0) and (1), Ker φ = (0) or Ker φ = (1) = R. We also know that φ is
not the trivial homomorphism since φ(1) = 1. Thus Ker φ = (0), in other words if
x ∈ Ker φ, x = 0. So φ is injective.
(c)⇒(a) Now assume that every homomorphism of R into a non-zero ring B is
injective. We must show that R is a field. Let a ∈ R such that a 6= 0. Consider the
canonical map
φ :R → R/(a)
x 7→ x
Since φ is a homomorphism by (c) it is injective. However φ(a) = 0 which implies
that a = 0, a contradiction since a was assumed to be nonzero. This means that
R/(a) = 0 and so R = (a). Now, since R is a ring there exists 1 ∈ R so 1 ∈ (a),
meaning that 1 is a multiple of a, in other words ax = 1 for some x ∈ R. Therefore
a has a multaplicative inverse and R is a field.
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1.9)
Let f : R → S be a ring homomorphism.
(a) Prove that Ker f is an ideal of R.
Solution. Note that Ker φ is non-empty since 0 ∈ Ker φ. Let a ∈ Ker φ, and let
r ∈ R. Then φ(ar) = φ(a)φ(r) = 0·r = 0, so ar ∈ Ker φ. By a similar argument,
ra ∈ Ker φ. Now let r, s ∈ Ker φ. Then φ(r + s) = φ(r) + φ(s) = 0 + 0 = 0.
Therefore, r + s ∈ Ker φ.
(b) Prove that if J is an ideal of S then f −1 (J) is an ideal of R that contains Ker f .
Solution. Note that φ−1 (J) = r ∈ R|φ(r) ∈ J. Let a, b ∈ φ−1 (J). Then a and
b are also contained in R, and φ(a + b) = φ(a) + φ(b) ∈ J, so a + b ∈ φ−1 (J).
Now suppose a ∈ φ−1 (J), r ∈ R. Then φ(ar) = φ(a)φ(r) where φ(a) ∈ J and
φ(r) ∈ S. Thus, φ(ar) ∈ J and ar ∈ φ−1 (J). Similarly, ra ∈ φ−1 (J). Therefore,
φ−1 (J) is an ideal of R.
To show Ker φ is contained in φ−1 (J), let ar ∈ Ker φ. Then φ(ar) = 0. Since
0 ∈ J, then ar ∈ φ−1 (J). Therefore, Ker φ is contained in φ−1 (J).
(c) Prove that if P is a prime ideal of S then f −1 (P ) is a prime ideal of R.
Solution. Consider φ−1 (P ) = x ∈ R|φ(x) ∈ P . We know that φ−1 (P ) is nonempty since φ : 0 → 0, so let ab ∈ φ−1 (P ). We want to show that a ∈ φ−1 (P )
or b ∈ φ−1 (P ). Since ab ∈ φ−1 (J), φ(ab) ∈ P and φ(ab) = φ(a)φ(b). Since P
is prime, either φ(a) or φ(b) ∈ P . So without loss of generality, let φ(b) ∈ P .
Then b ∈ φ−1 (P ), and therefore, φ−1 (P ) is prime.
1.10) Let p be a prime and consider the ring of polynomials in x with coefficients in Fp . This
ring is denoted by Fp [x]. Let ϕ : Fp [x] → Fp [x] be the map given by ϕ(f ) = f p . Prove
that ϕ is an endomorphism. This map is called the Frobenius endomorphism.
Solution.:
Let f , g ǫ Fp [x].
a)Then ϕ(f ∗ g) = (f ∗ g)p = f p ∗ g p = ϕ(f ) ∗ ϕ(g).
p . By the binomial theorem (f + g)p = p f p +
b)
Then
ϕ(f
+
g)
=
(f
+
g)
0
p−1
p p−1
p
g + ... + p−1
fg
+ pp g p . Since p0 = 1 and pp = 1 we can rewrite
1 f
p−1
p
f g . We know that in
(f + g)p = f p + S + g p where S = p1 f p−1 g + ... + p−1
(p−1)!
p!
p
general for p 6= 0, p then k = k!(p−k)! = p ∗ k!(p−k)! . And since p is prime, p has no
factors. Thus k! does not divide p and (p − k)! does not divide p. Thus there will
always be a factor of p out in front of every coefficient in S. And since our coefficients
are coming from Fp [x] then all coefficients of S will be equal to 0 in Fp [x]. Then
S = 0 ∗ f p−1 g + ... + 0 ∗ f g p−1 = 0. Therefore in Fp [x], (f + g)p = f p + f g . Thus we
have ϕ(f + g) = (f + g)p = f p + g p = ϕ(f ) + ϕ(g). Thus ϕ is an endomorphism. 1.11) Let S ⊆ R and let I be an ideal of R. Prove that the following statements are
equivalent:
(a) S ⊆ I.
(b) (S) ⊆ I.
This fact is useful when you want to show that one ideal is contained in another.
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1.12) Prove the following equalities in the polynomial ring R = Q[x, y]:
(a) (x + y, x − y) = (x, y).
(b) (x + xy, y + xy, x2 , y 2 ) = (x, y).
(c) (2x2 + 3y 2 − 11, x2 − y 2 − 3) = (x2 − 4, y 2 − 1).
This illustrates that the same ideal can have many different generating sets and that
different generating sets may have different numbers of elements.
1.13) Let R be a ring and let I, J and K be ideals of R.
(a) Prove that I ∩ J is an ideal of R.
Proof: First, note that 0 ∈ I and 0 ∈ J. Therefore, 0 ∈ I ∩ J. Next, let
x, y ∈ I ∩ J, then x, y ∈ I and x, y ∈ J. So, since I and J are both ideals,
x + y ∈ J and x + y ∈ I. Therefore, x + y ∈ I ∩ J. Finally, let x ∈ I ∩ J and let
r ∈ R. Then x ∈ I and x ∈ J and since I and J are ideals, xr ∈ I and xr ∈ J
so xr ∈ I ∩ J
Therefore, I ∩ J is an ideal of R.
(b) Prove that I(J + K) = IJ + IK.
P
+ zi )|xi ∈ I, yi ∈
Proof: (⊆) Let a ∈ I(J + K). In other words, a ∈ { si xi (yi P
s
J,
s is the size
the set I(J +P
K). However,
Pn i xi (yi + zi ) =
Pof
Pzs i ∈ K} where P
m
s
s
i xi zi )|xi ∈ I, yi ∈
i xi zi . So a ∈ { i xi yi +
i xi yi +
i xi yi + xi zi =
J, zi ∈ K}. So I(J + K) ⊆ IJ + IK.
Pn
P
(⊇) Let a ∈ IJ + IK, then a ∈ { m
i xi zi |xi ∈ I, yi ∈ J, zi ∈ K}.
i xi yi +
However, 0 is in both IJ and IK so we can add zeroes P
to each ofP
the sums
t
t
x
y
+
and
make
them
the
same
size
of
t
=
m
+
n.
So
then
i xi zi =
i i i
Pt
Pt
Pt
i xi (yi + zi ). So a ∈ { i xi (yi + zi )|xi ∈ I, yi ∈ J, zi ∈ K}.
i xi yi + xi zi =
So, I(J + K) ⊇ IJ + IK.
Therefore, I(J + K) = IJ + IK.
(c) Prove that if either J ⊆ I or K ⊆ I then I ∩ (J + K) = I ∩ J + I ∩ K. (modular
law)
Proof: Without loss of generality, assume that J ⊆ I.
(⊆) Let x ∈ I ∩ (J + K) then x ∈ I and x ∈ J + K i.e. x ∈ {s + t|s ∈ J, t ∈ K}.
So x ∈ s + t for some s, t. Note also s ∈ I and s ∈ J. Since x = s + t,
t = x − s and x, s ∈ I so t ∈ I. So, s ∈ I ∩ J and t ∈ I ∩ K. This implies that
x ∈ {s + t|s ∈ I ∩ K and t ∈ I ∩ K}. Hence, x ∈ I ∩ J + I ∩ K. Therefore,
I ∩ (J + K) ⊆ I ∩ J + I ∩ K.
(⊇) Let x ∈ I ∩ J + I ∩ K i.e. x ∈ {m + n|m ∈ I ∩ J, n ∈ I ∩ K}. So m ∈ I and
m ∈ J and n ∈ I and n ∈ K. Since m, n ∈ I, x = m + n for some m, n must
also be an element of I i.e. x ∈ I. So, x ∈ I ∩ (J + K). Therefore I ∩ J ⊇ I ∩ K.
Therefore I ∩ (J + K) = I ∩ J + I ∩ K.
1.14) In the ring of integers Z compute the ideals:
(a) (2) + (3),
(b) (2) + (4),
6
(c) (2)((3) + (4)),
(d) (2)(3) ∩ (2)(4),
(e) (6) ∩ (8),
(f) (6)(8)
1.15) Let Q[x, y]. Compute the ideals:
(a) (x) ∩ (y),
(b) (x + y)2 ,
(c) (x, y)2 ,
(d) (x2 ) ∩ (x, y),
(e) (x2 + xy) ∩ (xy + y 2 ),
(f) (x) + (y),
(g) (x + 1) + (x),
(h) (x2 + xy)(x − y),
(i) (x2 ) ∩ ((xy) + (y 2 )),
(j) (x − y)((x) + (y 2 ))
√
1.16) Let R√be a ring. The nilradical 0 of R√
is the set of nilpotent elements of R. Prove
that 0 is an ideal, and if xn = 0 in R/ 0 for some n then x = 0.
Proof: Let x, y ∈
theorem,
√
0. Then ∃n, m ∈ N such that xn = y m = 0. By the binomial
n+m
X
n + m i n+m−i
(x + y)
=
xy
i
i=0
i n+m−i
For i > n, xi = xn xn−i = 0 · xn−i = 0. Thus, n+m
xy
= 0. For i < n,
i
n + m − i > m. Thus, y n+m−i = y m y n−i = 0 · y n−i = 0, so n+m
xi y n+m−i = 0.
i
√
Therefore, (x + y)n+m = 0, so x + y ∈ 0.
√
Let x ∈ 0, r ∈ R. Then ∃n ∈ N such that xn = 0.
√ Since multiplication is
commutative, (rx)n = rn xn = rn · 0 = 0. Therefore, rx ∈ 0.
√
√
Finally, 01 = 0, so 0 ∈ 0. Therefore, 0 is an ideal of R.
√
Let x ∈ R and x be the image of x under the canonical map R → R/ 0. Suppose
√
xn = 0 for some n ∈ N. That is, x · · · x = x · · · x = xn = 0. Therefore xn ∈ √0.
Thus, ∃m ∈ N such that (xn )m = 0. Since (xn )m = xnm and nm ∈ N, then x ∈ 0.
Therefore, x = 0.
n+m
√
√
1.17) Let R be a ring, and I ⊆ 0 an ideal, where 0 is the nilradical of R. Prove that if
x is a unit of R/I then x is a unit of R.
Proof: Assume that x is a unit of R/I, thus there exists y ∈ R/I such that xy = 1.
It follows that:
xy = i + 1 for some i ∈ I, and thus xy − 1 ∈ I.
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By the definition of I, (xy − 1)n = 0 for some n ≥ 1. The binomial expansion gives
that
n X
n
n
(xy − 1) =
(xy)i (−1)n−i = 0 for some n ≥ 1.
i
i=0
Every term of this series contains an x except for the case when i = 0, which gives
the term ±1. Subtract 1 from both sides and factor an x out of the remaining terms,
which yields:
X
n n i−1 i
n−i
x ±
x y (−1)
=1
i
i=1
P
Let r = ± ni=1 ni xi−1 y i (−1)n−i . Thus r ∈ R since it is a linear combination of
ring elements. Therefore xr = 1 for some r ∈ R, so x is a unit of R.
1.18) Let R be a ring and P a prime ideal of R. Let I be the ideal generated by all the
idempotent elements of P . Prove that R/I has no non-trivial idempotents.
IMMERSE 2008: Extras 1
1.19) A (not neccessarily commutative) ring R is boolean if x2 = x for all x ∈ R. If R is a
boolean ring show that
(a) 2x = 0 for all x ∈ R,
(b) R is commutative,
(c) every prime ideal p is maximal, and R/p is a field with two elements, and
(d) every finitely generated ideal is principal.
1.20) Let R be a ring in which every ideal of R except (1) is prime. Prove that R is a
field.
Solution. We first note that R is an integral domain because zero divisors would
contradict the primality of h0i. Now assume there exists a non-unit r ∈ R. Then r
must be in hr2 i. We write r = ar2 for some a ∈ R. Then (ar − 1)r = 0, and since
ar 6= 1, r = 0. Thus the only non-unit in R is 0 and R is a field.
1.21) For ideals I and J in a ring R their ideal quotient is
(I :R J) = {x ∈ R | xJ ⊆ I}.
Let R be a ring and let P be a finitely generated prime ideal of R with (0 :R P ) = 0.
Prove that (P :R P 2 ) = P .
1.22) Let K be a field and let R be the ring of polynomials in x over K subject to the
condition that they contain no terms in x or x2 . Let I be the ideal in R generated
by x3 and x4 . Prove that x5 6∈ I and x5 I ⊆ I 2 . (This shows that the assumption
that P is prime in 1.21 is necessary.)
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1.23) Let F be a field and let E = F × F . Define addition and multiplication in E by the
rules:
(a, b) + (c, d) = (a + c, b + d)
and
(a, b)(c, d) = (ac − bd, ad + bc)
Determine conditions on F under which E is a field.
1.24) Find all the monic irreducible polynomials of degree less than or equal to 3 in F2 [x],
and the same in F3 [x].
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1.25) Construct fields of each of the following orders:
(a) 9
(b) 49
(c) 8
(d) 81
1.26) Exhibit all the ideals in the ring F [x]/(p(x)), where F is a field and p(x) is a polynomial in F [x] (describe them in terms of the factorizations of p(x)).
1.27) An element x of a ring is idempotent if x2 = x. Prove that a local ring contains no
non-trivial idempotents. (The trivial idempotents are 0 and 1.)
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