Math 0430
Homework #10
Sec. 6.2: #4: (a) Let [a]n denote the congruence class of the integer a modulo n. Show that the
map f: Z 12  Z 4 that sends [a]12 to [a]4 is a well-defined surjective homomorphism.
Proof: Let [a]12 = [b]12. Then 12|(a - b), so 4|(a - b) and f([a]12) = f([b]12). Thus f is a welldefined mapping. f([a]12 + [b]12) = f([a + b]12) = [a]4 + [b]12 = f([a]12) + f([b]12) and f([a]12 [b]12) =
f([ab]12) =[a]4 [b]4 = f([a]12) f([b]12), so f is a homomorphism. Clearly, if [a]4  Z 4 , a is an integer,
[a]12  Z 12 and f([a]12) = [a]4, so f is surjective. 
(b) Find the kernel of f.
Proof: Ker(f) = {[a]12 Z 12 | f([a]12) = [a]4 =[0]4} = {[a]12 Z 12 : 4|a} = ([4]12]) =
{[0]12, [4]12 , [8]12}. 
Sec. 6.2: #8: (a) Let I = {0,3} in Z6 . Verify that I is an ideal in Z6 and show that Z6 I  Z3 .
Proof: Define f: Z6  Z3 by f([a]6 =[a]3. Then f is clearly surjective. For [a]6 ,[b]6  Z 6 ,
f([a]6 + [b]6) = f([a + b]6) = [a + b]3 = [a]3 + [b]3 = f([a]6) + f([b]6) and f([a]6 [b]6) = f([ab]6) =
[ab]3 = [a]3 [b]3 = f([a]6) f([b]6); so f is also a homorphism. The kernel of f is I. So I is an ideal
in Z6 and by the First Isomorphism Theorem, Z6 I  Z3 . 
Sec. 6.3: #7: Let R be a commutative ring with identity. Prove that R is a field if and only if (0R)
is a maximal ideal.
Proof: Assume that R is a field. Let (0R)  M  R. If M ≠(0R), then there is a M with a ≠0R.
This means that a is a unit, 1R  M and M = R. Therefore, M is a maximal ideal of R.
Conversely, assume that (0R) is a maximal ideal. Let a ≠ 0R and consider (a). (0R)  (a)  R,
so (a) = R. It follows that 1R (a) and there is b  R such that 1R = ba, and a is a unit.
Therefore, R is a field.
Sec. 6.3: #10: Let p be a fixed prime and let J be the set of polynomials in Z[x] whose constant
terms are divisible by p. Prove that J is a maximal ideal in Z[x].
Proof: Since p is prime, Zp is a field, Define : Z[x]  Zp by (f(x)) = [a0] where f(x) = anxn +
an-1xn-1 + …+a1x + a0. Let g(x) = bmxm + bm-1xm-1 + …+b1x + b0. Then (f(x) + g(x)) = [a0 +
b0] = [a0]+ [b0] = (f(x)) + (g(x)) and (f(x)g(x)) = [a0b0] = [a0][b0] = (f(x))(g(x)), so  is a
homomorphism. If [a]  Zp, then f(x) = a  Z[x], and (a) = [a]; so  is surjective.
Furthermore, the kernel of  is {f(x)  Z[x] | (f(x)) = [a0] = [0]} = {f(x)  Z[x] | p|a0} = J. By the
First Isomorphism Theorem,
Z[ x]
 Zp . Since Zp is a field, J is a maximal ideal of Z[x]. ■
J
Sec. 6.3: 11: Show that the principal ideal (x-1) in Z[x] is prime but not maximal.
Proof: Define : Z[x]  Z by (f(x)) = f(1) where f(x) = anxn + an-1xn-1 + …+a1x + a0. Let g(x)
= bmxm + bm-1xm-1 + …+b1x + b0. Then (f(x) + g(x)) = f(1) + g(1) = (f(x)) + (g(x)) and
(f(x)g(x)) = f(1)g(1) = (f(x))(g(x)), so  is a homomorphism. If a  Z, then f(x) = a  Z[x],
and (f(x)) = f(1) = a; so  is surjective. Furthermore, the kernel of  is {f(x)  Z[x] | (f(x)) =
f(1) = 0} = {f(x)  Z[x] | x-1|f(x)} = (x-1). By the First Isomorphism Theorem,
Z[ x ]
 Z . Since
( x  1)
Z is an integral domain and not a field, (x-1) is a principal ideal, but not a maximal ideal. ■
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Sec. 6.3: #13: Find an ideal in Z  Z that is prime, but not maximal.
Proof: Define f: Z  Z  Z by f((z1,z2)) = z1. F is a surjective homomorphism and the
kernel of f is I = (0)  Z . So the First Isomorphism Theorem implies that
ZZ
 Z . Since Z is an integral domain but not a field, (0)  Z is a prime ideal that is not
(0)  Z
maximal. ■
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