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Homework #4 Solutions
p 286, #8 Let φ : Zn → Zn be a ring homomorphism. Let a = φ(1). Then for any 0 6 = r ∈ Zn = { 1 , 2 ,... , n − 1 } we have
φ(r) = φ(1 + 1 +︸ ︷︷ · · · + 1︸ r times
) = φ(1) + φ(1) + · · · + φ(1) ︸ ︷︷ ︸ r times
= r · φ(1) = r · a = ra mod n.
Since a = φ(1) = φ(1 · 1) = φ(1)φ(1) = a^2
we’re finished.
p 286, #10 Let I = 〈x^2 +1〉 and let f (x) ∈ Z 3 [x]. By including zero coefficients if necessary we can write
f (x) =
∑^ n
i=
a 2 ix^2 i^ +
∑^ m
j=
a 2 j+1x^2 j+1,
for some ai ∈ Z 3 , i.e. we can write f (x) as the sum of its even degree and odd degree terms. In Z 3 [x]/I we have x^2 + I = −1 + I so that
f (x) + I =
∑^ n
i=
(a 2 i + I)(x^2 i^ + I) +
∑^ m
j=
(a 2 j+1 + I)(x^2 j+1^ + I)
∑^ n
i=
(a 2 i + I)(x^2 + I)i^ +
∑^ m
j=
(a 2 j+1 + I)(x + I)(x^2 + I)j
∑^ n
i=
(a 2 i + I)(−1 + I)i^ +
∑^ m
j=
(a 2 j+1 + I)(x + I)(−1 + I)j
( (^) n ∑
i=
((−1)ia 2 i + I)
( (^) m ∑
j=
((−1)j^ a 2 j+1 + I)
( (^) n ∑
i=
(−1)ia 2 i + x
∑^ m
j=
(−1)j^ a 2 j+
+ I
or, more succinctly, f (x) + I = a + bx + I
for some a, b ∈ Z 3. Moreover, if a + bx + I = c + dx + I for some a, b, c, d ∈ Z 3 then (a−c)+(b−d)x ∈ I, which means that x^2 +1 divides the linear polynomial (a−c)+(b−d)x, an obvious impossibility unless a − c = b − d = 0. That is, a + bx + I = c + dx + I implies that a + bx = c + dx. Hence, every element in Z 3 [x]/I can be expressed uniquely in the form a + bx + I, a, b ∈ Z 3. We will use this fact below.
Now define φ : Z 3 [i] → Z 3 [x]/I by φ(a + bi) = a + bx + I. This is a homomorphism since for any a, b, c, d ∈ Z 3 we have
φ((a + bi)(c + di)) = φ((ac − bd) + (ad + bc)i) = (ac − bd) + (ad + bc)x + I = (a + bx + I)(c + dx + I) − (bd + I)(x^2 + 1 + I) = (a + bx + I)(c + dx + I) = φ(a + bi)φ(c + di)
and
φ((a + bi) + (c + di)) = φ((a + c) + (b + d)i) = (a + c) + (b + d)x + I = (a + bx) + (c + dx) + I = (a + bx + I) + (c + dx + I) = φ(a + bi) + φ(c + di).
Moreover, the result of the preceding paragraph implies that this function is one-to-one and onto, hence provides an isomorphism between Z 3 [i] and Z 3 [x]/I.
p 286, #12 Define φ : Z[
2] → H by
φ(a + bi) =
a 2 b b a
This is obviously one-to-one and onto so to prove it is an isomorphism it suffices to show that it preserves addition and multiplication. Addition is easy: for any a, b, c, d ∈ Z
φ((a + b
2)) = φ((a + c) + (b + d)i) =
a + c 2(b + d) b + d a + c
a 2 b b a
c 2 d d c
= φ(a + b
Multiplication is no more difficult, just more interesting: for a, b, c, d ∈ Z we have
φ((a + b
2)(c + d
2)) = φ((ac + 2bd) + (ad + bc)
ac + 2bd 2(ad + bc) ad + bc ac + 2bd
a 2 b b a
c 2 d d c
= φ(a + bi)φ(c + di)
and we’re finished!
p 287, #18 Let φ : Zn → Zn be an isomorphism of rings. According to exercise 8 there is an a ∈ Zn satisfying a^2 = a so that φ(x) = ax for all x ∈ Zn. In order for φ to be an isomorphism we must also have a = a · 1 = φ(1) = 1. Therefore φ(x) = x for all x, i.e. φ must be the identity homomorphism.
and
φ(x + y) = (x + y)p^ =
∑^ p
k=
p k
xkyp−k^ = xp^ + yp^ = φ(x) + φ(y).
The middle terms in the last expression vanish because, according to the lemma, all the binomial coefficients are divisible by p, the characteristic of R. Hence, φ is a homomorphism. This homomorphism figures prominently in the Galois theory of finite fields.
p 288, #40 Let F be a field, R be a ring and φ : F → R be an onto homomorphism. According to the first isomorphism theorem F/ ker φ ∼= R. If R has more than one element then we cannot have ker φ = F. However, since the kernel is an ideal and F is a field, the only other option we have is ker φ = { 0 }. Hence, φ is also one-to-one and is therefore an isomorphism.
p 288, #46 Let φ : R → R be an isomorphism of rings. We can argue exactly as in Exercise 36 to conclude that φ(r) = r for all r ∈ Q.^1 Let x, y ∈ R with x < y. Then y − x > 0 so there is a z ∈ R+^ so that x − y = z^2. Then
φ(y) − φ(x) = φ(y − x) = φ(z^2 ) = φ(z)^2 > 0
since φ(z) 6 = 0 as φ is one-to-one. That is, if x < y then φ(x) < φ(y), i.e. φ preserves the natural order on R. Let x ∈ R and suppose that x < φ(x). Since Q is dense in R we can find an r ∈ Q with x < r < φ(x). But then φ(x) < φ(r) = r < φ(x), an impossibility. We have a similar contradiction if x > φ(x) and so we conclude that φ(x) = x. Since x was an arbitrary element of R we conclude that φ is the identity map.
p 289, #60 a. Let
a b b a
c d d c
∈ R. Then
φ
a b b a
c d d c
= φ
a + c b + d b + d a + c
= (a + c) − (b + d)
= (a − b) + (c − d) = φ
a b b a
c d d c
and
φ
a b b a
c d d c
= φ
ac + bd ad + bc ad + bc ac + bd
= (ac + bd) − (ad + bc)
= (a − b)(c − d) = φ
a b b a
φ
c d d c
proving that φ is a homomorphism. (^1) We have seen that any field of characteristic 0 contains Q as a subfield and that any field of characteristic p contains Zp as a subfield. In each case, these fields are called the prime subfields and it is a general fact that any automorphism of a field must fix its prime subfield element-wise.
b.
a b b a
∈ R is in the kernel of φ if and only if a − b = 0 or a = b. Thus
ker φ =
a a a a
∣ a^ ∈^ Z
c. Since φ : R → Z is a homomorphism and is clearly onto, the first isomorphism theorem tells us that R/ ker φ ∼= Z. d. Since R is a commutative ring with identity and R/ ker φ ∼= Z is an integral domain, we can apply Theorem 14.3 to conclude that ker φ is indeed a prime ideal. e. Since R is a commutative ring with identity and R/ ker φ ∼= Z is not a field, we can apply Theorem 14.4 to conclude that ker φ is not a maximal ideal.
