3.4 Homomorphisms
1.
a. No. θ is a general ring homomorphism, because 42 = 4 in 
But θ(1) = 4, and 4 ≠ 1 in 
But θ(1) = 4, and 4 ≠ 1 in
c. No. θ[(r, s) · (r′, s′)] = rr′ + ss′ need not equal
e. Yes. θ(fg) = (fg)(1) = f(1)g(1) = θ(f) · θ(g). Similarly for f ± g. The unity of 
is 
given by 
. Thus 
is given by . Thus
2.
a. Write θ(1) = e, and let s 
S. Then s = θ(r) for some r R (θ is onto) so
S. Then s = θ(r) for some r R (θ is onto) so
Hence e is the unity of S.
3. If 
is a general ring homomorphism, let θ(1) = e. Then e2 = e so either e = 1 (θ a ring homomorphism) or e = 0. In the last case
is a general ring homomorphism, let θ(1) = e. Then e2 = e so either e = 1 (θ a ring homomorphism) or e = 0. In the last case
for all 
, so θ is trivial.
, so θ is trivial.
5. If z Z(R) and s R1, write s = θ(r), r R. Then
Z(R) and s R1, write s = θ(r), r R. Then
Thus θ(z) Z(R1). Any non-onto ring homomorphism θ : R → R1 where R1 is commutative shows this need not be equality.
Z(R1). Any non-onto ring homomorphism θ : R → R1 where R1 is commutative shows this need not be equality.
7. If 
are ring homomorphisms then ϕθ is a group homomorphism and ϕθ(1R) = ϕ[θ(1R)] = ϕ(1S) = 1T,
are ring homomorphisms then ϕθ is a group homomorphism and ϕθ(1R) = ϕ[θ(1R)] = ϕ(1S) = 1T,
9. If R is a division ring and θ : R → S is a ring homomorphism, then ker θ is an ideal of R so ker θ = 0 or ker θ = R. If ker θ = 0 then θ is one-to-one so θ(R) ≅ R. If ker θ = R then θ(R) = 0. Thus the images of R are R and 0 up to isomorphism.
10. Clearly θ(r0) = θ(1) = 1 = [θ(r)]0. If θ(rn) = [θ(r)]n then
Hence (4) follows by induction.
11. In 
this is x3 + x − 1 = 0. If x = 0, 1, 2, − 1 then x3 + x − 1 = − 1, 1, 1, 1, so there is no solution in 
and hence none in 
this is x3 + x − 1 = 0. If x = 0, 1, 2, − 1 then x3 + x − 1 = − 1, 1, 1, 1, so there is no solution in and hence none in
13. In 
this is 4n2 = 2 and this has a solution (n = 2) in 
. However, in 
it is 7m2 = 9, or m2 = 8 · 9 = 72 = 6. But m2 = 0, 1, 3, 4, 5, 9 in 
, so there is no solution in 
and hence no solution in 
this is 4n2 = 2 and this has a solution (n = 2) in . However, in it is 7m2 = 9, or m2 = 8 · 9 = 72 = 6. But m2 = 0, 1, 3, 4, 5, 9 in , so there is no solution in and hence no solution in
15. The inverse map is a group isomorphism by group theory. Given s and s1 in S then
Since σ is one-to-one, σ−1(s) · σ−1(s1) = σ−1(ss1).
17. R ≅ R because 1R : R → R is an automorphism. If R ≅ S, say σ : R → S in an isomorphism, then σ−1 : S → R is an isomorphism by Exercise 15, so S ≅ R. Finally if τ : S → T is an isomorphism, so is τσ : R → T so R ≅ T.
19.
a. θ(A) is an additive subgroup of S by Theorem 1 §2.10. If x θ(A) and y S = θ(R), let x = θ(a), y = θ(r), a A, r R. Then
θ(A) and y S = θ(R), let x = θ(a), y = θ(r), a A, r R. Then
Similarly yx θ(A), so θ(A) is an ideal of S.
θ(A), so θ(A) is an ideal of S.
21. Let 
be a ring homomorphism. Then ker θ is an ideal of the field 
so 
or ker θ = 0. But ker θ = 0 means 
. Let θ(i) = a. Then a2 = [θ(i)]2 = θ(i2) = θ(− 1) = − 1, a contradiction as a is real. Since 
, there is no such θ.
be a ring homomorphism. Then ker θ is an ideal of the field so or ker θ = 0. But ker θ = 0 means . Let θ(i) = a. Then a2 = [θ(i)]2 = θ(i2) = θ(− 1) = − 1, a contradiction as a is real. Since , there is no such θ.
23.
a. If 
exists, then 
. Define 
this way. Then 
is well defined because
exists, then . Define this way. Then is well defined because
It is clearly a ring homomorphism and
for all r R. Thus 
R. Thus
24.
a. Define θ : R∗ → aut R by θ(u) = σu for all u R∗. Then
R∗. Then
for all r, so 
; that is, θ is a group homomorphism. Since θ(R∗) = inn R, this shows inn R is a subgroup of aut R. In fact inn R 
aut R. For if τ aut R and u R∗, then 
and 
. Hence 
is in inn R, as required.
; that is, θ is a group homomorphism. Since θ(R∗) = inn R, this shows inn R is a subgroup of aut R. In fact inn R aut R. For if τ aut R and u R∗, then and . Hence is in inn R, as required.
25.
a. e2 = baba = b1a = ba = e. Note that e(bra) = babra = b1ra = bra and, similarly, (bra)e = bra. This σ is a mapping R → eRe. Now σ(r + s) = σ(r) + σ(s) is onto and σ(r) · σ(s) = bra · bsa = br1sa = σ(rs). Finally σ(1) = ba = e shows σ is a ring homomorphism. If σ(r) = 0 then bra = 0 so r = 1r1 = (ab)r(ab) = a · 0 · b = 0. Thus σ is one-to-one. Finally, if r eRe then r = ere = (ba)r(ba) = σ(arb). So σ is a ring isomorphism.
eRe then r = ere = (ba)r(ba) = σ(arb). So σ is a ring isomorphism.
27. We use the isomorphism theorem by finding an onto ring homomorphism θ : R → S × S that has A = ker θ. If 
then θ is clearly onto, and the reader can verify that it is a ring homomorphism. Finally 
as required.
then θ is clearly onto, and the reader can verify that it is a ring homomorphism. Finally as required.
29.
a. Define 
by 
, 
. This is an onto R-homomorphism and ker θ = A(ω). Done by the isomorphism theorem.
by , . This is an onto R-homomorphism and ker θ = A(ω). Done by the isomorphism theorem.
31. Note that e = (1, 0) is the unity of 
. Define θ : R = S × T → T by θ(s, t) = t. Then θ is an onto ring homomorphism and 
. Done by the isomorphism theorem.
. Define θ : R = S × T → T by θ(s, t) = t. Then θ is an onto ring homomorphism and . Done by the isomorphism theorem.
33. Define θ : R/A → R/B by θ(r + A) = r + B. This is well defined because r + A = s + A ⇒ r − s A ⊆ B ⇒ r − s B ⇒ r + B = s + B. It is clearly an onto ring homomorphism and ker θ = {r + A 
r + B = 0} = B/A.
A ⊆ B ⇒ r − s B ⇒ r + B = s + B. It is clearly an onto ring homomorphism and ker θ = {r + A r + B = 0} = B/A.
35.
a. Define 
by 
where 
. This is an onto ring homomorphism and 
by where . This is an onto ring homomorphism and
c. Observe first that r + sη is a unit in R(η) if and only if r ≠ 0 (then the inverse is r−1 − r1sr−1η). Let A ≠ 0 be an ideal of R(η) ; we show that A = Rη = {rη r R}. As A ≠ 0 let 0 ≠ a + bη A. Then a = 0 because A ≠ R(η), so b ≠ 0. But then η A, whence Rη ⊆ A. If Rη ⊂ A let p + qη A, p ≠ 0, whence A = R(η), as required.
r R}. As A ≠ 0 let 0 ≠ a + bη A. Then a = 0 because A ≠ R(η), so b ≠ 0. But then η A, whence Rη ⊆ A. If Rη ⊂ A let p + qη A, p ≠ 0, whence A = R(η), as required.
36.
a. Define 
by 
where 
. This is an onto ring homomorphism and 
by where . This is an onto ring homomorphism and
c. Let A ≠ 0, R(γ) be an ideal of R(γ). Since A ≠ 0 there exists 0 ≠ r + sγ A, hence 1 + cg A for some c. Since (1 + cγ)(1 − cγ) = 1 − c2, the fact that A ≠ R(γ) means that c2 = 1. Hence c = ± 1 because R is a division ring. So either R(1 + γ) ⊆ A or R(1 − γ) ⊆ A.
A, hence 1 + cg A for some c. Since (1 + cγ)(1 − cγ) = 1 − c2, the fact that A ≠ R(γ) means that c2 = 1. Hence c = ± 1 because R is a division ring. So either R(1 + γ) ⊆ A or R(1 − γ) ⊆ A.
Now consider the maps θ : R(γ) → R and ϕ : R(γ) → R given by θ(r + sγ) = r + s and ϕ(r + sγ) = r − s for all r, s R. These are onto ring homomorphisms, and ker θ = R(1 − γ) and ker ϕ = R(1 + γ). It follows from the isomorphism theorem that R(1 − γ) and R(1 + γ) are both maximal ideals. Hence A = R(1 − γ) or A = R(1 + γ).
R. These are onto ring homomorphisms, and ker θ = R(1 − γ) and ker ϕ = R(1 + γ). It follows from the isomorphism theorem that R(1 − γ) and R(1 + γ) are both maximal ideals. Hence A = R(1 − γ) or A = R(1 + γ).
37. Define 
by 
. This is a ring homomorphism and 
and 
. Since m|t and n|t, 
. Let 
. Then m|a and n|a so t|a (because t = lcm(m, n)). Thus 
and so 
; a subring of 
by . This is a ring homomorphism and and . Since m|t and n|t, . Let . Then m|a and n|a so t|a (because t = lcm(m, n)). Thus and so ; a subring of
39. By Example 10 and induction, all ideals of R = R1 × 
× Rn look like A = A1 × × An, Ai an ideal of R. Hence 
. Thus A is maximal if and only if exactly one of the Ak is maximal in Rk, and Ai = Ri if i ≠ k.
× Rn look like A = A1 × × An, Ai an ideal of R. Hence . Thus A is maximal if and only if exactly one of the Ak is maximal in Rk, and Ai = Ri if i ≠ k.
41. Put 
. Thus 
. Then 
by Theorem 8 if we can show 
and 
. Define 
by σ(r) = re. This satisfies σ(r + s) = σ(r) + σ(s), and σ(rs) = σ(r)σ(s), and σ is one-to-one (σ(r) = 0 means 
, so r = 0). If a, b R then
. Thus . Then by Theorem 8 if we can show and . Define by σ(r) = re. This satisfies σ(r + s) = σ(r) + σ(s), and σ(rs) = σ(r)σ(s), and σ is one-to-one (σ(r) = 0 means , so r = 0). If a, b R then
where r = a + 2bu. Thus σ is onto so 
as rings. Similarly 
.
as rings. Similarly .
43.
a. Define θ : S → R by θ(f) = f(x). Then
The unity 
of S is the constant function onto 1, so
of S is the constant function onto 1, so
Thus θ is a ring homomorphism, onto because, if r R, r = θ(f) where f(t) = r for all t X. Since ker θ = {f S f(x) = 0} we are done by the isomorphism theorem because R is simple.
R, r = θ(f) where f(t) = r for all t X. Since ker θ = {f S f(x) = 0} we are done by the isomorphism theorem because R is simple.
44.
a. The map θ : R → R/A1 × × R/An with θ(r) = (r + A1, . . ., r + An) is a ring homomorphism with kernel A.
× R/An with θ(r) = (r + A1, . . ., r + An) is a ring homomorphism with kernel A.
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