MATH 521A: Abstract Algebra
Homework 6 Solutions
1. Let R, S be rings. Consider the embedding map f : R → R × S given by f : r 7→ (r, 0S ).
Prove that f is a homomorphism.
We have f (r + r0 ) = (r + r0 , 0S ) = (r + r0 , 0S + 0S ) = (r, 0S ) + (r0 , 0S ) = f (r) + f (r0 ). We
also have f (rr0 ) = (rr0 , 0S ) = (rr0 , 0S 0S ) = (r, 0S )(r0 , 0S ).
2. Let R, S be rings. Consider the projection map f : R × S → R given by f : (r, s) 7→ r.
Prove that f is a homomorphism.
We have f ((r, s) + (r0 , s0 )) = f ((r + r0 , s + s0 )) = r + r0 = f ((r, s)) + f ((r0 , s0 )), and
f ((r, s)(r0 , s0 )) = f ((rr0 , ss0 )) = rr0 = f ((r, s))f ((r0 , s0 )).
3. We call a ring element x idempotent if x2 = x. Let R, S be rings, and f : R → S a
homomorphism. Suppose x ∈ R is idempotent. Prove that f (x) is idempotent.
Suppose that x is idempotent, i.e. x = x2 . We have f (x) = f (x2 ) = f (xx) = f (x)f (x), so
f (x)2 = f (x).
4. We call a ring element x nilpotent if there is some n ∈ N such that xn = 0. Let R, S be
rings, and f : R → S a homomorphism. Suppose x ∈ R is nilpotent. Prove that f (x) is
nilpotent.
Recall that f (0R ) = 0S . Let x ∈ R be nilpotent, i.e. xn = 0R . We have f (x)n =
f (x)f (x) · · · f (x) = f (xx · · · x) = f (xn ) = f (0R ) = 0S .
5. Let R, S be rings, and f : R → S a homomorphism. Define the kernel of f , Kerf = {r ∈
R : f (r) = 0S }. Prove that Kerf is a subring of R.
Because f (0R ) = 0S , we have 0R ∈ Kerf . Suppose that a, b ∈ Kerf . Then f (a) = 0S , f (b) =
0S . We have f (a + b) = f (a) + f (b) = 0S + 0S = 0S , so a + b ∈ Kerf ; this proves additive
closure. We also have f (ab) = f (a)f (b) = 0S 0S = 0S , so ab ∈ Kerf ; this proves multiplicative closure. By a theorem, −f (a) = f (−a), so 0S = f (a) + f (−a). But f (a) = 0S since
a ∈ Kerf , so 0S = 0S + f (−a) = f (−a). Hence (−a) ∈ Kerf .
6. Let R, S be rings, and f : R → S a homomorphism. Prove that f is injective (one-to-one)
if and only if Kerf = {0R }.
First, suppose that Kerf = {0R }. Let a, b ∈ R such that f (a) = f (b). But then
f (a − b) = f (a) − f (b) = 0S , so a − b ∈ Kerf . Since Kerf = {0R }, in fact a − b = 0R , so
a − b + b = 0R + b, so a = b.
Next, suppose that c ∈ Kerf with c 6= 0R . Then f (c) = f (0R ) = 0S , so f is not injective.
7. Let R, S be rings, and f : R → S a homomorphism. Suppose that S1 is a subring of S.
Prove that f −1 (S1 ) = {r ∈ R : f (r) ∈ S1 } is a subring of R.
First, 0S ∈ S1 since every subring contains zero. Since f (0R ) = 0S , we have 0R ∈ f −1 (S1 ).
Next, suppose a, b ∈ f −1 (S1 ). Then f (a), f (b) ∈ S1 . Since S1 is a subring, it is closed so
f (a + b) = f (a) + f (b) ∈ S1 , and f (ab) = f (a)f (b) ∈ S1 . Hence a + b, ab ∈ f −1 (S1 ). Since
S1 is a subring, it contains additive inverses, so −f (a) ∈ S1 . By a theorem f (−a) = −f (a),
so −a ∈ f −1 (S1 ).
8. Let R, S, T be rings, and f : R → S, g : S → T two homomorphisms. Prove that g ◦ f :
R → T is a homomorphism.
Let a, b ∈ R. We have g(f (a + b)) = g(f (a) + f (b) = g(f (a)) + g((f (b)), because f , g are
homomorphisms (respectively). Similarly, g(f (ab)) = g(f (a)f (b)) = g(f (a))g(f (b)).
9. Let R, S be rings, and f : R → S an isomorphism. Let g = f −1 , i.e. for all r ∈ R,
g(f (r)) = r and for all s ∈ S, f (g(s)) = s. Prove that g : S → R is an isomorphism.
First, g is a bijection because the inverse of a bijection is a bijection (or we can prove it
if we like). We have g(a + b) = g(f (g(a)) + f (g(b))) = g(f (g(a) + g(b))) = g(a) + g(b),
where we use the homomorphism property of f for the second equality. Similarly, we have
g(ab) = g(f (g(a))f (g(b))) = g(f (g(a)g(b))) = g(a)g(b).
10. Let S = {( ab 2b
a ) : a, b ∈ Z}, which is a subring
√ (two-by-two matrices with integer
√ of M2,2 (Z)
entries). Prove that S is isomorphic to Z[ 2] = {a + b 2 : a, b ∈ Z}, a subring of R.
√
√
The notation gives a big hint;define f : S → Z[ 2] via f : ( ab 2b
a ) 7→ a + b 2. We
√
√
a+a0 2(b+b0 )
a0 2b0
check f ( ab 2b
= (a + a0 ) + (b + b0 ) 2 = (a + b 2) + (a0 +
=f
a ) + b0 a0
b+b0 a+a0
√
0 +2bb0 2ab0 +2a0 b 0
0
a0 2b0
) ab0 2b
. We also have f ( ab 2b
= f aa
=
b0 2) = f (( ab 2b
0 a0
0
0 b+b0 a 2bb0 +aa0
a )) + f
a
b
a
a
√
√
√
0 2b0
0
0
0
0
0
0
a
2b
a
(aa + 2bb ) + (ab + ba ) 2 = (a + b 2)(a + b 2) = f (( b a )) f b0 a0 . To prove in√
√
0
0
)) = f ab0 2b
jection, suppose f (( ab 2b
. Then a + b 2 = a0 + b0 2, which rearranges as
0
a
a
√
0
0
a − a0 = (b0 − b) 2. If b0 = b, then a = a0 and ( ab 2b
) = ab0 2b
. If b0 6= b, then we divide by
0
a
a
√
2√
is rational, a contradiction. √Surjection is “obvious” due to the
b0 − b and discover that
√
definitions: let a + b 2 ∈ Z[ 2], then f (( ab 2b
a )) = a + b 2.
11. Recall the ring from HW4 #6: R has ground set Z and operations ⊕, defined as:
a ⊕ b = a + b − 1,
a b = a + b − ab
Prove that R is isomorphic to Z.
The hard part of this problem is finding the isomorphsim, which is f : R → Z via f (x) =
1 − x. We first prove a bijection; if f (x) = f (x0 ) then 1 − x = 1 − x0 so x = x0 . Also, for
a ∈ Z we have f (1 − a) = 1 − (1 − a) = a.
Now, f (a ⊕ b) = f (a + b − 1) = 1 − (a + b − 1) = 2 − a − b = (1 − a) + (1 − b) = f (a) + f (b),
and f (a b) = f (a + b − ab) = 1 − (a + b − ab) = 1 − a − b + ab = (1 − a)(1 − b) = f (a)f (b).
12. Recall the ring from HW4 #7: R has ground set Z and operations ⊕, defined as:
a ⊕ b = a + b − 1,
a b = ab − a − b + 2
Prove that R is isomorphic to Z.
The hard part of this problem is finding the isomorphsim, which is f : R → Z via f (x) =
x − 1. We first prove a bijection; if f (x) = f (x0 ) then x − 1 = x0 − 1 so x = x0 . Also, for
a ∈ Z we have f (a + 1) = (a + 1) − 1 = a.
Now, f (a ⊕ b) = f (a + b − 1) = a + b − 2 = (a − 1) + (b − 1) = f (a) + f (b), and
f (a b) = f (ab − a − b + 2) = ab − a − b + 2 − 1 = ab − a − b + 1 = (a − 1)(b − 1) = f (a)f (b).