Mathematics 2215: Rings, fields and modules
Solutions to tutorial exercise sheet 1
Let R and S be rings.
1. If R ≈ S, show that R is unital if and only if S is unital.
Solution Since R ≈ S, there is an isomorphism θ : R → S.
Suppose R is unital. We claim that θ(1R ) is the unity of S. For if y ∈ S then y = θ(x)
for some x ∈ R, since θ is surjective, so θ(1R )y = θ(1R )θ(x) = θ(1R x) = θ(x) = y and
yθ(1R ) = θ(x)θ(1R ) = θ(x1R ) = θ(x) = y. So ∀y ∈ R, θ(1R )y = y = yθ(1R ), so θ(1R ) = 1S
and S is unital.
Conversely, suppose that S is unital. Then θ−1 is an isomorphism, so applying the argument
above to θ−1 instead of θ shows that θ−1 (1S ) is the unity of R, so R is unital. [Alternatively,
we could argue that R ≈ S ⇐⇒ S ≈ R, and apply the first part of the proof directly (using
a different isomorphism)].
2. If R ≈ S, show that R is commutative if and only if S is commutative.
Solution Since R ≈ S, there is an isomorphism θ : R → S.
Suppose R is commutative. Let y1 , y2 ∈ S. Then, since θ is surjective, there are x1 , x2 ∈ R so
that θ(x1 ) = y1 and θ(x2 ) = y2 . Because R is commutative we have x1 x2 = x2 x1 , so y1 y2 =
θ(x1 )θ(x2 ) = θ(x1 x2 ) = θ(x2 x1 ) = θ(x2 )θ(x1 ) = y2 y1 . This proves that ∀y1 , y2 ∈ S, y1 y2 = y2 y1 ,
so S is commutative.
Conversely, suppose that S is commutative. Then θ−1 is an isomorphism, so applying the
argument above to θ−1 instead of θ shows that R is commutative. [Alternatively, we could
argue that R ≈ S ⇐⇒ S ≈ R, and apply the first part of the proof directly (using a different
isomorphism)].
3. If R ≈ S, show that R has zero-divisors if and only if S has zero-divisors.
Solution Since R ≈ S, there is an isomorphism θ : R → S.
Suppose R has a zero-divisor x. Then x 6= 0R , and there is a y ∈ R with y 6= 0R and xy = 0R .
Hence θ(x)θ(y) = θ(xy) = θ(0R ) = 0S . Because θ is injective and x, y 6= 0R and θ(0R ) = 0S ,
we have θ(x) 6= 0S and θ(y) 6= 0S . Hence θ(x) is a zero-divisor in S, so S has a zero-divisor.
Conversely, suppose that S has a zero-divisor. Then θ−1 is an isomorphism, so applying the
argument above to θ−1 instead of θ shows that R has a zero-divisor. [Alternatively, we could
argue that R ≈ S ⇐⇒ S ≈ R, and apply the first part of the proof directly (using a different
isomorphism)].
4. If R ≈ S, show that R is a division ring if and only if S is a division ring.
Solution Since R ≈ S, there is an isomorphism θ : R → S.
Suppose R is a division ring. Then R is unital, so S is unital by Question 1, and as shown in
that solution, we have θ(1R ) = 1S . If y ∈ S with y 6= 0S then y = θ(x) for some x ∈ R, since θ
is surjective, and since θ(0R ) = 0S 6= y we have x 6= 0R . Since R is a division ring, x is invertible
in R. Moreover, θ(x−1 )θ(x) = θ(x−1 x) = θ(1R ) = 1S and θ(x)θ(x−1 ) = θ(xx−1 ) = θ(1R ) = 1S ,
so θ(x−1 ) is the inverse of y = θ(x) in the ring S, so y is invertible in S. We’ve proven that
∀y ∈ S × , y is invertible in S, so S is a division ring.
Conversely, suppose that S is a division ring. Then θ−1 is an isomorphism, so applying the
argument above to θ−1 instead of θ shows that R is a division ring. [Alternatively, we could
argue that R ≈ S ⇐⇒ S ≈ R, and apply the first part of the proof directly (using a different
isomorphism)].
5. If R ≈ S, show that R is a field if and only if S is a field.
Solution Since a ring is a field if and only if (by definition) it is both a commutative ring
and a division ring, this is immediate from Questions 2 and 4.
6. Find all the ideals of M (2, R).
Solution See the next homework solutions.
7. Find all the ideals of R.
Solution R is a field, so it only has the trivial ideals {0} and R. [You might want to prove
this without using the general result about fields; I guess you’d wind up repeating the general
proof, but for real numbers only instead of a general field.]
8. Find all the ideals of Z20 .
Solution Here’s a way to do this question without the correspondence theorem (which we’ll
prove, I think, before long). Using the correspondence theorem, it becomes much easier!
If k ∈ Z and [`] ∈ Z20 , let us write kZ20 = {[k`] : ` ∈ Z}. We claim that the ideals are
1Z20 = Z20 ,
2Z20 ,
4Z20 ,
5Z20 ,
10Z20
and 20Z20 = {[0]}.
These sets have sizes 20, 10, 5, 4, 2 and 1 respectively, so they’re all different. For example,
4Z20 = {[0], [4], [8], [12], [16], [20], [24], [28], [32], [36], [40], [44], [48], [52], [56], [60], [64], [68], [72], [76]}
= {[0], [4], [8], [12], [16]}
since we have [20] = [0], [24] = [4], etc.
Now we show that kZ20 C Z20 , for any k ∈ Z. Let’s write R = Z20 and I = kZ20 for clarity.
Then
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• I 6= ∅ since [0] ∈ I, and if x, y ∈ I then x = [km] and y = [kn] for some m, n ∈ Z, so
x − y = [k(m − n)] ∈ I, so I is a subgroup of R under addition; and
• if x = [km] ∈ I and z = [`] ∈ R then xz = [km`] ∈ I and zx = xz ∈ I, so the ideal
property is satisfied.
Finally, let’s show that the five ideals we listed above that the only ideals of Z20 . If I is an
ideal of Z20 , the it is an additive subgroup. Z20 is a cyclic group, so its subgroup I is cyclic,
which means that there is some x ∈ I so that I = {nx : n ∈ Z}. We have x = [k] for some
k ∈ {1, 2, . . . , 20}, so I = {[kn] : n ∈ Z} = kZ20 .
Now, it’s a nice exercise to prove that kZn = gZn where g = gcd(k, n). This uses the fact that
you can always find integers k 0 , n0 so that gcd(k, n) = kk 0 + nn0 . I suppose I’m cheating a bit
here, assuming you don’t know that fact. But we hinted at it in the groups course when we
discussed Euclid’s algorithm, and I expect we’ll prove it properly later in this course.
Since gcd(k, 20) is a divisor of 20, we can assume that k ∈ {1, 2, 4, 5, 10, 20} without changing
the ideal I = kZ20 . But then I = kZ20 is in the list above.
9. If I C R, show that M (2, I) C M (2, R) and M (2, R)/M (2, I) ≈ M (2, R/I).
a12
I+a11 I+a12
Solution Let θ : M (2, R) → M (2, R/I) be given by θ([ aa11
21 a22 ]) = [ I+a21 I+a22 ]. Then θ is a
b11 b12
11 a12
homomorphism, because if A = [ aa21
a22 ] and B = [ b21 b22 ] are in M (2, R) then the i, j entry of
θ(A + B) is I + (aij + bij ) = (I + aij ) + (I + bij ) (by the definition of addition in R/I) which
is the i, j entry of θ(A) + θ(B), so θ(A + B) = θ(A) + θ(B), and the i, j entry of θ(AB) is
I + (ai1 b1j + ai2 b2j ) = (I + ai1 )(I + b1j ) + (I + ai2 )(I + b2j ), which is the i, j entry of θ(A)θ(B),
so θ(AB) = θ(A)θ(B).
So θ is a homomorphism. It’s obviously surjective (why?).
I+0
The zero element of M (2, R/I) is [ I+0
I+0 I+0 ] so the kernel of θ consists of the matrices A so that
all of the entries aij of A have I + aij = I + 0 ⇐⇒ aij ∈ I, so ker θ = M (2, I). By the first
isomorphism theorem, M (2, R)/M (2, I) = M (2, R)/ ker θ ≈ θ(M (2, R)) = M (2, R/I).
10. If R is unital and I C R, show that the following are equivalent:
(a) I = R
(b) 1R ∈ I
Deduce that the only ideals of a field F are {0} and F .
Solution (a) =⇒ (b): Since R is unital we have 1R ∈ R. If I = R then 1R ∈ R = I, so
1R ∈ I.
(b) =⇒ (a): If 1R ∈ I then for any x ∈ R we have x1R = x ∈ I by the ideal property. So
R ⊆ I. Since I C R we have I ⊆ R. Hence I = R.
So (a) and (b) are equivalent.
If F is a field then it is a unital ring, and if I is an ideal of F apart from {0}, then I contains
some nonzero element a of F . Since F is a field and a ∈ F × , the element a is invertible in F .
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So a ∈ I and a−1 ∈ F , so by the ideal property for I we have a−1 a = 1F ∈ F . Hence I = F by
(b) =⇒ (a) above.
[Actually, we haven’t used the fact that F is commutative above. So the same proof will work
for any division ring.]
11. If R is commutative and x ∈ R, show that the set
(x) = {xr : r ∈ R}
is an ideal of R. Is this true if R is not assumed to be commutative?
Solution We have x0R = 0R ∈ (x), so (x) 6= ∅. If a, b ∈ (x) then a = xr and b = xs for some
r, s ∈ R, so a − b = x(r − s) ∈ (x). So (x) is an additive subgroup of R. If a ∈ (x) and y ∈ R
then a = xr for some r ∈ R, so ay = x(ry) ∈ (x) and ya = ay ∈ (x). So the ideal property
holds for (x). Hence (x) is an ideal of R.
This doesn’t always work if R is not commutative. Here’s an example. Suppose that R =
M (2, R) and X = [ 10 00 ]. Let S = {XA : A ∈ M (2, R)}; then S is the set of all matrices in
M (2, R) whose second row is zero. In particular, X ∈ S, so if S is were an ideal in M (2, R),
then BX ∈ S for every B ∈ M (2, R); but taking B = [ 01 00 ] we see that BX = B, so BX 6∈ S.
So S cannot be an ideal in M (2, R).
12. Prove that R × S ≈ S × R.
Solution Let θ : R × S → S × R, (r, s) 7→ (s, r). Then θ is clearly a bijection, and if x = (r, s)
and x0 = (r0 , s0 ) are in R×S, then θ(x+x0 ) = θ((r+r0 , s+s0 )) = (s+s0 , r+r0 ) = (s, r)+(s0 , r0 ) =
θ(x) + θ(x0 ), and θ(xx0 ) = θ((rr0 , ss0 )) = (ss0 , rr0 ) = (s, r)(s0 , r0 ) = θ(x)θ(x0 ). Hence θ is an
isomorphism.
13. Prove that R × {0} ≈ R and {0} × R ≈ R.
Solution Let θ : R → R × {0}, x 7→ (x, 0). Then θ is a bijection [why?] and if x, x0 ∈ R then
θ(x + x0 ) = (x + x0 , 0) = (x, 0) + (x0 , 0) = θ(x) + θ(x0 ), and θ(xx0 ) = (xx0 , 0) = (x, 0)(x0 , 0) =
θ(x)θ(x0 ), so θ is an isomorphism.
By the previous question, {0} × R ≈ R × {0}, and we’ve shown that R × {0} ≈ R. Hence
{0} × R ≈ R, by the transitivity of ≈.
14. Prove that R/{0} ≈ R.
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Solution Let θ : R → R be the identity map. Then θ is an isomorphism so θ(R) = R and
ker θ = {0}, so by the first isomorphism theorem, R/{0} ≈ θ(R) = R.
Or, you can prove this directly by checking that the map φ : R → R/{0} defined by φ(x) =
{0} + x is an isomorphism. This is a tiny bit more fiddly, but it’s not hard and you should do
it.
15. Prove that R/R ≈ {0}.
Solution Let θ : R → R be the zero homomorphism. Then θ(R) = {0} and ker θ = R, so
R/R ≈ {0} by the first isomorphism theorem.
Actually, this really is overkill in this case. A direct proof goes like this: R/R = {R} is a ring
with one element [why? I suppose this is what you need to convince yourself of] and the map
φ : R/R → {0} given by φ(R) = 0 is an isomorphism, because it’s a map between sets with one
element each so it’s bijective, and it’s a homomorphism because it’s a zero homomorphism.
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