4CCM121A, Abstract Algebra
Solution Sheet 9
1. Suppose that R is a ring with additive identity 0R and multiplicative
identity 0R . Recall that if r ∈ R and n ∈ Z, then n · r denotes the nth
multiple of r in the (additive) abelian group (R, +), and that nR = n · 1R .
(a) Prove that n · (rs) = (n · r)s = r(n · s) for all n ∈ Z, r, s ∈ R.
Solution: First we prove the case n = 0. By definition, 0·x = 0R for
all x ∈ R, so we have to prove that 0R = 0R s = r0R for all r, s ∈ R,
and this was part of Proposition 5.3.1.
Now we prove the formula for n > 0 by induction. For n = 1, we
have 1 · x = x for all x ∈ R (by definition), so the equations just say
rs = rs = rs. Now assume the equations holds for n, and we prove
them with n replaced by n + 1. By definition (n + 1) · x = n · x + x
for all x ∈ R, so
(n + 1) · (rs)
=
=
=
=
n · (rs) + rs
(n · r)s + rs
(n · r + r)s
((n + 1) · r)s
(by
(by
(by
(by
definition)
the induction hypothesis)
the distributive axiom)
definition).
Similarly
n · (rs) + rs = r(n · s) + rs = r(n · s + s) = r((n + 1) · s).
Finally we prove the formula for n < 0, in which case n·x is defined as
−((−n) · x). We know the formula holds with n replaced by −n > 0,
so
(−n) · (rs) = ((−n) · r)s = r((−n) · s).
The second part of Proposition 5.3.1 stated that −(xy) = (−x)y =
x(−y) for all x, y ∈ R. Therefore
n · (rs) = −((−n) · (rs)) = −((−n) · r)s = (−(−n) · r)s = (n · r)s
(the reasons for the equalities being: the definition, the case of −n,
the proposition, and the definition again). Similarly
n · (rs) = −((−n) · (rs)) = −r((−n) · s) = r(−(−n) · s) = r(n · s).
(b) Prove that n · r = (nR )r = r(nR ) for all n ∈ Z, r ∈ R.
Solution: Part (a) with s = 1 gives n · r = r(n · 1) = r(nR ) for all
r ∈ R. Similarly with r = 1 it gives n · s = (nR )s for all s ∈ R.
2. Suppose that R is a ring and S is a subset of R such that
• −1R ∈ S, and
• if x, y ∈ S, then x + y ∈ S and xy ∈ S.
Prove that S is a subring of R.
Solution: Since −1R ∈ S, we can apply the second assumption with
x = −1R and y = −1R to deduce that 1R = (1R )(1R ) = (−1R )(−1R ) ∈ S.
Now apply the second assumption again, but adding x = 1R and y = −1R
to conclude that 0R = 1R + (−1)R ∈ S. Finally note that if x ∈ S, then
since −1R ∈ S, the second assumption also implies that −x = (−1R )x ∈ S.
Therefore (S, +) is a subgroup of (R, +) that contains 1R and is closed
under multiplication, so S is a subring of R.
3. Which of the following subsets of Q are subrings?
(a) S = { a2 | a ∈ Z }
Solution: Not a subring since 1/2 ∈ S, but 1/4 = (1/2)(1/2) 6∈ S.
(b) S = { ab | a, b ∈ Z, b odd }
Solution: A subring since −1 ∈ S, and if x = a/b and y = c/d are
in S (where a, b, c, d ∈ Z with b and d odd), then
x + y = (ad + bc)/bd and xy = ac/bd
1
are also in S (since bd is odd). So by the criteria from the preceding
problem, S is a subring.
(c) S = { 2an | a, n ∈ Z, }
Solution: A subring since −1 ∈ S, and if x = a/2n and y = c/2m
are in S (where a, b, n, m ∈ Z), then
x + y = (2m a + 2n c)/2m+n
and xy = ac/2m+n
are also in S. So by the preceding problem, S is a subring.
n
(d) S = {0} ∪ { 2b | b, n ∈ Z, b odd }
Solution: Not a subring since 1 ∈ S and 2 ∈ S, but 1 + 2 = 3 6∈ S.
4. For each S in the preceding problem that is a subring, find S × .
Solution: (b) If x = a/b ∈ S × , then b is odd and x−1 = b/a ∈ S, which
requires a to be odd. So
S× = {
a
| a, b ∈ Z, a, b odd }.
b
(c) If x = a/2n ∈ S × , then x−1 = 2n /a ∈ S, which requires a = 2m for
some m ∈ Z. Therefore x = 2n−m = 2k for some k ∈ Z, and
S × = { 2k | k ∈ Z, }.
5. Suppose that (R, +R , ∗R ) and (S, +S , ∗S ) are rings. Define binary operations + and ∗ on R × S by
(r, s) + (r0 , s0 ) = (r +R r0 , s +S s0 )
and
(r, s) ∗ (r0 , s0 ) = (r ∗R r0 , s ∗S s0 ).
(Omit the symbols for the multiplicative operations and the subscripts on
the additive operations if you like.)
(a) Show that R × S is a ring with these operations.
Solution: We have already seen that (R × S, +) is a group, abelian
by Exercise 8.1.
In the proof of Prop. 4.11.2 (that the product of two groups is a
group), we saw that if ∗R and ∗S are associative, then so is the
operation ∗ on R × S, and that if these operations have identity 1R
and 1S , then (1R , 1S ) is an identity element for ∗.
All that’s left to prove is the distributive laws. Suppose (r, s), (r0 s0 )
and (r00 , s00 ) are elements of R × S. Then the definitions of the binary
operations on R × S give
(r, s) ∗ ((r0 , s0 ) + (r00 , s00 ))
=
(r, s) ∗ (r0 +R r00 , s0 +S s00 )
=
(r ∗R (r0 +R r00 ) , s ∗S (s0 +S s00 )),
and
(r, s) ∗ (r0 , s0 ) + (r, s) ∗ (r00 , s00 ) = (r ∗R r0 , s ∗S s0 ) + (r ∗R r00 , s ∗S s00 )
= ((r ∗R r0 ) +R (r ∗R r00 ) , (s ∗S s0 ) +S (s ∗S s00 )).
Since the distributive laws holds for the operations on R and S, we
have
r ∗R (r0 +R r00 ) = (r ∗R r0 ) +R (r ∗R r00 )
and s ∗S (s0 +S s00 ) = (s ∗S s0 ) +S (s ∗S s00 ).
Therefore (r, s)∗((r0 , s0 )+(r00 , s00 )) = ((r, s)∗(r0 , s0 ))+((r, s)∗(r00 , s00 )).
The proof that ((r0 , s0 )+(r00 , s00 ))∗(r, s) = ((r0 , s0 )∗(r, s))+((r00 , s00 )∗
(r, s)) is similar.
(b) Show that (R × S)× = R× × S × .
Solution: If (r, s) ∈ (R × S)× , then there is an element (r0 , s0 ) ∈
R × S so that
(r, s) ∗ (r0 , s0 ) = 1R×S = (r0 , s0 ) ∗ (r, s)
2
where 1R×S is the identity element (1R , 1S ) for ∗. This means that
(r ∗R r0 , s ∗S s0 ) = (1R , 1S ) = (r0 ∗R r, s0 ∗S s),
which means that r ∗R r0 = 1R = r0 ∗R r and s ∗ Ss0 = 1S = s0 ∗S s,
i.e., r ∈ R× and s ∈ S × , so (r, s) ∈ R× × S × . Conversely if (r, s) ∈
R× × S × , then there exist r0 ∈ R and s0 ∈ S so that
(r ∗R r0 , s ∗S s0 ) = (1R , 1S ) = (r0 ∗R r, s0 ∗S s),
i.e., (r, s) ∗ (r0 , s0 ) = 1R×S = (r0 , s0 ) ∗ (r, s), so (r, s) ∈ (R × S)× .
6. Let R be a commutative ring and let M2 (R) denote the ring of 2 × 2
matrices with coefficients
in R (under matrix addition and multiplication).
a b
For A =
∈ M2 (R), let det(A) = ad − bc ∈ R.
c d
(a) Prove that if A, B ∈ M2 (R), then
det(AB)= det(A)
det(B).
r s
w x
Solution: Let A =
and B =
be elements of
t u
y z
M2 (R). Then
rw + sy rx + sz
det(AB) = det
tw + uy tx + uz
= (rw + sy)(tx + uz) − (rx + sz)(tw + uy)
= rwuz + sytx − rxuy − sztw
= (ru − st)(wz − xy)
= det(A) det(B).
(b) Prove that the group of units in M2 (R) is
GL2 (R) = { A ∈ M2 (R) | det(A) ∈ R× }.
r s
∈ M2 (R). Let d = det(A) =
t u
ru − st. We must show that A has a multiplicative inverse in M2 (R)
if and only if d ∈ R× .
ud−1 −sd−1
If d ∈ R× , then d−1 ∈ R and we let B =
. Using
−td−1 rd−1
the axioms and basic properties of R (including commutativity), we
find that
rud−1 + s(−td−1 ) r(−sd−1 ) + srd−1
AB =
−1
−1
−1
) t(−sd−1 ) +
tud + u(−td
urd
−1
−1
(ru − st)d
(−rs + sr)d
=
−1
−1
(tu
−
ut)d
(−ts
+ ur)d
−1
−1
dd
0R d
=
−1
−1
0
d
R
dd
1R 0R
=
= I.
0R 1R
Solution: Suppose A =
Similarly we find that BA = I, so A is a unit in R× .
Conversely suppose that A is a unit in R, so AB = BA = I for
some B ∈ M2 (R). Then det(A) det(B) = det(AB) = det(I) = 1R ,
and det(B) det(A) = det(I) = 1R , so d = det(A) has an inverse
det(B) ∈ R. Therefore d ∈ R× .
(c) Prove that det : GL2 (R) → R× is a homomorphism.
Solution: We have already shown that det(AB) = det(A) det(B),
so det is a homomorphism.
(d) Prove that GL2 (Z2 ) is isomorphic to S3 .
Solution: The elements of GL2 (Z2 ) are as follows:
[1] [0]
[1] [1]
I=
, A=
, B=
[0] [1] [1] [0] [1] [1]
[0] [1]
C=
, D=
, E=
[0] [1]
[1] [0]
3
[0] [1]
,
[1] [1] [1] [0]
.
[1] [1]
The multiplication table is then:
I
A
B
C
D
E
I
I
A
B
C
D
E
A
A
B
I
D
E
C
B
B
I
A
E
C
D
C
C
E
D
I
B
A
D
D
C
E
A
I
B
E
E
D
C
B
A
I.
Comparing this with the multiplication table for S3 , we see that the
bijection defined by
I ←→ e,
A ←→ (123),
C ←→ (23), D ←→ (13),
is an isomorphism.
4
B ←→ (132),
E ←→ (12),