114 Groups and Rings 2008–09
Suggested solutions to exercise set 5
April 1, 2009
25.2 If k ∈ Z then [2k] [5] = [10k] = [0] in Z10 , so [2], [4], [6], [8] and [5] are zero
divisors. On the other hand, if 0 < n < 10 and [n] [m] = [0] in Z10 but n is
not even or divisible by 5, then 10|nm and (10, n) = 1 so 10|m, so [m] = [0];
so [n] is not a zero divisor. So the zero divisors in Z10 are [2], [4], [5], [6], [8].
25.6 Let R = { 2mn : m, n ∈ Z, n ≥ 0} and let S be the smallest subring of R with
1
∈ S.
2
Let m, n ∈ Z with n ≥ 0. Since S is a ring with 12 ∈ S, it follows that
( 12 )n = 21n ∈ S and so S also contains m( 21n ) = 2mn . So R ⊆ S.
Clearly
1
2
=
1
21
∈ R. Moreover, R is a subring of R by Theorem 25.2, since
– R is nonempty.
t
n
s
∈R
– if m, n, s, t ∈ Z with n, t ≥ 0 then 2mn + 2st = 2 m+2
2n+t
m s
ms
and 2n · 2t = 2n+t ∈ R. So R is closed under addition and multiplication.
– if m, n ∈ Z with n ≥ 0 then − 2mn =
negation.
(−m)
2n
∈ R, so R is closed under
So R is a subring of R containing 12 ; so S ⊆ R by the definition of S.
So S = R.
25.23 If x = [ ac db ] is in the centre of M (2, Z) then
[ ac 00 ] = [ ac db ][ 10 00 ] = x[ 10 00 ] = [ 10 00 ]x = [ 10 00 ][ ac db ] = [ a0 0b ]
so b = c = 0 and x = [ a0 d0 ]. Similarly,
[ 00 a0 ] = [ a0 d0 ][ 00 10 ] = x[ 00 10 ] = [ 00 10 ]x = [ 00 10 ][ a0 d0 ] = [ 00 d0 ]
so a = d. So x = [ a0 a0 ] = aI where I = [ 10 01 ] is the 2 × 2 identity matrix. All
such matrices are in the centre of M (2, Z) since for any y ∈ M (2, Z),
(aI)y = a(Iy) = ay = y(aI).
So the centre of M (2, Z) is {aI : a ∈ Z}.
1
26.2 No such example exists: any finite integral domain is a field (Theorem 26.1).
26.6 Z10 is a commutative ring and [2] is a zero divisor in Z10 .
nh
i
o
0 a b
0 0 c : a, b, c ∈ Z
26.10 T0 (3, Z) =
is a noncommutative ring without unity
0 0 0
[with operations inherited from M (3, Z)].
Alternatively, observe that if R is noncommutative and S is any nonunital
ring then the direct sum R × S is noncommutative and nonunital. So the
ring M (2, Z) × E (where E is the [nonunital] ring of even integers) will do.
26.18 Let F1 and F2 be fields with unit elements e1 and e2 , respectively. The unit
of F1 × F2 is then (e1 , e2 ). However, for any (f1 , f2 ) ∈ F1 × F2 ,
(e1 , 0)(f1 , f2 ) = (e1 f1 , 0f2 ) = (f1 , 0) 6= (e1 , e2 ).
So (e1 , 0) is a non-zero element of F1 × F2 with no multiplicative inverse in
F1 × F2 . So F1 × F2 is not a field.
27.2 Since R ≈ S, there is an isomorphism θ : R → S. Let a, b ∈ S. Then
a = θ(x) and b = θ(y) for some x, y ∈ R, since θ is onto, and xy = yx
since R is commutative. So
ab = θ(x)θ(y) = θ(xy) = θ(yx) = θ(y)θ(x) = ba.
Since a, b ∈ S were arbitrary, S is commutative.
27.6 If n = m = 1 then clearly (mn)e = e = (me)(ne).
Suppose that (mn)e = (me)(ne) for some m, n ∈ Z with m, n ≥ 1. Then
(m + 1)n e = (mn + n)e
= (mn)e + ne [by the “laws of exponents”, p. 77]
= (me)(ne) + e(ne) [since (mn)e = (me)(ne) and e(ne) = ne]
= (me + e)(ne) [by distributivity]
= (m + 1)e (ne) [by the laws of exponents].
Similarly, m(n + 1) e = (mn+ m)e = (mn)e + me = (me)(ne) + (me)e =
(me)(ne + e) = (me) (n + 1)e .
By induction, (mn)e = (me)(ne) for all m, n ∈ Z with m, n ≥ 1.
If m < 0 < n then by Theorem 24.2, (mn)e = −((−m)n)e = −((−m)e)(ne) =
(me)(ne). Similarly, if n < 0 < m then (mn)e = −(m(−n))e = −(me)((−n)e) =
(me)(ne). If m, n < 0 then (mn)e = ((−m)(−n)e) = ((−m)e)((−n)e) =
(−me)(−ne) = (me)(ne). Finally, if m = 0 or n = 0 then me = 0 or ne = 0
so (mn)e = 0e = 0 = (me)(ne).
2
27.21 In any ring we have x2 = (−x)2 . So if x ∈ R then x = x2 = (−x)2 = −x, so
2x = x + x = 0. If x, y ∈ R then x + y = (x + y)2 = x2 + xy + yx + y 2 =
x + xy + yx + y, so subtracting x + y from each side gives xy + yx = 0, so
xy = −yx = yx (using the fact that z = −z for any z ∈ R). Since x, y ∈ R
were arbitrary, R is commutative.
For example, R = Z2 is such a ring.
Another example is in Exercise 24.22: the ring P(S) satisfies A2 = A∩A = A
for any A ⊆ S.
27.25 (a) Let S = Z × R with the given operations. To check that S is a ring:
– (S, +) is the direct product of the Abelian groups (Z, +) and (R, +), so
it is an Abelian group.
– if x = (m, a), y = (n, b) and z = (p, c) are elements of S then
x(yz) = (m, a)(np, pb + nc + bc)
= (mnp, m(pb + nc + bc) + npa + a(pb + nc + bc))
= (mnp, mpb + mnc + mbc + npa + pab + nac + abc)
and
(xy)z = (mn, na + mb + ab)(p, c)
= (mnp, p(na + mb + ab) + mnc + (na + mb + ab)c)
= (mnp, mpb + mnc + mbc + npa + pab + nac + abc) = x(yz).
So multiplication is associative.
– if x = (m, a), y = (n, b) and z = (p, c) are elements of S then
x(y + z) = (m, a)(n + p, b + c)
= (m(n + p), (n + p)a + m(b + c) + a(b + c))
= (mn + mp, na + mb + ab + pa + mc + ac) = xy + xz,
and similarly (x + y)z = xz + yz. So the distributivity laws hold.
(b) For (m, a) ∈ S,
(m, a)(1, 0) = (m, 1a + m0 + a0) = (m, a) and
(1, 0)(m, a) = (m, m0 + 1a + 0a) = (m, a).
So (1, 0) is a unit for S.
(c) Let θ : R → R0 , θ(a) = (0, a). If θ(a) = θ(b) then (0, a) = (0, b) so a = b;
so θ is one-to-one. Moreover, θ(a+b) = (0, a+b) = (0, a)+(0, b) = θ(a)+θ(b)
and (0, a)(0, b) = (0, 0a+0b+ab) = (0, ab), so θ(ab) = θ(a)θ(b). So θ preserves
both ring operations, so its image, R0 = θ(R), is a subring of S and (since θ
is one-to-one), R ≈ θ(R) = R0 .
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