114 Groups and Rings 2008–09
Suggested solutions to exercise set 10
May 6, 2009
38.2 This is not a homomorphism since θ(1 + 1) = 4 6= 2 = θ(1) + θ(1).
38.4 This is not a homomorphism since θ(−1) = 1 6= −1 = −θ(1).
38.6 This is not a homomorphism since
θ([ 00 10 ][ 01 00 ]) = θ([ 10 00 ]) = [ 10 00 ] 6= [ 00 01 ] = [ 01 00 ][ 00 10 ] = θ([ 00 10 ])θ([ 01 00 ]).
38.10 The statement is true, since if e is a unit element for R and θ : R → S is a
any ring homomorphism then for each x ∈ θ(S) we can write x = θ(a) for
some a ∈ R, and
xθ(e) = θ(a)θ(e) = θ(ae) = θ(a) = x and θ(e)x = θ(e)θ(a) = θ(ea) = θ(a) = x.
So θ(e) is a unit element for θ(S), which is a typical homomorphic image
of R.
38.11* (a) This is the additive cyclic group generated by 21 , which is { k2 : k ∈ Z}.
(b) This is the multiplicative cyclic group generated by 21 ,
which is { 21n : n ∈ Z}.
(c) As in Exercise 25.6, this is { 2kn : k, n ∈ Z}.
(d) Any ideal containing 12 is a subring, so contains the subring in (c). So it
contains 210 = 1, the unit of Q. An ideal containing the unit is equal to the
whole ring, so the answer is Q.
(e) Let F this subfield. Then F contains the additive subgroup in (a), so F
contains Z. Since the inverse of a non-zero element of F is in F , we see that
F contains { k1 : k ∈ Z, k ≥ 1}, so F contains { nk : n, k ∈ Z, k ≥ 1} = Q. So
F = Q.
38.12 Every subring of Z is equal to (n) for some n ∈ Z, so is a principal ideal of Z.
38.16 Consider the principal ideals (a) = {ra : r ∈ Z12 } for a ∈ Z12 . Observe that
(−a) = (a), so if we employ the usual abuse of notation and write k instead
of [k] for k ∈ Z then these are
(0) = {0},
(1) = (5) = (7) = Z12 ,
(3) = (9) = {0, 3, 6, 9},
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(2) = (10) = {0, 2, 4, 6, 8, 10},
(4) = {0, 4, 8}, (6) = {0, 6}.
There are no other ideals of Z12 since every ideal is an additive subgroup,
and we just have written down all of the additive subgroups of Z12 .
38.20 By Theorem 38.1, ker θ is an ideal in F . Since F is a field, Example 38.4
shows that the only possibilities are
Thm 38.1
ker θ = {0} ⇐⇒ θ is one-to-one, or
ker θ = F ⇐⇒ θ(a) = 0 for all a ∈ F .
38.22* We computed the centre of M (2, Z) in Exercise 25.23. The centre of M (2, Z)
is not an ideal of M (2, Z) since, for example, [ 10 01 ] is in the centre and [ 10 00 ]
is in M (2, Z) but [ 10 01 ][ 10 00 ] = [ 10 00 ] is not in the centre.
39.6 Recall that I + x = I + y ⇐⇒ x − y ∈ I. (This is a consequence of
Lemma 16.1). So
R/I is commutative ⇐⇒ ∀a, b ∈ R : (I + a)(I + b) = (I + b)(I + a)
⇐⇒ ∀a, b ∈ R : I + ab = I + ba
⇐⇒ ∀a, b ∈ R : ab − ba ∈ I.
39.7 Let n ∈ Z, n > 0. Observe that
(n) = {k ∈ Z : n|k}.
If (n) is a prime ideal and n = ab for some a, b ∈ Z with a, b ≥ 1 then
n = ab ∈ (n), so a ∈ (n) or b ∈ (n), so n|a or n|b. Since 1 ≤ a, b ≤ n, this
forces one of a, b to be 1 and the other to be n. So the only factors of n are
1 and n; so n is prime.
Conversely, if n is prime then (n) 6= Z and if a, b ∈ (n) then n|ab. Since n is
prime, this implies that n|a or n|b, so either a ∈ (n) or b ∈ (n). So (n) is a
prime ideal.
39.8 Suppose that P is a prime ideal of R with P 6= R. Since R is unital, R/P
is unital, and since P 6= R the ring R/P contains more than one element, so
the unit is not equal to the zero element in this ring. For a, b ∈ R we have
(P + a)(P + b) = 0R/P = P ⇐⇒ P + ab = P ⇐⇒ ab ∈ P.
Since P is a prime ideal, this gives a ∈ P or b ∈ P , that is, P + a = P or
P + b = P . Since P = 0R/P , this shows that R/P has no zero divisors. So it
is an integral domain.
Conversely, if R/P is an integral domain and a, b ∈ R then ab ∈ P ⇐⇒
(P + a)(P + b) = P so P + a = P or P + b = P (since R/P is an integral
domain with zero element P ), i.e. a ∈ P or b ∈ P . So P is a prime ideal.
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