M a 5b HOMEWORK 3 SOLUTION
WINTER 10
The exercises are taken from the text, Abstract Algebra (third edition) by Dummit and Foote.
Page 267,1. Note (1−e)2 = 1−2e+e2 = 1−e. Take r, s ∈ R, then re+
se = (r + s)e, −(re) = (−r)e,r(se) = (rs)e, (se)r = (sr)e. Therefore
Re is an ideal. r(1 − e) + s(1 − e) = r + s − (r + s)e = (r + s)(1 − e),
−(r(1 − e)) = (−r)(1 − e),r(s(1 − e)) = (rs)(1 − e).(s(1 − e))(r) =
s((1 − e)r) = s(r − er) = s(r − re) = sr(1 − e). So R(1 − e) is an ideal.
Now ere = re2 = re, and r(1 − e)(1 − e) = r(1 − e), i.e e and 1 − e are
identities for Re and R(1 − e).
Since (re)(s(1−e)) = 0, and e+(1−e) = 1, we have (Re)(R(1−e)) =
{0}, and Re and R(1 − e) are coprime. By the Chinese Remainder
Theorem, R ∼
= R/Re × R/R(1 − e). By the Second Isomorphism
Theorem, R/Re = (Re + R(1 − e))/Re ∼
= Re/(Re ∩ R(1 − e)) = Re.
∼
Similarly, R/R(1 − e) ∼
Re.
Hence
R
=
= Re × R(1 − e)
Page 268,6. Let aij be the coefficient of degree j in fi (x) for i =
1, 2, . . . , k, j = 0, 1, . . . , d. By the Chinese Remainder Theorem, we
have a surjective map Z −→ Z/n1 Z × Z/n2 Z × · · · × Z/nk Z, so there
exists some aj ∈ Z such that aj ≡ aij mod ni . Therefore the coefficients
of f (x) = a0 + a1 x + · · · + an xn with the coefficients of fi (x) mod ni .
If fi (x) is monic, then aid = 1 for all i, so we can choose ad = 1, and
f (x) is monic.
Page 278,10. Since Z[i] is ED, and I is a nonzero ideal, we can write
I = (α) for some α ∈ Z[i]. Now for any γ ∈ Z[i], we can write
γ = αη + β with η, β ∈ Z[i] and N (β) < N (α)/2. Therefore, the set of
representatives in Z[i]/I are all β with N (β) < N (α)/2, which is finite.
Page 279,11. (a)If e is a least common multiple, then e ∈ (a), and
e ∈ (b). Since the intersection of two ideals is an ideal, e ∈ (a) ∩ (b).
If e0 ∈ (a) ∩ (b), then a|e0 , and b|e0 . So we have e|e0 , and (e0 ) ⊆
(e). Therefore e is a generator for the unique largest principal ideal
contained in (a) ∩ (b)
1
2
M A 5B HOMEWORK 3 SOLUTION
WINTER 10
(b) A Euclidan Domain is a PID, so (a)∩(b) = (e0 ) for some (e0 ) ∈ R.
If a|f and b|f , then (f ) ⊆ (a) ∩ (b) = (e0 ), and e0 |f . Hence e0 is the
least common multiple up to a unit.
(c) Let d = (a, b), and e = ab
. Now a|e and b|e. Take any e0 such
d
that a|e0 , b|e0 , then e0 = bk for some k, and a|bk. By problem 4, ad |k.
|bk = e0 Hence e is the least common multiple.
Therefore e = ba
d
Page 279,12. Suppose (M, N ) = 1, then M is a unit in Z/N Z, and
M ϕ(N ) ≡ 1(mod N ). If d and d0 are inverses in mod ϕ(N ), we can write
0
dd0 = 1 + ϕ(N )k for some interger k. Now M1d ≡ M dd ≡ M ϕ(N )k+1 ≡
1k M ≡ M (mod N ).