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Homework 3 Solutions
Problem 1. Suppose F is a field and f is a polynomial with coefficients in F . Let a be the leading coefficient of
f . Prove that there exists an integer m and irreducible monic polynomials p1 , . . . , pm such that f = ap1 · · · pm .
Proof. Suppose for a contradiction that there is a polynomial f which cannot be factored as above. Then the
set X of polynomials with degree greater than 1 which cannot be factored as above would be nonempty and
thus has an element of minimal degree. Let p ∈ X have minimal degree. Note that p ∈ X implies that p is not
irreducible. Therefore p = ab for some polynomials a, b ∈ F [x] with degree less than p. Since p was of minimal
degree, a, b 6∈ X and therefore have a factorizations as above which implies the product of the factorizations
would be a factorization for p which contradicts p ∈ X. Hence X must be empty and the result follows.
Problem 2. With the notation from problem 1, prove that if q1 , . . . , qr are irreducible monic polynomials such
that f = aq1 · · · qr , then r = m and there exists a permutation s such that, for all i, qi = ps(i) .
Proof. We proceed by strong induction on the degree of f . For deg f = 1 the result holds since f is irreducible.
Now suppose that the result holds for polynomials with degree less than n. If q1 · · · qr = f /a = p1 · · · pm then
q1 | p1 · · · pm ⇒ q1 | pi for some i, by relabeling if necessary we can assume that i = 1. Since p1 is monic and
irreducible this implies that q1 = p1 . Thus q2 · · · qr = p2 · · · pm and by the inductive hypothesis r = m and
there is permutation s such that qi = ps(i) for i ≥ 2. Extend the permutation by s(1) = 1 and the induction is
complete.
Problem 4.4.3. Find all integer roots of the following equations
(b) p(x) = y 3 − 9y 2 − 24y + 216
(d) q(x) = x5 − 34x3 + 29x2 + 212x − 300
Solution. (b) By proposition 4.4.1, if c ∈ Z is a root of p(x) then c | 216. Moreover, for any n ∈ Z (c − n) | p(n)
(c.f. example 4.4.2). In particular c − 1 | f (1) = 8 · 23. Thus c has possible values: 2, 3, 5, 9, 24, 47, 93, 185, −1,
−3, −7, −22, −45, −91, −183. Among the list the only divisors of 216 are: 2, 3, 9, 24, −1, −3. One can check
that the only root in the list is 9.
(d) Similar to (b), if c ∈ Z is a root of q(x) then c | 300 and c − 1 | 92. Thus c has possible values: 5, −3, −1,
3, 2. One can check that −3 and 2 are the only roots among the list. Moreover q(x) = (x + 3)(x − 2)g(x) where
g(x) = x3 − x2 − 27x + 50 and has g(2) = 0 and g(−3) 6= 0. Thus 2 and -3 are the only roots of q(x) where 2 is
a repeated root.
Problem 4.4.5. Use Eisenstein’s criterion to show that each of these polynomials is irreducible over the field of
rational numbers.
(b) f (x) = x6 + x3 + 1
(d) g(x) = x3 − 3x2 + 9x − 10.
Proof. (b) One can check that f (x + 1) = x6 + 6x5 + 15x4 + 21x3 + 18x2 + 9x + 3. Thus f (x + 1) satisfies
Eisenstein’s criterion for p = 3 and hence f (x + 1) is irreducible which implies that f (x) is irreducible.
(d) Similarly, one can verify that g(x + 1) = x3 + 6x − 3. Hence g(x + 1) satisfies Eisenstein’s criterion for
p = 3 and thus g(x + 1) is irreducible which implies that g(x) is also irreducible.
Problem 4.4.14. Let m and n be positive integers. Prove that xm − 1 is a factor of xn − 1 in Q[x] if and only
if m | n.
Proof. First if m | n then n = mk for some k ∈ Z and hence xn − 1 = (xm − 1)(xm(k−1) + xm(k−2) + · · · + xm + 1).
For the other implication consider the ring R = Q[x]/hxm − 1i. By the division algorithm we can write
n = mq + r where 0 ≤ r < m. Since xm − 1 | xn − 1 it follows that [0] = [xn − 1] = [xmq+r − 1] = [xmq · xr − 1]
in R. However [xm ] = [1] in R so the previous equality becomes [0] = [xr − 1]. But this would imply that
xm − 1 | xr − 1 which is a contradiction unless xr − 1 = 0, i.e. r = 0. Therefore m | n.
Problem 5.1.7. An element a of a commutative ring R is called nilpotent if an = 0 for some positive integer n.
Prove that if u is a unit in R and a is nilpotent, then u − a is a unit in R.
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Proof. First consider 1 + a. Recall that x − a | xk − ak for any positive integer k (lemma 4.1.8), i.e. xk − ak =
(x − a)fk (x) for some polynomial fk (x) ∈ R[x]. Thus 1 − ak = (1 − a)fk (1). Since a is nilpotent an = 0 for some
positive integer n and therefore 1 = 1 − an = (1 − a)fn (1) and fn (1) ∈ R. Hence 1 − a ∈ R× where R× is the
group of units of R.
Now if u is a unit then u−1 a is nilpotent as R is commutative. Thus 1 − u−1 a ∈ R× and u − a = u(1 − u−1 a) ∈
R× since R× is a group under multiplication.
Problem 5.1.20. An element e of a ring R is said to be idempotent if e2 = e. Find all idempotent elements of
the following rings.
(a) Z8 and Z9
(b) Z10 and Z12
Proof. (a) One can check that the idempotents of Z8 are 0 and 1 and the idempotents of Z9 are 0 and 1
(b) One can check that the idempotents of Z10 are 0, 1, 5 and 6 while the idempotents of Z12 are 0, 1, 4 and 9
m n
Problem
5.2.6. Show that the set of all matrices over Z of the form [ 2n
m ] is a ring isomorphic to the ring
√
Z[ 2].
m n
Proof. Let M = {[ 2n
m ] | m, n ∈ Z} ⊆ M2 (Z). To check that M is a ring it suffices to show its a subgring of
M2 (Z) which is easily checked.√
√
m n
Define a map f : M → Z[ 2] where f ([ 2n
m ]) = m + n 2. This
√ map is clearly a bijection and it’s easily
verified to be a ring homomorphism. Hence M is isomorphic to Z[ 2].
Problem 5.3.14. Let R be a commutative ring with ideals I, J. Define the product of the two ideals by
n
X
ai bi | ai ∈ I, bi ∈ J, n ∈ Z+ }
IJ = {
i=1
(a) show that IJ is an ideal that is contained in I ∩ J.
(d) Determine (nZ)(mZ) in the ring of integers.
P
Proof. (a) First
ai bi and r ∈ R
P IJ is nonempty since 0 ∈ IJ. If α, β ∈ IJ then clearly α + β ∈ IJ. Also if α =
then rα = (rai )bi ∈ IJ since ai ∈ I ⇒ rai ∈ I. Thus IJ is an ideal.
If ai ∈ I andP
bi ∈ J then since I is an ideal ai bi ∈ I and since J is an ideal ai bi ∈ J. Thus ai bi ∈ I ∩ J.
Thus given α =
ai bi ∈ IJ, since ai bi ∈ I ∩ J for any i and I ∩ J is an ideal it follows that α ∈ I ∩ J. Hence
IJ ⊆ I ∩ J.
P
P
(b) I claim that (nZ)(mZ) = nmZ. First if α ∈ (nZ)(mZ) then α = (ai n)(bi m) = nm ai bi ∈ nmZ thus
(nZ)(mZ) ⊆ nmZ. On the other hand it’s clear that nmZ ⊆ (nZ)(mZ) and hence the claim is shown.
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