Math 418 Spring 2017
Homework 4
9.3.2. Suppose f (x), g(x) ∈ Q[x], and f (x)g(x) ∈ Z[x]. Then the product of any coefficient
of f (x) with any coefficient of g(x) is an integer.
Proof: Let f (x) = an xn +· · ·+a1 x+a0 , and g(x) = bm xm +· · ·+b1 x+b0 where ai , bi ∈ Q.
By Gauss’ Lemma, if the product f (x)g(x) factors in Q[x], then there exist r, s ∈ Q so
that rf (x), sg(x) ∈ Z[x] and rs = 1. Hence, for any i, j, ai bj = (rs)ai bj = (rai )(sbj ) ∈ Z.
9.4.2. Prove that the following polynomials are irreducible over Z.
(a) x4 − 4x3 + 6.
Proof: This polynomial is irreducible by Eisenstein’s Criterion with p = 2.
(b) x6 + 30x5 − 15x3 + 6x − 120.
Proof: This polynomial is irreducible by Eisenstein’s Criterion with p = 3.
(c) x4 + 4x3 + 6x2 + 2x + 1.
Proof: Way 1: If x4 + 4x3 + 6x2 + 2x + 1 is not irreducible, it factors as f (x)g(x)
and both f (x) and g(x) quadratic, or one is a degree 3 polynomial and the other is
linear. In fact, since the polynomial is monic, if it has a linear factor then it has a root
in Z. By Proposition, the only possible rational roots of this polynomial are ±1. By
plugging in, we see neither is a root. If it factors into quadratic polynomials, then
x4 + 4x3 + 6x2 + 2x + 1 = (x2 + ax + b)(x2 + cx + d)
for some a, b, c, d ∈ Z. Multiplying out, we get
x4 + 4x3 + 6x2 + 2x + 1 = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd.
Since bd = 1, b = d = 1 or b = d = −1. Then (ad + bc)x = ±(a + c)x = 2x, but
(a + c)x3 = 4x3 . This is impossible. Therefore the polynomial x4 + 4x3 + 6x2 + 2x + 1
is irreducible.
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Way 2: Suppose a polynomial f (x) factors as f (x) = g(x)h(x) in Z[x]. Then if we
substitute x − 1 for x, f (x − 1) = g(x − 1)h(x − 1). That is, f (x) is irreducible if and
only if x − 1 is irreducible.
If we substitute x − 1 into this polynomial, we get
(x − 1)4 + 4(x − 1)3 + 6(x − 1)2 + 2(x − 1) + 1 = x4 − 2x + 2.
We then apply Eisenstein’s Criterion with p = 2 to see it is irreducible.
(d)
(x+2)p −2p
x
where p is an odd prime.
Proof: We can use the Binomial Theorem to write (x + 2)p =
fore, the polynomial simplifies to
p (x + 2)p − 2p X p k−1 p−k
x 2 .
=
k
x
Pp
p
k=0 k
xk 2p−k . There-
k=1
p
is divisible by p. The constant term
This
is
a
monic
polynomial.
For
1
≤
k
≤
p
−
1,
k
p p−1
p−1
2
= 2 p is not divisible by p . Therefore we can apply Eisenstein’s Criterion
1 2
with prime p to see this polynomial is irreducible.
9.4.4. The polynomial (x − 1)(x − 2) · · · (x − n) + 1 is irreducible over Z for all n > 1.
Proof: Suppose by way of contradiction that (x − 1)(x − 2) · · · (x − n) + 1 = f (x)g(x)
was a factorization into smaller degree polynomials. The degree of the given polynomial
is n, so each of f (x) and g(x) must have degree smaller than n. At x = 1, 2, . . . , n,
f (x)g(x) = 1. Since these are integers, at each of those points f (x) = g(x) = 1 or
f (x) = g(x) = −1. Therefore, if we consider the polynomial f (x) − g(x), it has at least n
zeroes (at x = 1, 2, . . . , n).
But the polynomial f (x) − g(x) has degree lower than n. Hence f (x) − g(x) = 0.
Therefore, our polynomial factors as (x−1)(x−2) · · · (x−n)+1 = f (x)2 . The degree of f (x)2
is even, and the degree of (x − 1)(x − 2) · · · (x − n) + 1 is n, so if n is odd no such f (x) exists,
and the polynomial is irreducible. If n = 2, then (x−1)(x−2)+1 = x2 −3x+3 is clearly not
a square. Otherwise, there would exist a ∈ Z so that (x − a)2 = x2 − 2ax + a2 = x2 − 3x + 3.
Since 3 is not a square, this is impossible.
For n > 4 even, note that if the polynomial factors into f (x)2 in Z[x], it must also
factor as f (x)2 in Q[x]. For every x ∈ Q, f (x)2 ≥ 0, so the polynomial must be positive or
0. If we plug in n − 1/2 for x, however, we get
(x − 1)(x − 2) · · · (x − n) + 1 = (n − 3/2)(n − 5/2) · · · (1/2)(−1/2) + 1.
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This is negative for n > 4. Therefore, it cannot be written as a square, and the given
polynomial is irreducible.
9.4.9. The polynomial x2 −
√
√
2 is irreducible over Z[ 2].
√
√
√
√
√Proof: Consider the
√ element
√ 2 ∈ Z[ 2]. Suppose there√are elements a+b √2, c+d 2 ∈
Z[ 2] so that (a + b 2)(c + d 2) = (ac + 2bd) + (ad + bc) 2 is a multiple of 2. That is,
there exist x, y ∈ Z so that
√
√
√
√
2(x + y 2) = 2y + x 2 = (ac + 2bd) + (ad + bc) 2.
Then 2y = ac + 2bd, so ac√is even.
means a or c is√even. If a is even,
√ This √
√ then a = 2k
for some k√∈ Z, and a + b √2 = 2(b + k 2). That is, 2 divides a + b 2. Similarly,
√ if
c is even, √
2 divides c + d 2. Therefore sqrt2 is prime. This means the ideal√(− 2) is
prime
We can apply Eisenstein’s Criterion to the polynomial x2 − 2. Since
√ 2].
√ in Z[
2
− 2 6∈ (− 2) = (2), the polynomial is irreducible.
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