Download (Dummit-Foote 9.4 #1) (a)

Solutions to Homework 8
41. Throughout this problem, let f (x) denote the polynomial in question.
(Dummit-Foote 9.4 #1) (a) We have f (0) = 1 = f (1), hence f has no roots in F2 . As it is a
quadratic polynomial, this proves that it is irreducible.
(b) In F3 [x], f (x) = (x − 1)(x2 + x − 1). The quadratic factor has no roots in F3 , hence this is
a factorization of f (x) into irreducibles.
(c) In F5 [x], f (x) = (x2 + 2)(x2 − 2), and both of these quadratics have no roots.
(d) Neither 1 or −1 is a root of f (x) in Z. Since the constant term of f (x) is 1, it follows that
we cannot factor a linear term out of f (x). So suppose we had a factorization
x4 + 10x2 + 1 = (x2 + ax + b)(x2 + cx + d)
of f (x) into two quadratics in Z[x]. Expanding out the RHS and equating coefficients gives a = −c
and bd = 1, hence b = d = 1 or b = d = −1. Equating the coefficients of x2 gives 10 = ac + b + d.
But then we either have 8 = −a2 or 12 = −a2 , both clearly impossible.
(Dummit-Foote 9.4 #2) (a) Use the Eisenstein Criterion with p = 2.
(b) Use the Eisenstein Criterion with p = 3.
(c) The substition yields f (x − 1) = x4 − 2x + 2, now Eisenstein with f = 2 proves that f (x) is
irreducible.
(d) We have
(x + 2)p − 2p
=
f (x) =
x
Pp
i=0
p
i
p 2p−i xi − 2p X p p−i i−1
2 x .
=
x
i
i=1
So every coefficient of f (x) is divisible by p (excluding the leading coefficient). But it is clear from
the above expression that the constant term of f (x) is p2p−1 , and as p 6= 2, we have that p2 does
not divide p2p−1 . Thus Eisenstein with the prime p proves that f (x) is irreducible.
42. Again, let f (x) denote the polynomial in question.
(Dummit-Foote 9.4 #3) Suppose we had a factorization f (x) = g(x)h(x) in Z[x], with deg(g), deg(h) <
n. If k is an integer between 1 and n, then
f (k) = −1 = g(k)h(k)
so we have g(k) = 1 and h(k) = −1, or vice-versa. In any case g(k) + h(k) = 0 for n values of k,
which says that the polynomial g(x) + h(x) is identically zero. But then p(x) = −g(x)2 , and for
1
sufficiently large m we have p(m) > 0, a contradiction.
(Dummit-Foote 9.4 #4) As in the previous part, suppose that f (x) = g(x)h(x), and that n > 4.
Then for all k between 1 and n, we have f (k) = 1, so g(k) = h(k) for all such k. As g and h are
monic polynomials of degree less than n, this is enough to prove that g(x) = h(x).
Now we have f (x) = g(x)2 , which says that f is always positive, and that n is even. But
f (3/2) = (1/2)(an odd number of sufficiently large negative numbers) + 1 < 0
which is a contradiction. (The cases where n < 4 are all easy)
43. If char(F ) = 0, then Q is a subfield of F , and so Q× is a subgroup of F × . As Q× is not cyclic
(let p/q be a generator, then p/2q ∈
/ hp/qi), F × cannot be cyclic.
Now suppose char(F ) = p > 0. If F is infinite, then clearly F × is infinite, and if it is cyclic it
must be torsion-free. This immediately rules out any F where p > 2, as such an F must contain
F×
p which is torsion and non-trivial for p > 2.
We are reduced to the case where char(F ) = 2 and F2n * F for any n > 1. Let α be a generator
for F × . Then α must be transcendental over F2 . But α + 1 ∈ F × must be some power of α, say
αm = α + 1. Thus αm + α + 1 = 0, and α is algebraic over F2 , which is a contradiction.
44. If p = 2, then x4 + 1 = (x + 1)4 by the Freshman’s Dream Theorem.
Suppose that p ≡ 1 mod 4. Then −1 is a square mod p, so let b ∈ Fp be a square root of −1.
Then
x4 + 1 = (x2 + b)(x2 − b).
×
If in addition we have p ≡ 1 mod 8, then 8 divides p − 1, the order of F×
p . Now Fp is a cylic group,
×
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thus there exists some α ∈ Fp of order 8. We have α = −1, hence α is a root of x4 + 1, as are
α3 , α5 , and α7 . Thus
x4 + 1 = (x − α)(x − α3 )(x − α5 )(x − α7 ).
If p 6≡ 1 mod 8 (i.e. p ≡ 5 mod 8), then F×
p contains no elements of order 8, and thus we only have
the factorization into two quadratics as above.
Now suppose p ≡ 7 mod 8. Then 2 is a square mod p, so let a2 = 2. Then
x4 + 1 = (x2 + ax + 1)(x2 − ax + 1)
is a factorization of x4 + 1 into irreducibles.
Lastly, if p ≡ 3 mod 8, then −2 is a square mod p. If a2 = −2, then
x4 + 1 = (x2 + ax − 1)(x2 − ax − 1)
is a factorization into irreducibles.
(All of the information on congruences in this solution can be found, among other places, in
the Wikipedia page on Quadratic Reciprocity)
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45. Suppose that p/q ∈ F is a root of a monic polynomial with coefficients in R, with p and q
relatively prime. Then
(p/q)n + an−1 (p/q)n−1 + . . . + a0 = 0
for some ai ∈ R. Multiplying through by q n gives an equation
pn + an−1 pn−1 q + . . . + a0 q n = 0.
Moving the pn to the other side shows that q|pn , which says that q must be a unit. Thus p/q ∈ R.
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