Math 422/501 Homework 9 Solutions
Problem 1. Suppose that n 6= 2m ; then we can write n = kl for k ≥ 3 an odd number. When we do this
we find that the product
(1 + 2l )(1 − 2l + 22l − 23l + · · · + 2(k−1)l )
telescopes to 1 + 2kl , which shows 2kl + 1 is composite. (The key point here is that we need k − 1 to be even
so our product ends up as 1 + 2kl rather than 1 − 2kl !)
Problem 2. (a) If τ ∈ H we have
τ (αH ) =
X
(τ σ)(ζn ) =
σ∈H
X
σ(ζn ) = αH
σ∈H
since σ 7→ τ σ is a permutation of H.
(b) In the case where n is prime, we know that Q(ζp ) has degree p − 1 and thus the “obvious” basis is
1, ζp , . . . , ζpp−2 . It’s easy to then see you can substitute
ζpp−1 = −1 − ζp − · · · − ζpp−2
basis too. Equivalently, {σ(ζp ) : σ ∈ H} is a basis.
into this basis forP
1, so ζp , . . . , ζpp−2 , ζpp−1 is a P
Then, αH =
σ(ζ
)
and
τ
(α
)
=
p
H
σ∈H
σ∈H (τ σ)(ζp ) are actually written as sums of distinct basis
vectors, so for them to be equal all of the vectors involved have to be the same, i.e.
{σ(ζp ) : σ ∈ H} = {(τ σ)(ζp ) : σ ∈ H}.
Since σ(ζp ) = σ 0 (ζp ) iff σ = σ 0 this further means that
{σ : σ ∈ H} = {τ σ : σ ∈ H},
i.e. H = τ H as cosets. This forces τ ∈ H.
So we’ve established that the group of automorphisms fixing Q(αH ) is H itself; by the fundamental
theorem of Galois theory this means Q(αH ) is equal to the fixed field of H.
(c) In this case, −ζn = (ζn )n/2+1 is also a primitive n-th root of unity, so there is a unique automorphism
σ : Q(ζn ) → Q(ζn ) with σ(ζn ) = −ζn . Evidently σ 2 = id, so H = {1, σ} is a subgroup of order 2 and is
proper in Gal(Q(ζn )/Q) (by assumption n = 2m ≥ 8). Also, αH = ζn + σ(ζn ) = ζn − ζn = 0 as desired, so
Q(αH ) = Q is not equal to Q(ζn )H .
Problem 3. (a) There’s a few different ways to do this; one is to note that the automorphisms of
Gal(Q(ζ5 )/Q) which don’t fix ζ5 + ζ5−1 send it to ζ52 + ζ5−2 . Thus we expect the following polynomial it
satisfies to be its minimal polynomial over Q:
(x − (ζ5 + ζ5−1 ))(x − (ζ52 + ζ5−2 )).
Multiplying this out we can get that it’s
x2 − (ζ5 + ζ52 + ζ53 + ζ54 )x + (ζ5 + ζ52 + ζ53 + ζ54 ).
Going back to the definition of ζ5 , it’s a root of the polynomial Φ5 (x) = 0, i.e.
1 + ζ5 + ζ52 + ζ53 + ζ54 = 0;
so our polynomial satisfied by ζ5 + ζ5−1 is just
x2 + x − 1.
1
√
√
By the quadratic formula, the roots to this are 12 (−1 ± 5), so Q(ζ5 + ζ5−1 ) = Q( 5). Since ζ5 + ζ5−1 is a
√
positive real number (it’s 2 Re(ζ5 ), and ζ5 is in the first quadrant) it must be the positive root − 21 + 21 5.
(b) Again, the nontrivial automorphism of Q(ζ5 )/Q(ζ5 + ζ5−1 ) is the one taking ζ5 to ζ5−1 , so we take the
polynomial
(x − ζ5 )(x − ζ5−1 ) = x2 − (ζ5 + ζ5−1 )x − 1.
Problem 4. Let K = Fq (x, y) and F = Fq (xp , y p ); we know that K/F is a degree-p2 extension, and
moreover if c ∈ K then F (x + cy) is degree-p over F (it satisfies the polynomial T p − (xp + cp y p ) in F [T ]).
So again we want to show there are infinitely many c such that F (x + cy) are all distinct; but we can’t just
let c run over constants in our base field Fq because there are only finitely many.
However, we claim that if c 6= c0 are two distinct elements of F , then F (x + cy) and F (x + c0 y) are
distinct. Suppose not; then let F 0 be the common extension F (x + cy) = F (x + c0 y). We then have
(x + cy) − (x + c0 y) = (c − c0 )y ∈ F 0 , and since c0 − c is nonzero in F we can invert it and get y ∈ F 0 .
Since c, c0 , y ∈ F 0 we also conclude that x ∈ F 0 , and thus F 0 = F (x, y) = K. This contradicts the fact that
[K : F ] = p2 and [F 0 : F ] = p.
Q
Problem 5. (a) We have N (a − ζn ) = σ∈Gal(K/F ) (a − σ(ζn )) by definition, and since the set {σ(ζn ) :
σ ∈ Gal(K/F )} is exactly the set of primitive
n-th roots of unity we’ve really just rewritten the definition
Q
of the cyclotomic polynomial Φn (x) = ζ primitive (x − ζ).
(b) Since Q
xn − 1 is the product of (x − ζ) where ζ runs over all n-th roots of unity, we remember we have
n
x − 1 = d|n Φn (x). Dividing both sides by x − 1 = Φ1 (x) we have
xn−1 + xn−2 + · · · + x + 1 =
Y
Φd (x).
1<d|n
Evaluating at x = 1 gives the desired equality.
Q(ζ )
(c) By part (a) we have Φn (1) = NQ n (1 − ζn ), so we just need to prove that Φn (1) is p or 1 depending
on whether n is a prime power. We show this by induction on the number of prime divisors of n, using the
formula in (b). For the base case that n = p is prime, (b) gives Φp (1) = p directly. For the inductive step,
if n = pr is a prime power then (b) gives
r
p =
Y
Φd (1) =
1<d|pr
If n is not a prime power then we have
Y
n=
r
Y
Φps (1) = pr−1 Φpr (1).
s=1
Φd (1) ·
d|n:d a prime power
Y
Φd (1);
d|n:d not a prime power
the first product is exactly n by considering its prime factorization (and using induction) and the latter is
Φn (1) times a bunch of terms that are 1 by induction.
2