2πi
1.(a) Note that ω := e n is a root of the polynomial xn − 1 ∈ Q[x] and
hence is algebraic over Q. Now use the facts
ω + ω −1 = (cos
2π
2π
2π
2π
2π
+ i sin
) + (cos
− i sin
) = 2 cos
n
n
n
n
n
2π
1
= (ω + ω −1 ) ∈ Q(ω)
n
2
2π
1
2π
−1
ω − ω = 2i sin
⇒ sin
= (ω − ω −1 ) ∈ Q(ω, i)
n
n
2i
2π
too see that cos 2π
n , sin n are algebraic over Q.
⇒ cos
(b) Q(ω) is the splitting field of the polynomial xn − 1 over Q. In fact all
roots of this polynomial are in the form ω i for i = 1, 2, ..., n which are elements
of Q(ω). So the extension Q(ω)/Q is Galois and its Galois group G is Abelian.
It comes from the fact that if σ ∈ G then σ is completely determined by σ(ω).
But as ω is a root of the polynomial xn − 1 then so is σ(ω), this implies that
σ(ω) = ω i for some i. Now if we have two elements of G namely σ, τ and
σ(ω) = ω i
τ (ω) = ω j
then
στ (ω) = σ(ω j ) = ω ij
which is the same as τ σ(ω). So στ = τ σ which proves that G is Abelian. (It
can be seen from the fact that G ∼
= Z∗n . We use this fact for computing the
Galois group of the extension Q(cos( 2π
n ))/Q).
Now note that Q(cos( 2π
))/Q
is
Galois
iff
n
Gal(
Q(ω)
)EG
Q(cos 2π
n )
which is true because G is Abelian. In fact
Gal(Q(cos
2π
Z∗
)/Q) = n
n
Z2
Q(ω)
where Z2 = Gal( Q(cos
2π ) : for showing this note that
)
n
ω + ω −1 = 2 cos
2π
2π
⇒ ω 2 − 2ω cos
+1=0
n
n
this is the minimal polynomial of ω over Q(cos 2π
/ Q(cos 2π
n ) because ω ∈
n ) ⊆ R
(of course we should assume that n ≥ 3. for n = 1, 2 there is nothing to prove.)
It means that [Q(ω) : Q( 2π
n )] = 2, it is Galois and its Galois group is Z2 .
(c) Yes ! by the same way as we did in (b). By
sin
2π
1
= (ω − ω −1 ) ∈ Q(ω, i) ⊆ Q(ζ)
n
2i
1
2πi
πi
where ζ = e 4n = e 2n . The second inclusion is true because
i = ζ n ∈ Q(ζ)
ω = ζ 4 ∈ Q(ζ)
in this case we should check that Gal(Q(ζ)/Q(sin 2π
n )) is a normal subgroup of
Gal(Q(ζ)/Q(sin 2π
))
which
is
true
and
indeed
we
have
n
Gal(Q(sin
2π
Gal(Q(ζ)/Q)
)/Q) =
n
Gal(Q(ζ)/Q(sin 2π
n ))
2. First note that F is the splitting
field of the polynomial x4 − 2 over Q:
√
√
4
the roots of this polynomial are ± 2, ± 4 2i, so the splitting filed is
√
√
√
4
4
4
Q(± 2, ± 2i) = Q( 2, i)
As F is the splitting field of the separable polynomial x4 − 2 over Q the
field extension F/Q
It can be seen by
√ is Galois. We know that [F : Q] = 8. √
showing that [Q( 4 2) : Q] = 4 and the observation i ∈
/ Q( 4 2). So the group
G := Gal(F/Q) has 8 elements. In fact for any σ ∈ G we have
√
√
√
4
4
4
σ( 2) ∈ {± 2, ± 2i}
σ(i) ∈ {±i}
which means that we have at most 8 automorphisms in the Galois group. But
from |G| = 8 each one of these 8 maps gives us one element of G. This is very
important to note that it is not always the case that we can extend all of the
possible maps to the elements of the Galois group, in this case it follows by
knowing the |G|. For instance we have 2 elements σ, τ ∈ G defined by
√
√
4
4
σ( 2) = 2i
σ(i) = i
√
√
4
4
τ ( 2) = 2
τ (i) = −i
then it is a good exercise to see that σ has order 4, τ has order 2 and τ σ = σ −1 τ .
We can see the last case for example here:
√
√
√
√
√
4
4
4
4
4
τ σ( 2) = τ (σ( 2)) = τ ( 2i) = τ ( 2)τ (i) = − 2i
on the other hand
√
√
√
√
√
√
4
4
4
4
4
4
σ( 2) = 2i ⇒ σ −1 ( 2i) = 2 ⇒ σ −1 ( 2)σ −1 (i) = 2
but from σ(i) = i we have σ −1 (i) = i, so
√
√
√
√
4
4
4
4
σ −1 ( 2)i = 2 ⇒ σ −1 ( 2) = − 2i
it can also be found from σ −1 = σ 3 . So
√
√
√
4
4
4
σ −1 τ ( 2) = σ −1 ( 2) = − 2i
2
√
so τ σ −1 and σ −1 τ have the same value on 4 2. By showing that their value on
i is also the same we get the equality τ σ = σ −1 τ . But from what we saw the
group G is isomorphic to D4 and in fact σ, τ are the generators for G.
3. This polynomials are irreducible over Q. In fact both of them are x3 +x+1
mod 2 which is irreducible ( it has no root and of degree 3 ). For the first
polynomial we have
∆ = −4(−3)3 − 27(1)2 = 81 = 92
is square in Q. So the asked Galois group is A3 ⊆ S3 . For the second polynomial
∆ = −4(1)3 − 27(1)2 = −31
which is not square. So the answer is S3 .
3