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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