By Jürgen Müller

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**Extra info for Algebra**

**Example text**

Let G := Aut(L/Q), hence we have |G| = 6. We have L = M (ρ), and since [L : M ] = 3 we conclude that f = X 3 − 2 ∈ M [X] is irreducible, thus there is τ ∈ Aut(L/M ) given by τ : ζ → ζ, ρ → ζρ. Hence τ : ρ → ζρ → ζ −1 ρ → ρ shows |τ | = 3, thus Aut(L/M ) = τ ∼ = C3 and FixL (τ ) = M . We have L = K0 (ζ), and since [L : K0 ] = 2 we conclude that µζ = X 2 + X + 1 ∈ K0 [X] is irreducible, thus there is σ ∈ Aut(L/K0 ) given by σ : ρ → ρ, ζ → ζ −1 , hence Aut(L/K0 ) = σ ∼ = C2 and FixL (σ) = K0 . σ We have τ : ζ → ζ, ρ → ζ −1 ρ, hence τ σ = τ −1 , and thus G ∼ = S3 .

For n ∈ N let ζn := exp( 2π n−1 ) ∈ C. Then Q(ζn ) ⊆ C√is called the n-th cyclotomic field; we have ζ1√= 1 and ζ2 = −1 and ζ4 = −1, thus Q = Q(ζ1 ) = Q(ζ2 ), and Q(ζ4 ) = Q( −1) is called the Gaussian number field. The group ζn ≤ C∗ of n-th roots of unity has order n, its generators are called primitive n-th roots of unity. Thus we have X n −1 = k∈Z/nZ (X −ζnk ) ∈ C[X]. Running over the primitive roots yields the n-th cyclotomic polynomial Φn := k∈(Z/nZ)∗ (X − ζnk ) ∈ C[X], thus deg(Φn ) = ϕ(n) = |(Z/nZ)∗ |.

A) There is a Galois radical extension M/K such that f splits in M [X], where Aut(M/K) is soluble. There is a splitting field L/K for f such that L ⊆ M , hence Aut(L/K) ∼ = Aut(M/K)/Aut(M/L) is soluble. b) There is a Galois radical extension M/K such that L ⊆ M . 10) Symmetric polynomials. a) Let K be a field, let n ∈ N, and let n X = {X1 , . . , Xn } be commuting indeterminates. Then f := j=1 (X − Xj ) = n X n + k=1 (−1)k Sn,k X n−k ∈ K(X )[X], where the elementary symmetric k polynomials are Sn,k := 1≤i1

### Algebra by Jürgen Müller

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