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By Jürgen Müller

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Example text

Since HomK (L, L′ ) ≤ Maps(L, L′ ) as L′ -vector spaces, the set {ϕ1 , . . , ϕn } ⊆ HomK (L, L′ ) is L′ -linearly indepen′ dent, hence n ≤ dimL′ (HomK (L, L′ )). Since HomK (L, L′ ) ∼ = K [L : K]×[L : K] as K-vector spaces we have [L′ : K]·dimL′ (HomK (L, L′ )) = dimK (HomK (L, L′ )) = [L′ : K] · [L : K], thus dimL′ (HomK (L, L′ )) = [L : K]. 4) Proposition: Artin’s Theorem. Let L/K be a finite field extension, and let H ≤ Aut(L/K). Then we have [L : FixL (H)] = |H|. Proof. Let M := FixL (H). From ϕ|M = idM for all ϕ ∈ H we conclude n := |H| ≤ [L : M ].

For G ≤ Sn we have G = Since any finite group is isomorphic to a subgroup of some symmetric group, any finite group can be realised as a Galois group of a suitable Galois extension. b) The set S is algebraically independent over K, i. e. if Z := {Z1 , . . , Zn } are commuting indeterminates, the evaluation map K[Z] → K[X ] : Zk → Sn,k is injective: Let Y := {Y1 , . . , Yn } be commuting indeterminates and g := X n + nk=1 (−1)k Yk X n−k ∈ K(Y)[X] be the general polynomial of degree n. n Let L/K(Y) be a splitting field for g, hence we have g = j=1 (X − yj ) ∈ L[X], and the evaluation map K[X ] → L : Xj → yj implies Yk = Sn,k (y1 , .

Then the irreducible polynomial µi ∈ K[X] has the root ai ∈ L, thus f := ni=1 µi splits in L[X]. Since L = K(a1 , . . , an ) we conclude that L is a splitting field for f over K. Let conversely L be a splitting field for f ∈ K[X] \ K, and let g ∈ K[X] be irreducible having a root a ∈ L. Let M/L be a splitting field for g, and let b ∈ M be a root of g. Hence there is an isomorphism ϕ : K(a) → K(b) such that ϕ|K = idK . Since L(a) = L is a splitting field for f over K(a), and L(b) is a splitting field for f over K(b), there is an isomorphism ϕ : L(a) → L(b) extending ϕ.

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