An Introduction to Galois Theory by Andrew Baker

By Andrew Baker

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Un ∈ L are algebraic. Then K(u1 , . . , un )/K is a finite extension. Proof. 6(ii). We now come to an important notion for extensions. 14. Definition. The extension L/K is algebraic or L is algebraic over K if every element t ∈ L is algebraic over K. 15. Proposition. Let L/K be a finite extension. Then L/K is algebraic. Proof. Let t ∈ L. Since the K-vector space L is finite dimensional, when viewed as elements of this vector space, the powers 1, t, . . , tn , . . must be linearly dependent over K.

50 Proof. 34 applied to the case L = K. 1) K(u1 )/K, K(u1 , u2 )/K(u1 ), · · · , L = K(u1 , . . , uk )/K(u1 , . . , uk−1 ). So we can use the following to compute (L : K) = (K(u1 , . . , uk ) : K). 70. Proposition. Let L/K and M/L be finite extensions. Then (M : K) = (M : L)(L : K). Proof. For α ∈ MonoK (M, K) let αL ∈ MonoK (L, K) be its restriction to L. 49, each element of MonoK (L, K) extends to a monomorphism M −→ K, so every element β ∈ MonoK (L, K) has the form β = αL for some α ∈ MonoK (M, K).

The formal derivative ∂ : K[X] −→ K[X] has the following properties. (i) ∂ is K-linear. , for f (X), g(X) ∈ K[X], ∂(f (X)g(X)) = ∂(f (X))g(X) + f (X)∂(g(X)). (iii) If char K = 0, then ker ∂ = K and ∂ is surjective. (iv) If char K = p > 0, then ker ∂ = {h(X p ) : h(X) ∈ K[X]} and im ∂ is spanned by the monomials X k with p (k + 1). Proof. (i) This is routine. (ii) By K-linearity, it suffices to verify this for the case where f (X) = X r and g(X) = X s with r, s 0. But then ∂(X r+s ) = (r + s)X r+s−1 = rX r−1 X s + sX r X s−1 = ∂(X r )X s + X r ∂(X s ).

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