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    Separable splitting fields are Galois

    PR-ZCKLJ

    A finite separable splitting field is Galois, with degree equal to the number of base-field automorphisms.Suggest a slogan

    Let \(L/k\) be a finite field extension, with separable degree \([L:k]_s\).

    1. If \(L/k\) is separable, then \([L: k] = [L:k]_s\).

    2. If \(L/k\) is separable and \(L\) is the splitting field of a polynomial in \(k[x]\), then \[ [L:k] = \abs{\Aut_{\Fieldsover{k}}(L)}, \] so \(L/k\) is Galois with \(\Gal(L/k)=\Aut_{\Fieldsover{k}}(L)\).

    1. Fix an algebraic closure \(\bar k\) containing \(L\). Since \(L/k\) is a finite splitting field, it is normal (PR-TZN4M), so every \(k\)-embedding \(L\to\bar k\) has image \(L\) (PR-OZYUC) and is a \(k\)-automorphism of \(L\). Hence \([L:k]_s=\abs{\Aut_{\Fieldsover{k}}(L)}\), and (a) gives \([L:k]=\abs{\Aut_{\Fieldsover{k}}(L)}\).

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