8 Global Fields Definition 8 1 Global field A global field is a field which is either i An algebraic number field ii A global function field i e a finite extension of p t Lemma 8 2 Assuming that K is a complete discretely valued field L K a finite Galois extension with absolute value L extending Then for x L and Gal L K we have x L x L Proof Since x x L is another absolute value on L extending on K the result follows from uniqueness of L Lemma 8 3 Kummer s Lemma Assuming that K a complete discretely valued field f X K X a separable irreducible polynomial with roots 1 n K sep K sep is the separable closure of K K sep with 1 i for i 2 n Then 1 K Proof Let L K L L 1 n Then L L is a Galois extension Let Gal L L We have 1 1 1 using Lemma 8 2 Hence 1 1 so 1 K Proposition 8 4 Assuming that F is a complete discretely valued field f X i 0 n a i X i O K X a separable irreducible monic polynomial K sep a root of f Then there exists 0 such that for any g X i 0 n b i X i O K X monic with a i b i for all i there exists a root of g X such that K K Nearby polynomials define the same extensions Proof Let 1 n K sep be the roots of f which are necessarily distinct Then f 1 0 We choose sufficiently small such that g 1 f 1 2 and f 1 g 1 f 1 Then we have g 1 f 1 2 g 1 2 the equality is by Lemma 1 6 By Hensel s Lemma version 1 applied to the field K 1 there exists K 1 such that g 0 and 1 g 1 Then g 1 f 1 j 1 n 1 j 1 i for i 2 n Use 1 i 1 since i integral Since 1 1 i i using Lemma 1 6 we have that Kummer s Lemma gives that 1 K and hence K 1 K Theorem 8 5 Assuming that K is a local field Then K is the completion of a global field Proof Case 1 is archimedean Then is the completion of and is the completion of i with respect to Case 2 non archimedean equal characteristic Then K q t is the completion of q t with respect to the t adic valuation Case 3 non archimedean mixed characteristic Then K p with a root of a monic irreducible polynomial f X p X Since is dense in p we choose g X X as in Proposition 8 4 Then K with a root of g X Since dense in p K and K is complete we must have that K is the completion of