To prove it, let be an automorphism of fixing . The absolute value has a unique extension to every finite extension of the complete field , so it is invariant under . If , the ultrametric inequality and the displayed strict inequality giveBut and invariance gives , a contradiction. Every automorphism fixing therefore fixes , which proves the field inclusion by Galois correspondence.
Now let be finite. By the primitive element theorem, write with separable minimal polynomial . Approximate the coefficients of closely by those of a polynomial of the same degree. Continuity of roots over a non-Archimedean field gives a root of arbitrarily close to . Choose it close enough for Krasner's lemma; thenThe degree bound from forces equality. If and is the prime selected by the embedding , its completion isThus every finite extension of is the completion of a number field at a prime ideal.
Let and let lie over . These primes define two extensions to of the -adic absolute value on . The conjugacy of extensions of a valuation to a normal extension says that some carries the first extension to the second. Equivalently, . This proves the transitivity of the Galois action on primes.
The decomposition group is the stabilizerEach such automorphism extends continuously to the completions, and restriction gives the decomposition group of a prime and local Galois group isomorphism
For the splitting field of , the global Galois group is . Since , the field contains . The local splitting field is therefore , an Eisenstein, totally ramified cyclic extension of degree three. Hence there areprimes of above , and the decomposition group of each is the normal subgroup
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