Krasner's lemma says that if is complete, is separable over , and
then .
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 give
But 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; then
The degree bound from forces equality. If and is the prime selected by the embedding , its completion is
Thus 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 stabilizer
Each 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 are
primes of above , and the decomposition group of each is the normal subgroup

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