From: Robin Chapman <rjc@maths.ex.ac.uk>
Subject: Re: extensions of local fields
Date: Tue, 31 Aug 1999 08:47:15 GMT
Newsgroups: sci.math.research

In article <7qacpr$on8$1@nntp.Stanford.EDU>,
  hwatheod@leland.Stanford.EDU (theodore hwa) wrote:
> Let K be a local field, and A the valuation ring of K (the elements of
> absolute value < = 1).  Let L be a finite extension of K, and let B be the
> integral closure of A in L.  Let B' be the valuation ring of L.  We know
> that B is a subset of B' (even if the fields were, say, number fields
> instead), but does B = B' in the local case?
>
> I'm pretty sure I have a proof, but at least two algebraic number
> theory books I'm looking at don't mention this, so I'm wondering if I'm
> missing something here.

You are right. Indeed B = B'. Let x be in L, and let f(X) be its
minimal polynomial. Then x is integral over A iff the coefficients
of f all lie in A (this comes from Gauss's lemma). If x is integral
over A but |x| > 1 we get a contradiction to f(x) = 0 by noting
that the leading term strictly exceeds the rest in norm. If x
is not integral over A then some coefficient of f lies outside A.
From the theory of the Newton polygon (see e.g. Koblitz's book)
since f is irreducible then its constant term f(0) must lie outside A
too (all this theory depends ultimatiely on Hensel's lemma). But
the characteristic polynomial g(X) of x over K is a power of f(X)
and so g(0) is outside A. Then |x| = |g(0)|^{1/n} > 1 where
n = |L:K|.

> (Yes, this was posted on sci.math also earlier, but I don't seem to be
> getting a response so I thought I might reach a slightly different
> audience here...)

I'm sorry, I didn't see this on sci.math.

--
Robin Chapman
 http://www.maths.ex.ac.uk/~rjc/rjc.html
  "They did not have proper palms at home in Exeter."
    Peter Carey, _Oscar and Lucinda_


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