From: Robin Chapman 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_ Sent via Deja.com http://www.deja.com/ Share what you know. Learn what you don't.