请输入您要查询的字词:

 

单词 KrasnersLemma
释义

Krasner’s lemma


Krasner’s lemma (along with Hensel’s lemma) connects valuationsPlanetmathPlanetmath on fields to the algebraic structurePlanetmathPlanetmath of the fields, and in particular to polynomial roots.

Lemma 1.

(Krasner’s Lemma) Let K be a field of characteristic 0 completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath with respect to a nontrivial nonarchimedean absolute valueMathworldPlanetmathPlanetmath. Assume α,βK¯ (where K¯ is some algebraic closureMathworldPlanetmath of K) are such that for all nonidentity embeddingsMathworldPlanetmathPlanetmathPlanetmath σHomK(K(α),K¯) we have |α-β|<|σ(α)-α|. Then K(α)K(β).

This says that for any αK¯, there is a neighborhoodMathworldPlanetmathPlanetmath of α each of whose elements generates at least the same field as α does.

Proof.

It suffices to show that for every σHomK(β)(K(α,β),K¯), we have σ(α)=α, for then α is in the fixed field of every embedding of K(β), so αK(β). Note that

|σ(α)-β|=|σ(α)-σ(β)|=|σ(α-β)|=|α-β|

where the final equality follows since |σ()| is another absolute value extending ||K to K(α,β) and thus must be equal to ||. But then

|σ(α)-α|=|(σ(α)-β)+(β-α)|max(|σ(α)-β|,|α-β|)=|α-β|

But this is impossible by the bounds on α,β unless σ(α)=α.∎

The first application of Krasner’s lemma is to show that splitting fieldsMathworldPlanetmath are “locally constant” in the sense that sufficiently close polynomialsMathworldPlanetmathPlanetmathPlanetmath in K[X] have the same splitting fields.

Proposition 2.

With K as above, let P(X)K[X] be a monic irreducible polynomialMathworldPlanetmath of degree n with (distinct) roots α1,,αn. Then any monic polynomialMathworldPlanetmath Q(X)K[X] of degree n that is “sufficiently close” to P(X) will be irreduciblePlanetmathPlanetmath over K with roots β1,βn, and (after renumbering) K(αi)=K(βi).

Here “sufficiently close” means the following: consider the space of degree n polynomials over K as homeomorphic to Kn as a topological spaceMathworldPlanetmath; close then means close in the obvious metric induced by ||.

Proof.

Since P(X) has distinct roots, we may choose 0<γ<min(|αi-αj|) for ijn. Since the roots of a polynomial vary continuously with its coefficients, we say that a degree n polynomial Q(X)K[X] is sufficiently close to P(X) if Q(X) has roots β1,,βn with |αi-βi|<γ. But {αj}ji are all the Galois conjugates of αi, and |αi-βi|<γ<|αi-αj| by construction, so by Krasner’s lemma, K(αi)K(βi). But

[K(βi):K]degQ=degP=[K(αi):K]

so that K(βi)=K(αi). In additionPlanetmathPlanetmath, we see that degQ=[K(βi):K] and thus that Q(X) is irreducible.∎

We use this fact to show that every finite extensionMathworldPlanetmath of p arises as a completion of some number fieldMathworldPlanetmath.

Corollary 3.

Let K be a finite extension of Qp of degree n. Then there is a number field E and an absolute value || on E such that E^K.

Proof.

Let K=p(α) and let P be the minimal polynomial for α over p. Since is dense in p, we can choose Q(X)[X] (note: in [X], not p[X]), and β a root of Q(X), as in the propositionPlanetmathPlanetmath, so that p(α)=p(β). Let E=(β). Clearly E is a number field which, when regarded as embedded in p(β), has absolute value ||E, the restrictionPlanetmathPlanetmathPlanetmathPlanetmath of the absolute value on p(α)=p(β). Then E^ is a complete field with respect to that absolute value; p(β) is as well, and E is dense in both, so we must have E^=p(β)=p(α)=K.∎

Finally, we can prove the following generalizationPlanetmathPlanetmath of Krasner’s Lemma, which is also given that name in the literature:

Lemma 4.

Let K be a field of characteristic 0 complete with respect to a nontrivial nonarchimedean absolute value, and K¯ an algebraic closure of K. Extend the absolute value on K to K¯; this extensionPlanetmathPlanetmathPlanetmath is unique. Let K¯^ be the completion of K¯ with respect to this absolute value. Then K¯^ is algebraically closed.

Proof.

Let α be algebraic over K¯^ and P(X) its monic irreducible polynomial in K¯^[X]. Since K¯ is dense in K¯^, by proposition 2 we may choose Q(x)K¯[X] with a root βK¯^ such that K¯^(α)=K¯^(β). But K¯^(β)=K¯^ so that αK¯^.∎

随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 21:04:37