## Extensions of discrete valuations September 5, 2009

Posted by Akhil Mathew in algebra, algebraic number theory, commutative algebra, number theory.
Tags: , , ,

With the school year starting, I can’t keep up with the one-post-a-day frequency anymore. Still, I want to keep plowing ahead towards class field theory.

Today’s main goal is to show that under certain conditions, we can always extend valuations to bigger fields. I’m not aiming for maximum generality here though.

Dedekind Domains and Extensions

One of the reasons Dedekind domains are so important is

Theorem 1 Let ${A}$ be a Dedekind domain with quotient field ${K}$, ${L}$ a finite separable extension of ${K}$, and ${B}$ the integral closure of ${A}$ in ${L}$. Then ${B}$ is Dedekind.

This can be generalized to the Krull-Azizuki theorem.

I’ll sketch the proof. We need to check that ${B}$ is Noetherian, integrally, closed, and of dimension 1.

• Noetherian. Indeed, ${B}$ is a finitely generated ${A}$ module. The ${K}$-linear map ${(.,.): L \times L \rightarrow K}$, ${a,b \rightarrow \mathrm{Tr}(ab)}$ is nondegenerate since ${L}$ is separable over ${K}$. Let ${F \subset B}$ be a free module spanned by a ${K}$-basis for ${L}$. Then since traces preserve integrality and ${A}$ is integrally closed, we have ${B \subset F^*}$, where ${F^* := \{ x \in K: (x,F) \subset A \}}$. Now ${F^*}$ is ${A}$-free on the dual basis of ${F}$ though, so ${B}$ is a submodule of a f.g. ${A}$ module, hence a f.g. ${A}$-module.
• Integrally closed. It is an integral closure, and integrality is transitive.
• Dimension 1. Indeed, integral extensions preserve dimension by lying over, going up, and a corollary.

So, consequently the ring of algebraic integers (integral over ${\mathbb{Z}}$) in a number field (finite extension of ${\mathbb{Q}}$) is Dedekind.

Extensions of discrete valuations

The real result we care about is:

Theorem 2 Let ${K}$ be a field, ${L}$ a finite separable extension. Then a discrete valuation on ${K}$ can be extended to one on ${L}$

Indeed, let ${R \subset K}$ be the ring of integers. Then ${R}$ is a DVR, hence Dedekind, so the integral closure ${S \subset L}$ is Dedekind too (though in general it is not a DVR). Now as above, ${S}$ is a finite ${R}$-module, so if ${\mathfrak{m} \subset R}$ is the maximal ideal, then $\displaystyle \mathfrak{m} S \neq S$

by Nakayama. So ${\mathfrak{m} S}$ is contained in a maximal ideal ${\mathfrak{M}}$ of ${S}$ with, therefore, ${\mathfrak{M} \cap R = \mathfrak{m}}$. (This is indeed the basic argument behind lying over, which I could have just invoked.) Now ${S_{\mathfrak{M}} \supset R_{\mathfrak{m}}}$ is a DVR as it is the localization of a Dedekind domain at a prime ideal. So there is a discrete valuation on ${S_{\mathfrak{M}}}$. Restricted to ${R}$, it will be a power of the given ${R}$-valuation. This is ok, since a power of a discrete valuation is a discrete valuation too.

This completes the proof. Note that there is a one-to-one correspondence between extensions of the valuation on ${K}$ and primes of ${S}$ lying above ${\mathfrak{m}}$. Indeed, the above proof indicated a way of getting valuations from primes. For an extension of the valuation on ${K}$ to ${L}$, let ${\mathfrak{M} := \{ x \in S: \left| x \right| < 1\}}$.

Also, I think the above result can be extended to purely inseparable extensions (hence all algebraic extensions): if $L/K$ is purely inseparable of degree $p^i$, then define the valuation on $L$ by raising to the power $p^i$, taking the valuation in $K$, and raising the resulting real number to the power $1/p^i$.

Next up: Ramification.