Discrete Valuation Rings

November 22, 2009 by hilbertthm90 | 2 Comments

Maybe this title isn’t exactly what the post is about, but today will mostly be a hodgepodge attempt to get some more out there. I’m not sure what else to do with regular local rings (other than systems of parameters which I’m not too excited to post on), so I’ll move on. The next set of theorems in Hartshorne (that are not proved in the text) has to do with Noetherian local domains of dimension 1.

Before this is stated we need quite a bit of terminology. Let $k$ be a field and $G$ a totally ordered abelian group. A valuation is a map $v: k\setminus\{0\}\to G$ such that $v(xy)=v(x)+v(y)$ and $v(x+y)\geq \min\{v(x), v(y)\}$ .

We form a subring of $k$ by taking the set $R=\{x\in k : v(x)\geq 0\}\cup \{0\}$ which is called the valuation ring of $v$ . This ring is local with maximal ideal $\frak{m}=\{x\in k : v(x)>0\}\cup \{0\}$ .

Mostly we care about discrete valuation rings (DVR). These are the ones whose value group is the integers. Now we can state and prove the Theorem stated in Hartshorne:

Let $(R, \frak{m}, k)$ be a Noetherian local domain of dimension 1. Then the following are equivalent:

1) $R$ is a discrete valuation ring
2) $R$ is integrally closed
3) $\frak{m}$ is principal
4) $\dim_k(\frak{m}/\frak{m}^2)=1$
5) Every non-zero ideal is a power of $\frak{m}$
6) There exists $x\in R$ such that every non-zero ideal is of the form $(x^k)$ , $k\geq 0$ .

Proof: We’ll just go in the standard cyclic order for proof. If $R$ is a DVR, then we consider an integral element $x\in Frac(R)$ . If $x\in R$ then we are done. If $x=a/b\notin R$ , then the claim is that $b/a\in R$ . This is simply because $0=v(1)=v(xx^{-1})=v(x)+v(x^{-1})$ . Since $R=\{x\in Frac(R): v(x)\geq 0\}\cup \{0\}$ , and $v(x)=-v(x^{-1})$, we get that $x^{-1}\in R$ . Thus if $x^n+b_1x^{n-1}+\cdots + b_n=0$ we get by multiplying by $x^{1-n}$ that $x=-(b_1+b_2x^{-1}+\cdots + b_n)\in R$ and every integral element is in the ring.

For the next, assume $R$ is integrally closed. Let $r\in \frak{m}$ and $r\neq 0$ . Since $\frak{m}$ is the only non-zero prime ideal, $\sqrt{(r)}=\frak{m}$ . Thus there is some integer such that $\frak{m}^n\subset (r)$ such that $\frak{m}^{n-1}\not\subset (r)$ . Now let $a\in \frak{m}^{n-1}\setminus (r)$ . Let $x=r/b\in Frac(R)$ . Now since $b\notin (r)$ , we cannot reduce $b/r$ to a form $a/1$ , so $x^{-1}\notin R$ . By integrally closed, we get that $x^{-1}$ is not integral over $R$ . If $x^{-1}\frak{m}\subset \frak{m}$ , then $\frak{m}$ would be a finitely generated (as an $R$ module) faithful $R[x^{-1}]$ -module, and hence $x^{-1}$ would be integral. Thus $x^{-1}\frak{m}\not\subset \frak{m}$ . Clearly, $x^{-1}\frak{m}\subset R$ , so $x^{-1}\frak{m}=R$ . Thus $\frak{m}=(x)$ is principal.

Now if $\frak{m}$ is principal, we have $\dim_k(\frak{m}/\frak{m}^2)\leq 1$ . But since $\frak{m}/\frak{m}^2\neq 0$ by the Noetherian condition, we get that $\dim_k(\frak{m}/\frak{m}^2)=1$ .

Suppose $\dim_k(\frak{m}/\frak{m}^2)=1$ . Thus $\frak{m}=(x)$ is principal. Let $\frak{a}$ be any non-zero ideal. Since all ideals are $\frak{m}$ -primary we again get that $\frak{m}^n\subset \frak{a}$ for some n. Since $R/\frak{m}^n$ is Artinian we get that $\overline{\frak{a}}=\overline{\frak{m}}^r$ for some r. Thus $\frak{a}=\frak{m}^r$ .

Suppose every non-zero ideal is a power of $\frak{m}$ . By Noetherian we have $\frak{m}\neq \frak{m}^2$ , so we can pick $x\in \frak{m}\setminus\frak{m}^2$ . So there is some $r$ such that $(x)=\frak{m}^r$ . Thus $r=1$ or else we’d have a prime chain longer than 1. Now given any ideal, $\frak{a}=\frak{m}^n=(x^n)$ .

Suppose $x\in R$ such that every non-zero ideal has the form $(x^k)$ . Again, we must have $(x)=\frak{m}$ . So by Noetherian $(x^k)=\frak{m}^k\neq \frak{m}^{k+1}=(x^{k+1})$ . Thus if $r\in R$ is non-zero, there is a well-defined $k$ such that $(r)=(x^k)$ . Naturally we get a discrete valuation $v(r)=k$ and extend in the obvious way to the rest of the field by $v(a/b)=v(a)-v(b)$ . By putting everything in reduced form, we see that something in $Frac(R)$ that is not in $R$ has negative valuation, and hence $R$ is the valuation ring of $v$ .

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

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