# Discrete Valuation Rings

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 $v$.

# Properties of Completions

First note that taking an inverse limit is a functor. I won’t need the functorial properties in the immediate future, but it would be good to state some of them. First off, the functor is not exact, but it is left exact. So given an exact sequence of inverse systems $0\to \{A_n\}\to \{B_n\}\to \{C_n\}\to 0$ (it is exact at each level and all the diagrams commute) we get an exact sequence $\displaystyle 0\to \lim_{\longleftarrow}A_n\to \lim_{\longleftarrow}B_n\to \lim_{\longleftarrow} C_n$.

It turns out that if the first system $\{A_n\}$ has the property that the homomorphisms $\theta_n$ are surjective, then the inverse limit is exact. So in our case with completions, this always happens.

The properties I’d really like to prove are the ones listed in Hartshorne without proof. Suppose for the rest of this post that $(R, \frak{m})$ is a Noetherian local ring and $\hat{R}$ is its completion with respect to the $\frak{m}$-adic topology. The numbers will refer to Hartshorne numbering:

5.4A(a) $\hat{R}$ is a local ring with maximal ideal $\hat{\frak{m}}=\frak{m}\hat{R}$ and there is a natural injective homomorphism $R\to \hat{R}$.

We already discussed the second part, since the kernel of the hom is just $\cap \frak{m}^n=0$. Using right exactness of tensoring and exactness of completions, we get that for any finitely generated $R$-module $M$, $\hat{R}\otimes_R M\to \hat{M}$ is an iso (if we remove Noetherian on R, we only get surjective). This gives us that $\hat{R}\otimes_R \frak{m}\to \hat{\frak{m}}$ is an isomorphism and since the image is $\frak{m}\hat{R}$ we get the first part of the statement.

Now we need that it is a unique maximal ideal. But applying the above result to any ideal (which is finitely generated since R is Noetherian) we get that $\widehat{\frak{a}^n}=\frak{a}^n\hat{R}=(\hat{R}\frak{a})^n=(\hat{\frak{a}})^n$. Thus $R/\frak{a}^n\cong \hat{R}/\hat{\frak{a}}^n$. Taking inverse limits gives that $R/\frak{m}\cong \hat{R}/\hat{\frak{m}}$ and hence $\hat{\frak{m}}$ is a maximal ideal since the quotient is a field. But for any $x\in\hat{m}$, we can define an inverse for $(1-x)$ formally by $(1-x)^{-1}=1+x+x^2+\cdots$. Since we are in the completion, this converges in $\hat{R}$ and hence $x\in J(\hat{R})$. But a maximal ideal $\hat{\frak{m}}\subset J(\hat{R})$ means that it is the Jacobson radical and hence the unique maximal ideal.

5.4A (b) If $M$ is a finitely generated $R$-module, its completion $\hat{M}$ is isomorphic to $M\otimes_R \hat{R}$. Well, I needed this to prove (a) and briefly described how it would go, but since I didn’t prove the exactness properties, it seems needlessly detailed to do a full proof using them. For more details, see posts at delta epsilons.

5.4A (c) $\dim R=\dim \hat{R}$.

Let’s use some of the machinery we developed. The dimensions are equal to the degree of the Hilbert polynomial, but $R/\frak{m}\cong \hat{R}/\hat{\frak{m}}$ says precisely that $\chi_m(n)=\chi_{\hat{m}}(n)$. So the polynomials are actually the same.

5.4A (d) $R$ is regular if and only if $\hat{R}$ is regular.

We’ll hold off on this until I cover regularity (which will either be next time or the time after).

# Some Corollaries

Today will just be some quick results we get from this build up.

First, if we localize a polynomial ring at a maximal ideal, say $k[x_1, \ldots, x_n]$ at $\frak{m}=(x_1, \ldots, x_n)$, then $\dim R_\frak{m}=n=\dim R$. This is because $G_m(R)$ has Poincare series $(1-t)^{-n}$ so the order of the pole is $n$ which is the dimension by the last post.

This one will be really useful later: $\dim R\leq \dim_k(\frak{m}/\frak{m}^2)$. Let $\{x_i\}_1^r \subset\frak{m}$ such that $\overline{x_i}\in \frak{m}/\frak{m}^2$ are a basis for the vector space. Then by Nakayama’s Lemma the $x_i$ generate $\frak{m}$. Thus $\dim_k(\frak{m}/\frak{m}^2)=s\geq \dim R$.

This one is also useful in algebraic geometry. If $R$ is Noetherian, and $x_1, \ldots , x_r\in R$, then every minimal ideal $\frak{p}$ belonging to $(x_1, \ldots, x_r)$ has height $\leq r$. Unfortunately, we cannot push this to equality. Geometrically the example is that if $Y$ is the twisted cubic, then $I(Y)$ has height 2, but cannot be generated by less than 3 elements.

Lastly, we’ll do the famous Principal Ideal Theorem. If $R$ is Noetherian and $x\in R$ is neither a zero-divisor nor a unit, then every minimal prime ideal $\frak{p}$ of $(x)$ has height 1. By the last paragraph we know that $ht(p)\leq 1$. If $ht(\frak{p})=0$ then it belongs to $(0)$. Thus every element of $\frak{p}$ is a zero-divisor which is a contradiction since $x\in \frak{p}$.

# Beginning Dimension Theory

Recall the purpose of this development is to get some results on ring dimensions. All the hypothesis and notation from last time still hold (the important one to remember is that $(R, \frak{m})$ is a local ring).

Let’s introduce a new notation, which will disappear shortly. We call the characteristic polynomial of the $\frak{m}$-primary ideal $\frak{q}$, $\chi_q^M(n)=l(M/\frak{q}^nM)$. An immediate corollary to the last post is that for large $n$, $\chi_q(n)=l(R/\frak{q}^n)$ has degree $\leq s$ where $s$ is the least number of generators of $\frak{q}$.

Now we want to show that for our purposes the choice of $\frak{m}$-primary ideal doesn’t matter. The claim is that $\deg \chi_q(n)=\deg \chi_m(n)$.

We know that there is some integer $r$ such that $\frak{q}$ contains $\frak{m}^r$. i.e. $\frak{m}\supset \frak{q}\supset \frak{m}^r$. Thus $\frak{m}^n\supset \frak{q}^n \supset \frak{m}^{rn}$. Thus for large $n$, we get $\chi_m(n)\leq \chi_q(n)\leq \chi_m(rn)$. Since these are polynomials, we let $n$ tend to $\infty$ to get the claim.

Let’s denote the common degree $d(R)$. Thus $d(R)$ is the order of the pole at $t=1$ of the Hilbert function of $G_\frak{m}(R)$.

Since this is short so far, we will very briefly start our first goal of showing that if $\delta(R)$ is the least number of generators of an $\frak{m}$-primary ideal, and we impose Noetherian on $R$, then $\delta(R)=d(R)=\dim R$.

What we just showed above in this new notation is that $\delta(R)\geq d(R)$. The way we will eventually show the equality is to get $\delta(R)\geq d(R)\geq \dim R \geq \delta(R)$.

The next step is involved and needs the Artin-Rees Lemma, so I’ll hold off and do it next time.

# Applying the Hilbert Polynomial

Let’s start applying to some specific situations now. Suppose $R$ is a Noetherian local ring with maximal ideal $\frak{m}$. Let $\frak{q}$ be an $\frak{m}$-primary ideal. Let $M$ be a finitely-generated $R$-module, and $(M_n)$ a stable $\frak{q}$-filtration of $M$.

Don’t panic from the set-up. I think I haven’t talked about filtrations. All the stable $\frak{q}$-filtration means is that we have a chain of submodules $M=M_0\supset M_1\supset \cdots \supset M_n\supset \cdots$ such that $\frak{q}M_n=M_{n+1}$ for large $n$.

The goal for the day is to prove three things.

1) $M/M_n$ has finite length for all $n\geq 0$.

Define $G(M)=\bigoplus \frak{q}^n/\frak{q}^{n+1}$ and $G(M)=\bigoplus M_n/M_{n+1}$. We have a natural way to make $G(M)$ into a finitely-generated graded $G(R)$-module. The multiplication in the ring comes from the following. If $x_n\in\frak{q}^n$, then let the image in $\frak{q}^n/\frak{q}^{n+1}$ be denoted $\overline{x_n}$. We take $\overline{x_n}\overline{x_m}=\overline{x_nx_m}$. This does not depend on representative.

We’ll say $G_n(M)$ is the n-th grade: $M_n/M_{n+1}$. Now $G_0(R)=R/q$ is an Artinian local ring and each $G_n(M)$ is a Noetherian $R$-module annihilated by $\frak{q}$. Thus they are all Noetherian $R/\frak{q}=G_0(R)$-modules. So by the Artinian condition we get that each $G_n(M)$ is of finite length. Thus $l_n=l(M/M_n)=\sum_{r=0}^{n-1} l(G_r(M))<\infty$.

2) For large $n$, $l(M/M_n)$ is a polynomial $g(n)$ of degree $\leq s$ where $s$ is the least number of generators of $\frak{q}$.

Suppose $x_1, \ldots, x_s$ generate $\frak{q}$. Then $\{\overline{x_i}\}$ in $\frak{q}/\frak{q}^2$ generate $G(R)$ as an $R/\frak{q}$-algebra. But $l$ is an additive function on the filtration, so by last time we saw thatfor large $n$ there is some polynomial such that $f(n)=l(G_n(M))=l(M_n/M_{n+1})$, and each $\overline{x_i}$ has degree 1, so the polynomial is of degree $\leq s-1$.

Thus we get that $l_{n+1}-l_n=l(G_n(M))=f(n)$. So from two posts ago, we get for large $n$ that $l_n$ is some polynomial $g(n)$ of degree $\leq s$.

3) Probably the most important part is that the degree and leading coefficient of $g(n)$ depends only on $M$ and $\frak{q}$ and not on the filtration.

Let $(\overline{M_n})$ be some other stable $\frak{q}$-filtration with polynomial $\overline{g}(n)=l(M/\overline{M_n})$. Since any two stable $\frak{q}$-filtrations have bounded difference, there is an integer $N$ such that $M_{n+N}\subset \overline{M_n}$ and $\overline{M_{n+N}}\subset M_n$ for all $n\geq 0$. But this condition on the polynomials says that $g(n+N)\geq \overline{g}(n)$ and $\overline{g}(n+N)\geq g(n)$, which means that $\lim_{n\to\infty}\frac{g(n)}{\overline{g}(n)}=1$. Thus they have the same degree and leading coefficient.

That seems to be enough for one day. Unfortunately, I haven't quite got to the right setting that I want yet.

# Localization 2

Let’s figure out what “local” means and see if our construction somehow makes a local ring, i.e. is a “localization.”

Local: A ring is called local if there is a unique maximal ideal. This seems like a rather silly term, but it actually makes sense when you look at how rings arise in algebraic geometry or manifold theory. We won’t go there, though.

Sadly, it turns out that $S^{-1}R$ is not always a local ring. But this is where primality comes into play. If $P\subset R$ is a prime ideal then $S=R\setminus P$ is a multiplicative set. Suppose it weren’t, then there would be two elements $x,y\in S$ such that $xy\notin S$, i.e. $xy\in P$, but this is impossible, since by definition either $x\in P$ or $y\in P$. We now denote the localization of $R$ at $P$, to be $S^{-1}R=(R\setminus P)^{-1}R$ which we denote with the shorthand $R_P$. This does turn out to be local since by the property listed last time of the embedding $\phi^{-1}(S^{-1}P)=P$, so $S^{-1}P=\{r/s : r\in P, \ s\notin P\}$ is the unique maximal ideal in $R_P$.

Proof: Suppose $x\in R_P$, then $x=r/s$ with $r\in R$ and $s\notin P$. If $r\notin P$, then $r/s$ is a unit in $R_P$. So all nonunits are in $S^{-1}P$. Now if I is any ideal in $R_P$ that contains an element $r/s$ with $r\notin P$, then $I=R_P$. Thus every proper ideal in $R_P$ is contained in $S^{-1}P$. So $R_P$ is local with unique max ideal $S^{-1}P$. For notational purposes outside of this blog, people usually write the prime ideal as $\mathfrak{p}$ and the unique maximal ideal of $R_\mathfrak{p}$ as $\mathfrak{p}R_{\mathfrak{p}}$.

I guess I’ve been rather sparse on the examples. The first one that comes to mind is surely to take $\mathbb{Z}=R$. Then our prime ideals are just the principal ideals generated by the primes, so take $P=p\mathbb{Z}$ for some prime p. Then $\mathbb{Z}_P=\mathbb{Z}_{(p)}$.

I guess the importance of prime ideals leads us to explore some properties of prime ideals that could be useful.

Property 1: If $S\subset R$ is any multiplicative set (not containing 0) and if $P\subset R\setminus S$ is a maximal ideal, then $P$ is prime. Also, any ideal $I\subset R\setminus S$ is contained in such a $P$. I’ll omit proving this. The first part is fiddling with things until it works and the second statement uses Zorn’s Lemma.

OK, well I thought I had some other properties, but I can’t seem to find them/think of them now. I’m not sure where I’m going next. I’ll either move on to some related things to get at this better like the nilradical, or I’ll generalize this one more time to modules and do it using the categorical construction. If anyone has suggestions on which of these paths to take, just post. You probably have a few days as I’ll get busy again.