# Integrally Closed Noetherian Domain

I don’t feel like an all out post, but I did need to use this fact today and it relates to what I’ve been doing, so may as well post the proof. Let $R$ be an integrally closed Noetherian domain. Then $R=\cap_{ht(p)=1}R_p$.

We start with a Lemma: Every prime divisor of a non-zero principal ideal has height 1.

Suppose $P$ is a prime divisor of $aR$. Then there is an element $b\in R$ such that $(aR:b)=P$. Define $\frak{m}=PR_P$. So $(aR_P:b)=\frak{m}$. Thus $ba^{-1}=\frak{m}^{-1}$ and $ba^{-1}\notin R_P$.

If we had $ba^{-1}\frak{m}\subset \frak{m}$ then the standard determinant trick (see a proof a Nakayama’s Lemma) would give us that $ba^{-1}$ is integral over $R_P$ contradicting $R_P$ being integrally closed. So $ba^{-1}\frak{m}=R_P$ and hence $\frak{m}^{-1}\frak{m}=R_P$ and so $R_P$ is a DVR which means dimension of $R_P$ is 1 which means $ht(P)=1$. This gives the lemma.

We will prove that for $a,b\in R$ with $a\neq 0$ we have $b\in aR_p$ for all height 1 primes means that $b$ is actually in $aR$.

Let $P_1, \ldots, P_n$ be the prime divisors of $aR$ and let $q_1\cap \cdots \cap q_n$ be a primary decomposition of $aR$ where $q_i$ is $P_i$-primary. But each $P_i$ has height 1 by the Lemma above, so $b\in aR_{P_i}\cap R=q_i$. Thus $b\in \cap q_i=aR$.

Maybe a short discussion could be useful at this point. The use of this fact came up when trying to prove something about a normal point. This says that at a normal point on a variety, if you want to check if a function regular, you only need to check that it is regular on all codimension 1 subvarieties through that point. This almost immediately gives that if you have a regular function on $X\setminus P$ ($\dim X\geq 2$) where $P\in X$ is a normal point. Then the function actually extends to be regular on all of $X$. A function on a $\geq 2$ dim variety cannot blow up at just a single normal point.

# 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 $v$.

# Regular Local Rings I

We have defined and used the associated graded ring $G_a(R)=\bigoplus \frak{a}^n/\frak{a}^{n+1}$. Now we want to see how it behaves under completions.

By the last post, we have $\frak{a}^n/\frak{a}^{n+1}\cong \hat{\frak{a}}^n/\hat{\frak{a}}^{n+1}$, so we immediately get that $G_a(R)\cong G_{\hat{a}}(\hat{R})$.

A great theorem that I’ll skip proving is that if $R$ is Noetherian, and $\frak{a}$ is any ideal, then the completion with respect to the $\frak{a}$-adic topology is Noetherian. As a corollary we get that for any Noetherian ring $R$, $R[[x_1, \ldots, x_n]]$ is Noetherian by noting that the completion of the Noetherian ring $R[x_1, \ldots, x_n]$ with respect to the $(x_1, \ldots, x_n)$-adic topology is $R[[x_1, \ldots, x_n]]$.

After this brief excursion, we’ll come back to the dimension theory we left off from. The next natural place to go is to regular local rings. A local ring $(R, \frak{m}, k)$ is regular if $\dim_k(\frak{m}/\frak{m}^2)=\dim R$. (Recall that it is always true that $\dim_k(\frak{m}/\frak{m}^2)\geq \dim R$).

Suppose we have a Noetherian local ring such that $\dim R=d$. Then the following are equivalent definitions of regular: $G_m(A)\cong k[t_1, \ldots, t_d]$ or $\frak{m}$ can be generated by $d$ elements.

The first condition implies that $\dim_k(\frak{m}/\frak{m}^2)=d$, so it implies regular. Regular implies that $\frak{m}$ can be generated by $d$ elements, by this post. Lastly, if $\frak{m}$ can be generated by $x_1, \ldots, x_d$ (if you’ve seen the term, this is a system of parameters), then we have a surjective map of graded rings $\phi: k[x_1, \ldots, x_d]\to G_m(A)$ with kernel $\cap \frak{m}^n=0$. So it is an iso. This finishes up the equivalences.

Last time we saw without proof that (for Noetherian local rings) $R$ is regular if and only if $\hat{R}$ is regular. Now we can prove it.

By the equivalent definition of regular, $R$ is regular iff $G_m(R)\cong k[t_1, \ldots, t_n]$, but we proved that $G_m(R)\cong G_{\hat{m}}(\hat{R})$, so $G_{\hat{m}}(\hat{R})\cong k[t_1, \ldots, t_d]$ but this happens iff $\hat{R}$ is regular.

We’ll wrap up today with trying to keeping the geometric picture in mind. Regular means non-singular geometrically. So we see that passing to the completion doesn’t introduce any singularities. But since the dimension of the local ring at a point equals the dimension of the variety we actually get that completion of $\mathcal{O}_P$ where $P$ is non-singular is $k[[x_1, \ldots, x_n]]$ where $n$ is the dimension of the variety.

So if we interpret completion of $\mathcal{O}_P$ as the “analytically local” picture, then we see that locally all non-singular points on a variety are analytically isomorphic.

# 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).

# Completions II

We will call a topological group complete if $\phi: G\to \hat{G}$ is an isomorphism.

The case that we are particularly concerned with is when our group is a ring $R$ and we take for our inverse system some ideal $\frak{a}\subset R$ and $G_n=\frak{a}^n$. The topology that this determines is the “$\frak{a}$-adic topology”. This makes $R$ into a topological ring.

If we take the completion $\displaystyle \hat{R}=\lim_{\longleftarrow} R/G_n$, then the continuous ring homomorphism $\phi: R\to \hat{R}$ has kernel $\cap \frak{a}^n$.

Now we can also do all this with $R$-modules by taking the group to be $M$ and the inverse system $G_n=\frak{a}^nM$. The topology determined by this system is called the $\frak{a}$-topology on M. If we take the completion with respect to this topology (i.e. w.r.t this system), we get $\hat{M}$ which is a topological $\hat{R}$-module meaning the $\hat{R}$ action is continuous.

Rephrasing the motivating example from last time in this language we see that the $p$-adic integers are formed as the completion of the ring $\mathbb{Z}$ with respect to the $\frak{a}$-topology where $\frak{a}$ is the ideal $\frak{a}=(p)$.

The other really important example is to form the completion of $k[x]$ with respect to the $(x)$-adic topology. The completion is $\displaystyle \widehat{k[x]}=\lim_{\longleftarrow} k[x]/(x^n)=k[[x]]$ the ring of formal power series. Recall that by definition the inverse limit are all sequences $(a_0, \ldots, a_n, \ldots)$ such that $a_{n+1} \mod x^{n+1}\equiv a_n$. This just says that each $a_i$ is a polynomial, and it has to agree with the one before it up to the $x^i$ coefficient. So we can write each sequence $b_0+b_1x+\cdots +b_nx^n+\cdots$ where $b_i$ is the coefficient on the $x^i$ of the polynomial $a_i$. And for any power series we get a sequence in this way.

Recall our notion of $\frak{a}$-filtrations. We had a chain $M=M_0\supset M_1\supset \cdots$ such that $\frak{a}M_n\subset M_{n+1}$, and if equality held for all large $n$, then we called the filtration stable. Well, in our new language, these filtrations are inverse systems of modules, and hence determine a topology on $M$. A few posts ago we used the fact that any stable $\frak{a}$-filtrations have bounded difference. In this new language, this says precisely that all stable $\frak{a}$-filtrations determine the same topology on M, moreover this is the $\frak{a}$-topology.

Lastly, if we convert the Artin-Rees Lemma to this language, we get that if $R$ is Noetherian, $\frak{a}$ an ideal, $M$ a f.g. $R$-module, and $M'$ a submodule of $M$, then the $\frak{a}$-topology on $M'$ is actually just the subspace topology from the $\frak{a}$-topology on $M$.

We should probably do some properties of completions next time.

# Completions I

Today we’ll start a new section, but only because it is a tool we need when we come back to the stuff we just finished. We will look at completions.

To motivate the process take a Hausdorff abelian topological group $G$. Suppose there is a countable local basis at 0 (which implies countable basis, since the neighborhoods of 0 determine the entire topology). Since we assumed Hausdorff we have the usual notion of Cauchy sequences, so we can define the completion of $G$ to be completion in the usual sense $\hat{G}$. In particular, if $G=\mathbb{Q}$, then $\hat{\mathbb{Q}}=\mathbb{R}$.

Now suppose we have a local basis about 0 of subgroups (this rules out $\mathbb{Q}$), say $G=G_0\supset G_1\supset \cdots \supset G_n\supset \cdots$. If we are in this situation, then our topology is actually determined by a sequence of subgroups, so we will want to try to define the completion solely in terms of algebra.

Take any Cauchy sequence $(x_n)\subset G$. If we fix k, then at some $M(k)$, $\overline{x_n}\in G/G_k$ is constant for all $n\geq M(k)$. Note that $M$ really does depend on $k$. Set the limit $\overline{x_n}\to x_{M(k)}$.

If we make what we mod out by bigger, namely we go from $k+1$ to $k$, then projection $\theta_{k+1}: G/G_{k+1}\to G/G_k$ maps $x_{M(k+1)}\mapsto x_{M(k)}$. Thus our Cauchy sequence $(x_n)$ determined a “coherent sequence” $(x_{M(k)})$, where $\theta_{n+1}x_{M(n+1)}=x_{M(n)}$.

Conversely, we can define a Cauchy sequence corresponding to any coherent sequence by just picking an element in the equivalence class at each step. So we can now define the completion $\hat{G}$ to be the set of coherent sequences with group structure given entry-wise by the quotient group. The standard example here is the $p$-adic integers, where the group is $\mathbb{Z}$ and our fundamental system is $\mathbb{Z}\supset p\mathbb{Z}\supset p^2\mathbb{Z}\supset \cdots \supset p^n\mathbb{Z}\supset \cdots$. Coherent sequences are $(a_0, a_1, \ldots )$ where $a_{n+1}\mod p^n\equiv a_n$.

Whenever we have in general a sequence of groups $\{A_n\}$ and homomorphisms $\theta_{n+1} A_{n+1}\to A_n$ this is called an inverse system. The group of all coherent sequences is called the inverse limit of the system written $\displaystyle \lim_{\longleftarrow} A_n$. Thus our definition of completion can be written succinctly as $\displaystyle\hat{G}=\lim_{\longleftarrow} G/G_n$.

Next time we’ll transfer this to module language and get to a few results.

# 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 $x\in \frak{p}$.

# Finishing up Dimensions

We are now on the last inequality: $\dim R\geq \delta(R)$. Recall we’re supposing $(R, \frak{m})$ is Noetherian and local. Let $\dim R=d$, then the inequality is saying we can find an ideal, $\frak{q}$ that is an $\frak{m}$-primary ideal and is generated by $d$ elements: $x_1, \ldots, x_d$.

Let’s construct these elements inductively. The way we want to do it is so that at each step any prime ideal containing $(x_1, \ldots, x_i)$ has height bigger than or equal to $i$ to force the dimension of $R$ to be big.

Suppose the $x_1, \ldots, x_{i-1}$ have been constructed in the given way. Let $\{p_j\}$ be the minimal prime ideals of $(x_1, \ldots, x_{i-1})$ with height exactly $i-1$. But we have that $\frak{m}\neq p_j$ for any $j$ since $ht(\frak{m})=\dim R=d>i-1=ht(p_j)$. Thus $\frak{m}\neq \cup p_j$ (there is a well-known fact that if any ideal $\frak{a}\subset \cup p_j$, then in fact $\frak{a}\subset p_j$ for some $j$).

Now pick some element $x_i\in \frak{m}\setminus (\cup p_j)$, and let $\frak{q}$ be a prime ideal containing $(x_1, \ldots, x_i)$. We definitely have that $ht(\frak{q})\geq i$, since $\frak{q}$ contains a minimal prime ideal of $(x_1, \ldots , x_{i-1})$, say $\frak{p}$. If for some $j$, $\frak{p}=p_j$, then since $x_i\in\frak{q}\setminus \frak{p}$, we have strictly $\frak{q}\supset \frak{p}$ increasing the height. If $\frak{p}\neq p_j$ for any $j$, then since $\{p_j\}$ are all minimal primes of height $i-1$, we have $ht(\frak{q})>i-1$.

All that is left is to show that $\frak{q}=(x_1, \ldots, x_d)$ is $\frak{m}$-primary. Let $\frak{p}$ be any prime ideal of $\frak{q}$. Then if $\frak{p}\subsetneq \frak{m}$ we have that $ht(\frak{p})< ht(\frak{m})=d$. So this is impossible and we have $\frak{p}=\frak{m}$. Thus $\frak{q}$ is $\frak{m}$-primary.

Over the last couple of posts we have finally completed the first goal. We have $\delta(R)=d(R)=\dim R$. In other words, for Noetherian local rings we have an equivalence between the maximum length of chains of prime ideals, the degree of the Hilbert polynomial, and the least number of generators of an $\frak{m}$-primary ideal.

Next time I'll derive some results directly from this including the Principal Ideal Theorem. Then we'll move on to something different (only for awhile, then we'll return).

# The Next Inequality

Considering it has been at least a post removed, I’ll bring us back to our situation. We have a local Noetherian ring $(R, \frak{m})$. Our notation is that $\delta(R)$ is the least number of generators of an $\frak{m}$-primary ideal (which was shown to be independent of choice of ideal here). The goal for the day is to show that $d(R)\geq \dim R$.

Suppose $\frak{q}$ is $\frak{m}$-primary. We’ll prove something more general. Let $M$ be a finitely generated $R$-module, $x\in R$ a non-zero divisor in $M$ and $M'=M/xM$. Then the claim is that $\deg\chi_q^{M'}\leq \deg\chi_q^M -1$.

Since $x$ is not a zero-divisor, we have an iso as $R$-modules: $xM\cong M$. Define $N=xM$. Now take $N_n=N\cap \frak{q}^nM$. Since $\frak{q}^nM$ is a stable $\frak{q}$-filtration of $M$, by Artin-Rees we get that $(N_n)$ is a stable $\frak{q}$-filtration of $N$.

For each $n$ we have $0\to N/N_n \to M/\frak{q}^nM\to M'/\frak{q}^nM'\to 0$ exact.

Thus we get $l(N/N_n)-l(M/\frak{q}^nM)+l(M'/\frak{q}^nM')=0$. So if we let $g(n)=l(N/N_n)$, we get for large $n$: $g(n)-\chi_q^M(n)+\chi_q^{M'}(n)=0$.

But $(N_n)$ is also a stable $\frak{q}$-filtration of $M$, since $N\cong M$. We already showed that the degree and leading coefficient of $g(n)$ depends only on $M$ and $\frak{q}$ and not on the filtration. Thus $g(n)$ and $\chi_q^M(n)$ have the same degree and leading coefficient, so the highest powers kill eachother which gives $\deg\chi_q^{M'}\leq \deg \chi_q^M-1$.

In particular, we will need that $R$ as an $R$-module gives us $d(R/(x))\leq d(R)-1$.

Now we prove the goal for today. For simplicity, let $d=d(R)$. We will induct on $d$. The base case gives that $l(R/\frak{m}^n)$ is constant for large $n$. In particular, there is some $N$ such that $\frak{m}^n=\frak{m}^{n+1}$ for all $n>N$. But we are local, so $\frak{m}=J(R)$ and hence by Nakayama, $\frak{m}^n=0$. Thus for any prime ideal $\frak{p}$, we have $\frak{m}^k\subset \frak{p}$ for some $k$, so take radicals to get $\frak{m}=\frak{p}$. Thus there is only one prime ideal and we actually have an Artinian ring and hence have $\dim R=0$.

Now suppose $d>0$ and the result holds for $\leq d-1$. Let $p_0\subset p_1\subset \cdots \subset p_r$ be a chain of primes. Choose $x\in p_1\setminus p_0$. Define $R'=R/p_0$ and $\overline{x}$ be the image of $x$ in $R'$.

Note that since $R'$ is an integral domain, and $\overline{x}$ is not 0, it is not a zero-divisor. So we use our first proof from today to get that $d(R'/(\overline{x}))\leq d(R')-1$.

Let $\frak{m}'$ be the maximal ideal of $R'$. Then $R'/\frak{m}'$ is the image of $R/\frak{m}$, so $l(R/\frak{m}^n)\geq l(R'/\frak{m}'^n)$ which is precisely $d(R)\geq d(R')$. Plugging this into the above inequality gives $d(R'/(\overline{x}))\leq d(A)-1=d-1$.

So by the inductive hypothesis, $\dim(R'/\overline{x})\leq d-1$. Take our original prime chain. The images form a chain $\overline{p}_1, \ldots , \overline{p}_r$ in $R'/(\overline{x})$. Thus $r-1\leq d-1\Rightarrow r\leq d$. Since the chain was arbitrary, $\dim R\leq d(R)$.

A nice corollary here is that the dimension of any Noetherian local ring is finite. Another similar corollary is that in any Noetherian ring (drop the local) the height of a prime ideal is finite (and hence primes satisfy the DCC), since $ht(p)=\dim A_p$ which is local Noetherian.

# The Artin-Rees Lemma

We have a somewhat bumpy road to traverse today. I’ll start with the Artin-Rees lemma and see if we can get to a use of it to continue our set of inequalities we’re trying to prove.

First we’ll need some new ideas. Suppose $R$ is any old ring (in particular, we are dropping graded and Noetherian assumptions). Then if $\frak{a}$ is an ideal, we can form a new ring $R^*=\bigoplus_{n=0}^\infty \frak{a}^n$ which by construction is graded. Now for any $R$-module, say $M$ and an $\frak{a}$-filtration $M_n$ we can form a graded $R^*$-module, $M^*=\bigoplus M_n$.

Note that if $R$ is Noetherian in the situation above, then $\frak{a}=(x_1, \ldots, x_r)$, so $R^*=R[x_1, \ldots , x_r]$, so by Hilbert Basis Theorem, we get $R^*$ is Noetherian.

We’ll need that in the situation above the following two statements are equivalent: $M^*$ is finitely generated as an $R^*$-module, and that the filtration $M_n$ is stable.

Proof: Each $M_n$ is finitely generated, so $Q_n=\bigoplus_{r=0}^n M_r$ is finitely generated for all $n$. Let’s form $M_n^*=Q_n\oplus\left(\bigoplus_{k=1}^\infty \frak{a}^kM_n\right)$. We have that each $Q_n$ is finitely generated as an $R$-module, so we get that $M_n^*$ is finitely generated as an $R^*$-module.

Clearly, $M_0^*\subset M_1^*\subset \cdots$, so since $R^*$ is Noetherian we get that $M^*$ is finitely generated iff the ascending chain terminates iff $M_{n_0+r}=\frak{a}^r M_{n_0}$ for some $n_0$ and for all $r\geq 0$ iff the filtration is stable.

Now we can prove the Artin-Rees Lemma which says that if $R$ is a Noetherian ring, $\frak{a}$ an ideal, $M$ a finitely generated $R$-module, $M_n$ a stable $\frak{a}$-filtration and $M'$ a submodule of $M$, then $M'\cap M_n$ is a stable $\frak{a}$-filtration of $M'$.

The situation is fairly simple from the previous fact. Note that $\frak{a}(M'\cap M_n)\subset \frak{a}M'\cap \frak{a}M_n\subset M'\cap M_{n+1}$. So we do indeed get a filtration. But $M'^*$ is a graded $A^*$-submodule of $M^*$, so it is finitely generated. Now by the equivalence of finitely generated and stable we are done.

There are two important corollaries (both get referred to as the Artin-Rees Lemma as well). In the special case $M_n=\frak{a}^nM$ we get that the stable filtration condition says that there is some integer $N$ such that $(\frak{a}^nM)\cap M'=\frak{a}^{n-N}((\frak{a}^NM)\cap M')$ for all $n\geq N$.

The other result uses the bounded difference result from last time. Since $\frak{a}^nM'$ and $(\frak{a}^nM)\cap M'$ are both stable $\frak{a}$-filtrations, they have bounded difference, so the $\frak{a}$-topology of $M'$ coincides with induced topology from the $\frak{a}$-topology on $M$.

I think that is sufficient for today. Next time I’ll go ahead and knock off the next step of the inequalities: $d(R)\geq \dim R$.