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

## 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.

## 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.

## Hilbert Polynomial I

I’ve been fiddling around on here for a few weeks trying to figure out what my next major set of posts should be about. I’ve finally settled. It turns out that algebraic geometry requires knowledge of a ridiculously large amount of commutative algebra. Now I usually try to avoid repeat posting when I know that I’m doing it, but I don’t think I’m going to stick to that rule for this set of posts. For probably at least the next month I’m just going to try to vastly improve my commutative algebra knowledge.

The first topic will be the Hilbert polynomial. The motivation here is that we are looking for some invariants of projective algebraic sets.

Suppose $R=\oplus R_i$ is a graded ring. Then a graded R-module, M, is a module with an abelian group decomposition $\displaystyle M=\oplus_{-\infty}^\infty M_i$ such that $R_iM_j\subset M_{i+j}$.

Let $M$ be a finitely generated graded $k[x_1,\ldots, x_r]$-module (graded by degree of the polynomial). Then we define the Hilbert function of $M$ to be $H_M(s)=\dim_k M_s$. The function takes as input something from $\mathbb{Z}$ and outputs the dimension of that graded part.

Here is where the Hilbert polynomial enters in. It turns out that $H_M(s)$ actually agrees with a polynomial of degree less than or equal to $r$ for large $s$. We will denote this polynomial $P_M(s)$.

Let’s prove a general fact first. Suppose $f(s)\in\mathbb{Z}$ is defined for all natural numbers. Then if $g(s)=f(s)-f(s-1)$ agrees with a polynomial (with rational coefficients) of degree less than or equal to $n-1$ for all $s\geq s_0$, then $f(s)$ agrees with a polynomial (with rational coefficients) of degree less than or equal to $n$ for all $s\geq s_0$.

Suppose $Q(s)$ is a polynomial that satisfies the hypothesis of the preceding statement, i.e. $Q(s)=g(s)$ for $s\geq s_0$.

Set $P(s)=f(s)$ for $s\geq s_0$ and $\displaystyle P(s)=f(s_0)-\sum_{t=s+1}^{s_0} Q(t)$ for $s\leq s_0$.

Now just note that $P(s)-P(s-1)=Q(s)$ for all integers. So we are done since then $P(s)$ is a polynomial with rational coefficients of degree less than or equal to $n$.

As you may have guessed, this little fact was to set up an induction for the actual theorem. Let’s induct on the number of variables $r$. The base case just puts us in the case where our graded module is over a field and hence is a finite-dimensional vector space. Thus dimensions all have to be zero at some grading, so $H_M(s)=0$ for large $s$ and we are done.

Suppose the theorem holds in $r-1$ variables. Now let $K$ be the kernel of the multiplication map by $x_r$. This is a submodule of $M$, and we get an exact sequence $\displaystyle 0\to K(-1)\to M(-1)\stackrel{x_r}{\to} M\to M/(x_rM)\to 0$. Where the $(-1)$ means the grading is shifted by $-1$.

The exactness tells us something about the dimensions. So look at the $s$ part of the grading: $\dim_kK(-1)_s-\dim_k M(-1)_s+\dim_k M_s-\dim_k (M/x_rM)_s=0$. In terms of the Hilbert function, this says precisely that $\displaystyle H_M(s)-H_M(s-1)=H_{M/x_rM}(s)-H_K(s-1)$.

Since $K$ and $M/x_rM$ are f.g. graded modules over $k[x_1, \ldots, x_{r-1}]$ we can apply the inductive hypothesis to the right side. But since the right side is a polynomial for large $s$, so is the left side. Now the fact we proved before this gives us the full result.

There is much to say about Hilbert polynomials, so I’ll probably keep posting about them for awhile.

## Some Notions of Dimension

It is time to pull together some ideas we’ve built up, and show that they actually correlate how we want them to.

Recall that we have a notion of dimension for a ring called the Krull dimension. Review this if necessary, but essentially you just take the sup of the heights of all prime ideals in your ring.

For a topological space, we first define “irreducible.” Irreducible simply means that you can’t express the space as the union of two nonempty closed sets. To familiarize yourself with this definition some more you can try to prove that it is equivalent to the definition that any two non-empty open sets intersect non-trivially, or any non-empty open set is dense. So note right away that irreducible is a pretty rough condition. Almost none of the spaces I usually talk about are irreducible, since there are tons of non-dense open sets. And by the other criterion, any Hausdorff space is reducible.

Moving on, we now can define the dimension of a topological space to be $sup\{n : Z_0\supsetneq Z_1\supsetneq \cdots \supsetneq Z_n\}$ where the $Z_i$‘s are irreducible subspaces.

Naturally, we are now interested to see if the topological notion of dimension on the topological space $Spec(R)$ is the same as the Krull dimension of the ring R.

First, we’ll need a quick lemma:

A subspace $E\subset Spec(R)$ is irreducible if and only if $I(E)$ is a prime ideal. Note that this is really exactly what we wanted to happen, since prime ideals are points in Spec(R). When we say something is “irreducible,” what we are talking about are the smallest things that cannot be broken apart, i.e. points. I confess I am far oversimplifying this idea of “points” as we will see before the week is over if all goes as planned. (At this point you might want to review spec).

Proof of Lemma: I promise to fill this in later in the week. I just realized that it is an assignment to be turned in on Friday, and not all the readers of this blog that are in my commutative algebra class have this done yet.

Now let’s actually check dimensions. Theorem: $Krulldim(R)=dim(Spec(R))$. Just write it out now:

$dim(Spec(R))=sup\{n: Z_0\supsetneq Z_1\supsetneq \cdots \supsetneq Z_n , \ Z_i \ irreducible\}$
$=sup\{n: p_0\subsetneq p_1\subsetneq \cdots \subsetneq p_n , \ p_i=I(Z_i)\}$
$= Krulldim(R)$ where that switching comes from the Lemma. The correspondence is 1-1.

So for some future posts, I want to clarify some more on dimension and what “points” really are. I also want to do the Nullstellensatz and talk about why it is so important, but I may not. I’ll do some hunting to see what other math bloggers have done on it. I’m pretty sure it has been posted on extensively already, in which I’ll just point people in those directions and add some things that I personally find interesting.