# 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.

# 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.

# Hilbert Polynomial II

My overall goal has not changed, but I definitely have a much clearer picture of where my posts are headed for right now. I recently was working on what happens to dimension when you intersect varieties, and I needed a commutative algebra result that sort of surprised me. So that is my first benchmark on this front. Lucky for me, there is a nice clean way to prove it using the Hilbert polynomial, so I can just continue this course for now.

Let’s now reconstruct the Hilbert polynomial in a different way. As before let $M$ be a finitely generated graded $R$-module. Then $M_n$ is finitely generated as an $A_0$-module.

Let $\lambda$ be an additive funtion (in $\mathbb{Z}$) on the class of finitely generated $A_0$-modules. We define the Poincare series of $M$ to be the generating funciton of $\lambda(M_n)$. So we get a power series with coefficients in $\mathbb{Z}$: $P(M, t)=\sum \lambda(M_n)t^n$.

By a remarkably similar argument to the last post we can check by induction that $P(M, t)$ is a rational function in $t$ of the form $\displaystyle \frac{f(t)}{\prod_{t=1}^s (1-t^{k_i})}$ where $f(t)\in\mathbb{Z}[t]$.

Let’s suggestively call the order of the pole at $t=1$, $d(M)$.

We now simplify the situation by taking all $k_i=1$. Then the main idea for today is that $\lambda(M_n)$ is a polynomial of degree $d-1$. In fact, $\lambda(M_n)=H_M(n)$.

Our simplification gives that $P(M, t)=f(t)\cdot (1-t)^{-s}$. So $\lambda(M_n)$ is the coefficient of $t^n$. If we cancel factors of $(1-t)$ out of $f(t)$ we can assume $f(1)\neq 0$ and that $s=d$. Write $f(t)=\sum_{k=0}^N a_kt^k$. Then since $\displaystyle (1-t)^{-d}=\sum_{k=0}^\infty \left(\begin{matrix} d+k-1 \\ d-1 \end{matrix}\right) t^k$ we get that $\lambda(M_n)=\sum_{k=0}^N a_k \left(\begin{matrix} d+n-k-1 \\ d-1 \end{matrix}\right)$ for all $n\geq N$.

Thus we get a polynomial with non-zero leading term. Note the values at integers are integers, but the coefficients in general are only rationals.

Since $\lambda$ was any additive function, this is a bit more general. But taking $\lambda(M_n)=\dim M_n$ we get the Hilbert polynomial from last time.

Next time we’ll start using this to streamline some proofs about dimension.

# 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.