## 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: $d(R)\geq \dim R$.

## 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.

## Liouville’s Theorem for Projective Varieties?

Wow. I hate looking at the dates on old posts. I think that maybe a few days have gone by, and I’m horrified to find that 11 or 12 days have passed. It is hard to keep track of time in grad school.

The goal of this post is to prove the theorem: If V is an irreducible projective variety over an algebraically closed field k, then every regular function on V is constant. Note this says that $\mathcal{O}(V)\cong k$. Also, an exercise is to think about how this relates to Liouville’s Theorem if our field is $\mathbb{C}$.

Proof: Let V be an irreducible projective variety in $\mathbb{P}_k^n$. WLOG V is not contained in a hyperplane, since then we could just eliminate a variable and work in $\mathbb{P}_k^{n-1}$ and repeat this until it was not in any hyperplane.

Let $f\in\mathcal{O}(V)$. Consider the affine covering from last time $V=V_0\cup\cdots\cup V_n$. Note that $f\big|_{V_i}$ is regular as an affine morphism on $V_i$. So we can write this $f\big|_{V_i}$ as a polynomial in $x_j/x_i$ where $1\leq j\neq i\leq n$. i.e. we can factor out the homogeneous part of the denominator variable to get $\displaystyle f\big|_{V_i}=\frac{g_i}{x_i^{N_i}}$ where $g_i\in S(V)=k[x_0, \ldots, x_n]/I(V)$ is homogeneous of degree $N_i$.

But we assumed V irreducible, so $I(V)$ is prime and hence $S(V)$ is an integral domain. Let’s take the field of fractions then, $L=Frac(S(V))$. Then $\mathcal{O}(V)$, $k(V)$ and $S(V)$ are all embedded in L. So in L we can multiply by that denominator we had before to get $x_i^{N_i}f\in S_{N_i}(V)$.

Recall that $S(V)$ is a graded ring, so I just am denoting $S_{N_i}(V)$ to be the $N_i$-graded part. Thus if we take any integer $N\geq\sum N_i$, then $S_N(V)$ is a finite-dimensional k-vector space. Moreover, the monomials of degree N span the space.

Let $m\in S_N(V)$ be a monomial. Then it is divisible by $x_i^{N_i}$ for some i, so $mf\in S_N(V)$. Thus $S_N(V)f\subset S_N(V)$.

So we have a chain: $S_N(V)f^q\subset S_N(V)f^{q-1}\subset \cdots \subset S_N(V)f\subset S_N(V)$. So $x_0^Nf^q\in S_N(V)$ for any $q\geq 1$. Thus $S(V)[f]\subset x_0^{-N}S(V)\subset L$.

But $x_0^{-N}S(V)$ is Noetherian, since it is finitely generated as a $S(V)$-module, so $S(V)[f]$ is also finitely generated over $S(V)$. Thus $f$ is integral over $S(V)$.

i.e. there are $a_i\in S(V)$ such that $f^m+a_{m-1}+\ldots + a_0=0$. But this shows that $f$ is homogeneous of degree 0. i.e. $f\in S_0(V)\cong k$. So f is constant.