# 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.

## 2 thoughts on “Hilbert Polynomial I”

1. I think you meant “$K$ and $M/x_r M$ are f.g. graded modules over $k[x_1, \dots, x_{r-1}]$.”

2. hilbertthm90 says:

Thanks. Otherwise I wouldn’t be able to apply induction.