# 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.

### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

### 3 thoughts on “Hilbert Polynomial II”

1. Why would $\lambda (M_n) = H_M(n)$?

2. We can take $\lambda$ to be any additive function, and $H_M(n)$ is defined to be $\dim M_n$, so the additive function $\lambda(M_n)=\dim M_n$ is just one example of a $\lambda$ that works. This $\lambda$ is actually more general.