Hilbert Polynomial II

October 28, 2009 by hilbertthm90 | 3 Comments

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.

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Author: hilbertthm90

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

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Hilbert Polynomial II

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