A Mind for Madness

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Hilbert Polynomial II

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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.

3 thoughts on “Hilbert Polynomial II

  1. Pingback: Beginning Dimension Theory « A Mind for Madness

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

  3. 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.

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