8

I am trying to solve the following exercise from Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry:

Exercise 12.1: Let $f\in R = k[x,y,z]_{(x,y,z)}$ be a homogeneous form of degree $d$, monic in $x$. Show that $(y,z)$ is an ideal of finite colength on $M = R/(f)$. Compute the corresponding Hilbert-Samuel functions.

I am aware that a similar question has been asked here: Compute the Hilbert-Samuel function. Nonetheless, since there are no answers or comments at the linked post, I'm posting it again here.

Here is what I have tried: I have been able to show that $(y,z)$ has finite colength on $M$, as it's quite easy to see that some large power of $(x,y,z)$ is contained in $(y,z) + \text{ann } M$. For the second part however, I think I am stuck. To find the Hilbert Samuel polynomial, I need to compute the length of the module $M_n = (y,z)^nM/(y,z)^{n+1}M$. I am tempted to think that this length is $d(n+1)$, because as far as I can see the $M_n$ is a finite dimensional $k$ vector space, with basis $\{x^ay^bz^c |0\le a\le d - 1, b + c = n\}$. However, I'm not sure if this is useful, or how I can translate this to a statement about the length of $M_n$. I would be glad if someone could point out how to proceed.

user940160
  • 861

1 Answers1

3

You're almost there. A key thing to know is that the only simple module over $R/(f,y,z)=R/(x^d,y,z)\cong k[x]/(x^d)$ is $k$: any simple module is cyclic, so it's a quotient of $k[x]/(x^d)$; any such quotient is of the form $k[x]/(x^e)$ for $e\leq d$, and if $e>1$ then $x^{e-1}k[x]/(x^e)$ is a proper nonzero submodule. So the length of a module over $R/(f,y,z)$ equals its dimension as a $k$-vector space.

From here, your calculation of the dimension of $(y,z)^nM/(y,z)^{n+1}M$ is correct, so you have your answer.

KReiser
  • 74,746