# Computation of a Hilbert Samuel function

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.

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.