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Let $A=\bigoplus_{n \ge0} A_n$ be a Noetherian graded ring with $A_0$ Artinian. Suppose that $A=A_0[a_1,\dotsc,a_d]$ with $a_i$ having degree $1$. Let $M$ be a finitely-generated graded $A$-module. Then its Hilbert polynomial $\phi(n)_M$ is defined for $n\gg0$ and has degree $d-1$. Now let $N$ be a submodule of $M$.

My question is: how does the degree of $\phi_{M/N}$ behave as a function of $N$? What are the aspects of $N$ that affect this degree? In particular, if $N\neq 0$, can we show that $\operatorname{deg} \phi_M > \operatorname{deg} \phi_{M/N}$? Precise references in Matsumura or Eisenbud would be appreciated.

Remark: I can see that for $n\gg0$ we have $\phi_{M/N} = \phi_M - \phi_N$, however it seems to me that there is no reason in general that the leading coefficients would cancel.

Edit: Consider the special case where $A_0=k$, a field, $M=A=k[X_1,\cdots,X_n]$ and $N$ a non-zero homogeneous ideal. We have that $\operatorname{deg} \phi_M = n-1$. Question: will the degree of $\phi_{M/N}$ drop?

Manos
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  • @YACP: You are right, actually i have seen examples like that. What do you think of my edit? – Manos Jun 27 '13 at 18:54
  • @YACP: Also, as i understand the Krull dimension is 1 plus the degree of the Hilbert polynomial. In other words, the Krull dimension is equal to the degree of the Samuel function. Do you agree? – Manos Jun 27 '13 at 19:59
  • @YACP: Yes, that's what i am asking. Can you explain? It is not so clear to me, i must me missing something important. (you can make it an answer) – Manos Jun 27 '13 at 20:10
  • @YACP: because we can always extend a chain of prime ideals of $R/I$ to a chain of prime ideals of $R$ by appending the zero ideal in the end. I see...If you like, you can make these comments into an answer. I am also interested in this: can we say something in general about what would take for $N \neq 0$ to drop the dimension of the quotient $M/N$? – Manos Jun 27 '13 at 20:15
  • @YACP: thanks for the insight! – Manos Jun 27 '13 at 20:20
  • @YACP: interesting...i think this is enough food for though right now. I can accept and upvote the answerification of your comments :) – Manos Jun 27 '13 at 20:36

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Since the degree of the Hilbert polynomial of $M$ equals the Krull dimension of $M-1$, one can translate your question(s) into the dimension theory language. For instance, the inequality $\deg \phi_M > \deg \phi_{M/N}$ reduces to $\dim M>\dim M/N$. In general, this can be false: take $A=k[X,Y]$, $M=k[X,Y]/(X^2,XY)$ and $N=(\hat X)$; then $\dim M=\dim M/N=1$.

Furthermore, since $\dim M=\dim R/\operatorname{Ann}M$ and $I=\operatorname{Ann}M\subseteq\operatorname{Ann}M/N=J$ you have to compare $\dim A/I$ with $\dim A/J$ where $I\subseteq J$ are two (graded) ideals of $A$. Of course, $\dim A/J\le\dim A/I$, so you ask for the case when the inequality is strict. If $A$ is a polynomial ring over a field, then this reduces to compare $\operatorname{ht} I$ with $\operatorname{ht} J$: if $I=(0)$ and $J\neq 0$, then obviously $\dim A/J<\dim A$.

  • Dear YACP, in your first line i suppose you mean "Krull dimension of $M$-1"? Also, what is $(\hat{X})$? Is it the class of $X$ mod $(X^2,XY)$? Yes, that must be it. – Manos Jun 28 '13 at 14:10
  • Finally, if $I \subset J$ don't we have that $dim A/J \le dim A/I$? – Manos Jun 28 '13 at 14:16
  • One more thing: is it not true that $M/N \cong k$? – Manos Jun 28 '13 at 14:26