Let $S = k[x_1,\dots,x_n]$ be a polynomial ring over a field $k$ of characteristic zero. Let $M$ be a finitely generated graded $S$-module. The regularity of the Hilbert function $H(n,M)$ of $M$ is the smallest degree $r(M)$ such that for any $n \ge r(M)$ the Hilbert function agrees with the Hilbert polynomial $P_M(n)$. On the other hand, the Castelnuovo-Mumford regularity $reg(M)$ is the largest $i+j$ such that $(H_m^i(M))_j \neq 0$, where $H_m^i(M)$ is the $i$th graded local cohomology of $M$. By using Serre's formula \begin{align} H(n,M) - P(n,M) = \sum_{i= depth(M)}^{dim(M)} (-1)^i dim_k (H_m^i(M))_n, \end{align} we can show that $r(M) \le reg(M)-depth(M)+1$; just plug the upper bound $reg(M)-depth(M)+1$ into the formula and notice that the right hand side is zero.

Question: How can we show that $reg(M)-dim(M)+1 \le r(M)$?

  • $\begingroup$ Just out of curiosity: why do you write 'How can we show that' instead of 'Is it true that'? Did you already have a reason to believe it should be true? $\endgroup$ – R. van Dobben de Bruyn Mar 27 '17 at 5:19
  • $\begingroup$ @R.vanDobbendeBruyn: Yes, i had seen it quoted without proof. I will study your proof soon ;) $\endgroup$ – Manos Mar 27 '17 at 12:52

Notation. Let $k$ be a field, and let $R = k[x_1,\ldots,x_d]$, with maximal homogeneous ideal $\mathfrak m = (x_1,\ldots,x_d)$. Let $M$ be a finitely generated homogeneous $R$-module.

If $r(M)$ is the smallest number such that $H(n,M) = P_M(n)$ for all $n \geq r(M)$, then in fact the result is false, as shown by the following example. However, we will show below that the result is trivially true if we redefine $r(M)$; see the remark below.

Example. Let $d = 1$, and consider the module $M = R \oplus k(1) \oplus \ldots \oplus k(a)$ for some $a \in \mathbb Z_{> 0}$. Here, $k(n)$ is nonzero only in degree $-n$. Thus, $$H(n,M) = \dim M_n = \left\{\begin{array}{cc}1 & n \geq -a,\\ 0 & n<-a. \end{array}\right.$$ Thus, $P_M(n) = 1$ and $r(M) = -a$. We have an exact sequence $$0 \to \Gamma_{\mathfrak m}(M) \to M \to \bigoplus_{n \in \mathbb Z} H^0(\mathbb P^0, \tilde{M}(n)) \to H^1_{\mathfrak m}(M) \to 0;\label{Seq 1}\tag{1}$$ see Eidenbud's Commutative Algebra, Theorem A4.1. Thus, we compute \begin{align*} \Gamma_{\mathfrak m}(M) &= k(1) \oplus \ldots \oplus k(a),\\ H^1_{\mathfrak m}(M) &= \bigoplus_{d > 0} k(d). \end{align*} Thus, $\operatorname{reg}(M) = 0$, coming from $H^1_{\mathfrak m}(M)_{-1} \neq 0$. Since $a$ is arbitrary, we see that there can be no lower bound on $r(M)$ in terms of $\operatorname{reg}(M)$ and $\dim\operatorname{Supp}(M)$.

This example is not limited to $d = 1$: in any odd dimension one can add finite length modules to $M = R$ to extend the range where the Hilbert function equals the Hilbert polynomial. We need odd dimension because by Serre's formula and since $\operatorname{depth}(M) = \dim\operatorname{Supp}(M) = d$ (for $M = R$), the difference $H(n,M) - P_M(n)$ is positive if $d$ is even, so we cannot add something to $H(n,M)$ to make them equal in a larger range (note that finite length stuff does not affect $P_M(n)$).

Remark. If we let $r'(M)$ be the smallest integer such that $H^i_\mathfrak m(M)_n = 0$ for all $n \geq r'(M)$, then Serre's formula implies $r(M) \leq r'(M)$, and the above example shows that the difference can be arbitrarily large. Moreover, with this new definition, the inequalities $$\operatorname{reg}(M) - \dim\operatorname{Supp}(M) + 1 \leq r'(M) \leq \operatorname{reg}(M) - \operatorname{depth}(M) + 1$$ become obvious: if $H^i_\mathfrak m(M)_j \neq 0$, then $\operatorname{depth}(M) \leq i \leq \dim\operatorname{Supp}(M)$, so $$i+j - \dim\operatorname{Supp}(M) \leq j \leq i+j - \operatorname{depth}(M).\label{Eq 2}\tag{2}$$ The result follows from taking the supremum of (\ref{Eq 2}) over all $i,j$ for which $H^i_\mathfrak m(M)_j \neq 0$. (The $+1$ comes from the fact that $r'(M)$ is defined in terms of vanishing rather than nonvanishing.) $\square$

  • $\begingroup$ Hi R. There is a serious issue in the induction step: you are assuming that the irrelevant ideal is not an associated prime, thus there exists a linear form $f$ that is $M$-regular. However, it may so be the case that $\dim M>0$ but $depth(M)=0$, even when $ann(M)$ is proper ideal of the irrelevant ideal. I have tried fixing this, e.g., by replacing $M$ with $M/N$, where $N$ is the maximal submodule of $M$ of finite length, but now it is not clear how to get the inquality on $M$ from that of $M/N$. Alternatively we may try doing induction on $depth(M)$, but the base case... $\endgroup$ – Manos Apr 1 '17 at 22:35
  • $\begingroup$ ...$depth(M)=0$ seems somewhat hard. So the proof does not seem as easy after all. Let me know what you think... $\endgroup$ – Manos Apr 1 '17 at 22:35
  • $\begingroup$ @Manos: thanks for pointing that out. I actually found a counterexample for the version as stated, but I managed to pull the French trick of redefining my objects to make the statement trivial. $\endgroup$ – R. van Dobben de Bruyn Apr 2 '17 at 6:31
  • $\begingroup$ Interesting. I like $R=k[x], \, M = R(-\alpha) \oplus R/(x^{\alpha})$ better as a counterexample (which is definitely inspired by your counterexample). Can you please clean up your answer a little bit so that i can accept it? If you don't mind just include only the necessary information about the counterexample (also $depth(M)$ is not in general equal to $\dim Supp(M)$, perhaps you mean $\dim (M)$; anyway the discussion in the remark seems more appropriately placed in a comment). $\endgroup$ – Manos Apr 2 '17 at 16:37
  • $\begingroup$ @Manos: I've cleaned it up a bit, but left enough information to keep the answer accessible to non-experts (such as myself). Also, I don't know what you mean by $\dim(M)$ if not $\dim\operatorname{Supp}(M)$, but the only place where I asserted equality was for the specific module $M = R$. $\endgroup$ – R. van Dobben de Bruyn Apr 3 '17 at 0:14

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