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Given a specific ring $R$ (eg, $R=k[x_{1}, \cdots, x_{n}]/I)$ is there a (simple) way to determine whether or not $R$ satisfies Serre's condition $S_{n}$? In particular, is there a way to do this in Macaulay2?

($R$ satisfies $S_{n}$ if $depth R_{p} \geq \min(height(p), n)$ for all $p \in Spec R$.)

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  • $\begingroup$ So wait, can we assume that R is finitely presented (over a base ring with known properties) or no? (If we can't assume that, I am very doubtful that you'll be able to compute all that much with software). $\endgroup$ Commented Aug 27, 2010 at 21:40
  • $\begingroup$ sure, we can assume that R is finitely presented. Even if R is a quotient of a polynomial ring, I'm not sure how to go about determining if it's $S_{n}$. $\endgroup$
    – LAM
    Commented Aug 28, 2010 at 6:10

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I think there is a neat answer to this question.

Lemma: Let $S$ be regular local of dimension $d$, $M$ a f.g $S$-module. Then: $$\text{depth}(M)\geq n \Longleftrightarrow \text{Ext}^i(M,S)=0\ \text{for} \ i>d-n$$

Proof: LHS is equivalent to $e= \text{pd}_SM \leq d-n$. By using a minimal free resolution of $M$ to compute Ext, one sees that $\text{Ext}^i(M,S)$ is not $0$ for $i=e$ (Nakayama's lemma) and $0$ for $i>e$.

Now let $A=k[x_1,\cdots,x_d]$, $R=A/I$. Here is the main:

Claim: $R\ \text{is}\ (S_n) \Longleftrightarrow \text{dim}(\text{Ext}_A^i(R,A))\leq d-n-i \ \forall i>d- \text{dim}(R)$

(of course, we only need to check for values of $i$ up to $d$, as $A$ has finite global dimension $d$).

Proof: By Lemma one needs to check that for all $p\in \text{Spec} \ A$: $$\text{Ext}_{A_p}^i(M_p,A_p)=0\ \text{for} \ i>\text{dim}(A_p) -\min\{n,\text{\dim}(R_p)\}=\max\{\text{dim}(A_p)-n,\text{dim}(A_p)-\text{dim}(R_p)\}$$

This condition is equivalent to the fact that for all $i>0$ and each $p$ in the support of $\text{Ext}_A^i(R,A)$ we must have $i\leq \max\{\text{dim}(A_p)-n,d-\text{dim}(R)\}$. Note if $i < d-\text{dim}(R)$, $\text{Ext}^i(R,A)=0$, so the claim follows.

You can compute both Ext and dimension with Macaulay 2.

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  • $\begingroup$ I am unclear about several things: (1) What are $M$ and $S$ in the second part? (2) How do you pass between the Ext's of $A$-modules and $R$-modules? (3) Where does the dimension in the claim come from? $\endgroup$ Commented Aug 28, 2010 at 16:13
  • $\begingroup$ @Victor: In the second part, applying Lemma for $M=M_p$,$S=A_p$ and $n=\min\{n,ht(p)\}$. For (3), see my next to last sentence. $\endgroup$ Commented Aug 28, 2010 at 17:49
  • $\begingroup$ Oh, I see, I may messed up something about the dimensions. $\endgroup$ Commented Aug 28, 2010 at 18:25
  • $\begingroup$ The claim is correct whenever R is equidimensional. $\endgroup$ Commented Oct 7, 2022 at 21:53

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