ac.commutative algebra - Determining if a ring satisfies Serre's condition S_{n}

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 $pin text{Spec} A$:
$$text{Ext}_{A_p}^i(M_p,A_p)=0 text{for} i>text{dim}(A_p) -min{n,text{dim}(R_p)}=max{text{dim}(A_p)-n,text{dim}(A_p)-text{dim}(R_p)}$$

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 $ileq max{text{dim}(A_p)-n,d-text{dim}(R)}$. 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.