Let $k$ be a field, $S=k[x_1,\dots,x_r]$ and $M$ a finitely-generated, graded $S$-module.

Definition: We say that $M$ is weakly $m$-regular if $Ext^j(M,S)_n=0$ for all $j$ and $n=-m-j-1$. We say that $M$ is $m$-regular if $Ext^j(M,S)_n=0$ for all $j$ and $n \le-m-j-1$.

Theorem 20.17 in Eisenbud (CA with a view...): With the notation above, let $N$ be the maximal submodule of $M$ having finite length. If $M$ is weakly $m$-regular, then $M/N$ is $m$-regular.

The proof of the theorem proceeds by induction on $\dim M$. If $\dim M>0$, Eisenbud considers a short exact sequence $0 \rightarrow N \rightarrow M \rightarrow M/N \rightarrow 0$ and the corresponding long exact sequence of $Ext(-,S)$.

Question 1: Why is it true that $Ext^j(N,S)=0, \forall j<r$? As i understand, $r$ is the number of indeterminates, what does this have to do with $Ext^j(N,S)$? Eisenbud invokes proposition 18.4, but this proposition applied to $Ext^j(N,S)$ simply says that $Ext^j(N,S)=0$ for any $j$ less than the grade of $N$, which is a definition.

Question 2: Even if $Ext^j(M/N,S) \cong Ext^j(M,S), \forall j<r$, why does the weak $m$-regularity of $M$ imply weak $m$-regularity of $M/N$? How about $j \ge r$?

Question 3: Why is $(x_1,\dots,x_r)$ not inside any associated prime of $M/N$?


1) $\text{grade}(N) := \text{grade}(\text{ann}(N))$, but $N$ has finite length, so $\text{ann}(N)$ is primary to the maximal ideal $(x_1,...,x_r) =: \mathfrak{m}$, so $\text{grade}(\text{ann}(N)) = \text{grade}(\mathfrak{m})$.

2) What is the global dimension of $S$? What does this mean about $\text{Ext}^j_S(\_, \_)$ for $j > r$? For $j = r$, note that $\text{Ext}^r(M/N,S)$ injects into $\text{Ext}^r(M,S)$.

3) The maximal ideal $\mathfrak{m}$ is associated to $M/N$ iff $\text{depth}(M/N) = 0$. But by definition of $N$, $\text{depth}(M/N) > 0$. Alternatively, if $R/\mathfrak{m} \cong k \hookrightarrow M/N$, then $M/N$ would contain a finite-length submodule.

  • $\begingroup$ What a brilliant answer! Follow up on 3): how can you see that by definition of $N$ $depth(M/N) \neq 0$? $\endgroup$ – Manos Jan 22 '14 at 18:49
  • $\begingroup$ And a final question please: why is it true that $Ext^r(M/N,S)=0$? Eisenbud claims that in the conclusion of the proof, top of page 512. $\endgroup$ – Manos Jan 22 '14 at 19:49
  • $\begingroup$ Perhaps the phrase "by definition of $N$" was a bit confusing: what I meant was that by definition of $N$, $H^0_m(M/N) = 0$ (zeroth local cohomology), so $\text{depth}(M/N) > 0$. The alternative explanation is better though. As for your final question: one way to see that $\text{Ext}^r(M/N,S) = 0$ is localize and apply local duality (see e.g. Bruns-Herzog, 3.5.11) $\endgroup$ – zcn Jan 23 '14 at 2:16

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