Lemma 51.9.3. Let $X$ be a Noetherian scheme with dualizing complex $\omega _ X^\bullet $. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Let $k \geq 0$ be an integer. Assume $\mathcal{F}$ is $(S_ k)$. Then there is a finite number of points $x \in X$ such that

**Proof.**
We will prove this lemma by induction on $k$. The base case $k = 0$ says that $\mathcal{F}$ has a finite number of embedded associated points, which follows from Divisors, Lemma 31.2.5.

Assume $k > 0$ and the result holds for all smaller $k$. We can cover $X$ by finitely many affine opens, hence we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $\mathcal{F}$ is the coherent $\mathcal{O}_ X$-module associated to a finite $A$-module $M$ which satisfies $(S_ k)$. We will use Algebra, Lemmas 10.63.10 and 10.72.7 without further mention.

Let $f \in A$ be a nonzerodivisor on $M$. Then $M/fM$ has $(S_{k - 1})$. By induction we see that there are finitely many primes $\mathfrak p \in V(f)$ with $\text{depth}((M/fM)_\mathfrak p) = k - 1$ and $\dim (\text{Supp}((M/fM)_\mathfrak p)) > k - 1$. These are exactly the primes $\mathfrak p \in V(f)$ with $\text{depth}(M_\mathfrak p) = k$ and $\dim (\text{Supp}(M_\mathfrak p)) > k$. Thus we may replace $A$ by $A_ f$ and $M$ by $M_ f$ in trying to prove the finiteness statement.

Since $M$ satisfies $(S_ k)$ and $k > 0$ we see that $M$ has no embedded associated primes (Algebra, Lemma 10.157.2). Thus $\text{Ass}(M)$ is the set of generic points of the support of $M$. Thus Dualizing Complexes, Lemma 47.20.4 shows the set $U = \{ \mathfrak q \mid M_\mathfrak q\text{ is Cohen-Macaulay}\} $ is an open containing $\text{Ass}(M)$. By prime avoidance (Algebra, Lemma 10.15.2) we can pick $f \in A$ with $f \not\in \mathfrak p$ for $\mathfrak p \in \text{Ass}(M)$ such that $D(f) \subset U$. Then $f$ is a nonzerodivisor on $M$ (Algebra, Lemma 10.63.9). After replacing $A$ by $A_ f$ and $M$ by $M_ f$ (see above) we find that $M$ is Cohen-Macaulay. Thus for all $\mathfrak q \subset A$ we have $\dim (M_\mathfrak q) = \text{depth}(M_\mathfrak q)$ and hence the set described in the lemma is empty and a fortiori finite. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)