The Stacks project

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. $\square$