The Stacks project

Proof. By the short exact sequence $ 0 \to \mathcal{O}_ X \xrightarrow {a} \mathcal{O}_ X \to \mathcal{O}_ Z \to 0 $ we see that

Here $N[a] = \{ n \in N \mid an = 0\} $ for an $A$-module $N$. Thus if $a$ divides $b$, then $M_{X, a} \subset M_{X, b}$. Suppose that for some $c \in A$ the modules $M_{X, c}$ have bounded length. Then for every $X$ we have an exact sequence

where the second arrow is given by multiplication by $c$. Hence we see that $M_{X, c^2}$ has bounded length as well. Thus it suffices to find a $c \in A$ for which the lemma is true such that $a$ divides $c^ n$ for some $n > 0$. By More on Algebra, Lemma 15.125.6 we may assume $A/(a)$ is a reduced ring.

Assume that $A/(a)$ is reduced. Let $A/(a) \subset B$ be the normalization of $A/(a)$ in its quotient ring. Because $A$ is Nagata, we see that $\mathop{\mathrm{Coker}}(A \to B)$ is finite. We claim the length of this finite module is a bound. To see this, consider $f : X \to \mathop{\mathrm{Spec}}(A)$ as in the lemma and let $Z' \subset Z$ be the scheme theoretic closure of $Z \cap f^{-1}(U)$. Then $Z' \to \mathop{\mathrm{Spec}}(A/(a))$ is finite for example by Varieties, Lemma 33.17.2. Hence $Z' = \mathop{\mathrm{Spec}}(B')$ with $A/(a) \subset B' \subset B$. On the other hand, we claim the map

is injective. Namely, if $s \in H^0(Z, \mathcal{O}_ Z)$ is in the kernel, then the restriction of $s$ to $f^{-1}(U) \cap Z$ is zero. Hence the image of $s$ in $H^1(X, \mathcal{O}_ X)$ vanishes in $H^1(f^{-1}(U), \mathcal{O}_ X)$. By Lemma 54.7.5 we see that $s$ comes from an element $\tilde s$ of $A$. But by assumption $\tilde s$ maps to zero in $B'$ which implies that $s = 0$. Putting everything together we see that $M_{X, a}$ is a subquotient of $B'/A$, namely not every element of $B'$ extends to a global section of $\mathcal{O}_ Z$, but in any case the length of $M_{X, a}$ is bounded by the length of $B/A$. $\square$