The Stacks project

Proof. We can find a covering of $X$ by affine opens $U$ such that $f|_ U = s^{-1}a$ with $a \in \mathcal{O}_ X(U)$ and $s \in \mathcal{S}(U)$. Since $X$ is quasi-compact we can cover it by finitely many affine opens of this form. Thus it suffices to prove the lemma when $X = \mathop{\mathrm{Spec}}(A)$ and $f = s^{-1}a$. Note that $s \in A$ is a nonzerodivisor hence it suffices to prove the result when $f = a$. The condition $f|_{X_ h} = 0$ implies that $a$ maps to zero in $A_ h = \mathcal{O}_ X(X_ h)$ as $\mathcal{O}_ X \subset \mathcal{K}_ X$. Thus $h^ na = 0$ for some $n > 0$ as desired. $\square$

Proof. To prove this lemma we may assume $X$ is the spectrum of a Noetherian ring $A$. Say $x \in X$ corresponds to $\mathfrak p \subset A$.

Proof of (1). It is clear that $\mathcal{S}_ x$ is contained in the set of nonzerodivisors of $\mathcal{O}_{X, x} = A_\mathfrak p$. For the converse, let $f, g \in A$, $g \not\in \mathfrak p$ and assume $f/g$ is a nonzerodivisor in $A_{\mathfrak p}$. Let $I = \{ a \in A \mid af = 0\} $. Then we see that $I_{\mathfrak p} = 0$ by exactness of localization. Since $A$ is Noetherian we see that $I$ is finitely generated and hence that $g'I = 0$ for some $g' \in A$, $g' \not\in \mathfrak p$. Hence $f$ is a nonzerodivisor in $A_{g'}$, i.e., in a Zariski open neighbourhood of $\mathfrak p$. Thus $f/g$ is an element of $\mathcal{S}_ x$.

Proof of (2). Let $f \in \Gamma (X, \mathcal{K}_ X)$ be a meromorphic function. Set $I = \{ a \in A \mid af \in A\} $. Fix a prime $\mathfrak p \subset A$ corresponding to the point $x \in X$. By (1) we can write the image of $f$ in the stalk at $\mathfrak p$ as $a/b$, $a, b \in A_{\mathfrak p}$ with $b \in A_{\mathfrak p}$ not a zerodivisor. Write $b = c/d$ with $c, d \in A$, $d \not\in \mathfrak p$. Then $ad - cf$ is a section of $\mathcal{K}_ X$ which vanishes in an open neighbourhood of $x$. Say it vanishes on $D(e)$ with $e \in A$, $e \not\in \mathfrak p$. Then $e^ n(ad - cf) = 0$ for some $n \gg 0$ by Lemma 31.24.1. Thus $e^ nc \in I$ and $e^ nc$ maps to a nonzerodivisor in $A_{\mathfrak p}$. Let $\text{Ass}(A) = \{ \mathfrak q_1, \ldots , \mathfrak q_ t\} $ be the associated primes of $A$. By looking at $IA_{\mathfrak q_ i}$ and using Algebra, Lemma 10.63.15 the above says that $I \not\subset \mathfrak q_ i$ for each $i$. By Algebra, Lemma 10.15.2 there exists an element $z \in I$, $z \not\in \bigcup \mathfrak q_ i$. By Algebra, Lemma 10.63.9 we see that $z$ is not a zerodivisor on $A$. Hence $f = (zf)/z$ is an element of the total ring of fractions of $A$. This proves (2). $\square$

Proof. This lemma is a special case of Lemma 31.23.6 because in the locally Noetherian case weakly associated points are the same thing as associated points by Lemma 31.5.8. $\square$

Proof. For each generic point $\eta $ of $X$ pick a generator $s_\eta $ of the free rank $1$ module $\mathcal{L}_\eta $ over the artinian local ring $\mathcal{O}_{X, \eta }$. It follows immediately from the description of $\mathcal{K}_ X$ and $\mathcal{K}_ X(\mathcal{L})$ in Lemma 31.24.3 that $s = \prod s_\eta $ is a regular meromorphic section of $\mathcal{L}$. $\square$

Proof. Reduce to the affine case with $\mathcal{L} \cong \mathcal{O}_ X$, and $s = a/b$ with $a, b \in A$ both nonzerodivisors. Proof of reduction step omitted. Write $\mathcal{F} = \widetilde{M}$. Let $I = \{ x \in A \mid x(a/b) \in A\} $ so that $\mathcal{I} = \widetilde{I}$ (see proof of Lemma 31.23.9). Note that $T = V(I) \cup V((a/b)I)$. For any $A$-module $M$ consider the map $1 : IM \to M$; this is the map that gives rise to the map $1$ of the lemma. Consider on the other hand the map $\sigma : IM \to M_ b, x \mapsto ax/b$. Since $b$ is not a zerodivisor in $A$, and since $M$ has support $\mathop{\mathrm{Spec}}(A)$ and no embedded primes we see that $b$ is a nonzerodivisor on $M$ also. Hence $M \subset M_ b$. By definition of $I$ we have $\sigma (IM) \subset M$ as submodules of $M_ b$. Hence we get an $A$-module map $s : IM \to M$ (namely the unique map such that $s(z)/1 = \sigma (z)$ in $M_ b$ for all $z \in IM$). It is injective because $a$ is a nonzerodivisor also (on both $A$ and $M$). It is clear that $M/IM$ is annihilated by $I$ and that $M/s(IM)$ is annihilated by $(a/b)I$. Thus the lemma follows. $\square$