The Stacks project

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. $\square$