Lemma 31.24.5. Suppose given

1. $X$ a locally Noetherian scheme,

2. $\mathcal{L}$ an invertible $\mathcal{O}_ X$-module,

3. $s$ a regular meromorphic section of $\mathcal{L}$, and

4. $\mathcal{F}$ coherent on $X$ without embedded associated points and $\text{Supp}(\mathcal{F}) = X$.

Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal of denominators of $s$. Let $T \subset X$ be the union of the supports of $\mathcal{O}_ X/\mathcal{I}$ and $\mathcal{L}/s(\mathcal{I})$ which is a nowhere dense closed subset $T \subset X$ according to Lemma 31.23.9. Then there are canonical injective maps

$1 : \mathcal{I}\mathcal{F} \to \mathcal{F}, \quad s : \mathcal{I}\mathcal{F} \to \mathcal{F} \otimes _{\mathcal{O}_ X}\mathcal{L}$

whose cokernels are supported on $T$.

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. $\square$

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