The Stacks project

Proof. To prove this we may work locally on $X$. Hence we may assume $\mathcal{F}$ is a direct summand of a finite free module, see Lemma 17.18.3. Say $\mathcal{F} \oplus \mathcal{G} = \mathcal{O}_ X^{\oplus n}$. Set $R = \mathcal{O}_ X(X)$. Then we see $\mathcal{F}(X) \oplus \mathcal{G}(X) = R^{\oplus n}$ and correspondingly $\mathcal{F}(U) \oplus \mathcal{G}(U) = \mathcal{O}_ X(U)^{\oplus n}$ for all opens $U \subset X$. We conclude that $\mathcal{F} = \mathcal{F}_ M$ as in Lemma 17.10.5 with $M = \mathcal{F}(X)$ a finite projective $R$-module. In other words, we have $\mathcal{F}(U) = M \otimes _ R \mathcal{O}_ X(U)$. This implies that $\det (M) \otimes _ R \mathcal{O}_ X(U) = \det (\mathcal{F}(U))$ for all open $U \subset X$ with $\det $ as in More on Algebra, Section 15.118. By More on Algebra, Remark 15.118.1 we see that

is the annihilator of $\mathcal{F}(U)$. We conclude that $\det (\mathcal{F})$ as defined in the statement of the lemma is equal to $\mathcal{F}_{\det (M)}$. Some details omitted; one has to be careful as annihilators cannot be defined as the sheafification of taking annihilators on sections over opens. Thus $\det (\mathcal{F})$ is the pullback of an invertible module and we conclude. $\square$