Lemma 17.25.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a flat and finitely presented $\mathcal{O}_ X$-module. Denote

\[ \det (\mathcal{F}) \subset \wedge ^*_{\mathcal{O}_ X}(\mathcal{F}) \]

the annihilator of $\mathcal{F} \subset \wedge ^*_{\mathcal{O}_ X}(\mathcal{F})$. Then $\det (\mathcal{F})$ is an invertible $\mathcal{O}_ X$-module.

**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 15.116. By More on Algebra, Remark 15.116.1 we see that

\[ \det (M) \otimes _ R \mathcal{O}_ X(U) = \det (\mathcal{F}(U)) \subset \wedge ^*_{\mathcal{O}_ X(U)}(\mathcal{F}(U)) \]

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.
$\square$

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