# stacks/stacks-project

Say a bit more about det for flat fp modules

 @@ -3978,7 +3978,7 @@ \section{Rank and determinant} Thus the rank defines a homomorphism $$K_0(\textit{Vec}(X)) \longrightarrow \text{Map}_{cont}(X, \mathbf{Z}),\quad [\mathcal{E}] \longmapsto r_\mathcal{E} [\mathcal{E}] \longmapsto \text{rank}_\mathcal{E}$$ \medskip\noindent @@ -4043,26 +4043,73 @@ \section{Rank and determinant} \end{proof} \noindent {\bf Determinants, reprise.} Let $(X, \mathcal{O}_X)$ be a ringed space. Instead of looking at finite locally free $\mathcal{O}_X$-modules we could look at those $\mathcal{O}_X$-modules $\mathcal{F}$ which are locally on $X$ Let $(X, \mathcal{O}_X)$ be a ringed space. Instead of looking at finite locally free $\mathcal{O}_X$-modules we can look at those $\mathcal{O}_X$-modules $\mathcal{F}$ which are locally on $X$ a direct summand of a finite free $\mathcal{O}_X$-module. This is the same thing as asking $\mathcal{F}$ to be a flat $\mathcal{O}_X$-module of finite presentation, see Lemma \ref{lemma-flat-locally-finite-presentation}. For such a module in general the rank function is undefined; for If all the stalks $\mathcal{O}_{X, x}$ are local, then such a module $\mathcal{F}$ is finite locally free, see Lemma \ref{lemma-direct-summand-of-locally-free-is-locally-free}. In general however this will not be the case; for example $X$ could be a point and $\Gamma(X, \mathcal{O}_X)$ could be the product $A \times B$ of two nonzero rings and $\mathcal{F}$ could correspond to $A \times 0$. On the other hand, for $\mathcal{F}$ flat and of finite presentation we can still define $\det(\mathcal{F})$ and this will be an invertible $\mathcal{O}_X$-module in the sense of could correspond to $A \times 0$. Thus for such a module the rank function is undefined. However, it turns out we can still define $\det(\mathcal{F})$ and this will be an invertible $\mathcal{O}_X$-module in the sense of Definition \ref{definition-invertible} (not necessarily locally free of rank $1$). This is done using the construction of More on Algebra, Section \ref{more-algebra-section-determinants} on the values of $\mathcal{F}$ on sufficiently small opens of $X$. If we ever need this we will precisely state and prove the relevant lemmas here. free of rank $1$). Our construction will agree with the one above in the case that $\mathcal{F}$ is finite locally free. We urge the reader to skip the rest of this section. \begin{lemma} \label{lemma-determinant-as-socle} 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. \end{lemma} \begin{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 \ref{lemma-flat-locally-finite-presentation}. 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 \ref{lemma-construct-quasi-coherent-sheaves} 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 \ref{more-algebra-section-determinants}. By More on Algebra, Remark \ref{more-algebra-remark-determinant-as-socle} 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. \end{proof}