The Stacks project

Proof. Observe that a finite projective $R$-module is the same thing as a summand of a finite free $R$-module. The equivalence is given by the functor $\mathcal{E} \mapsto \Gamma (X, \mathcal{E})$. The inverse functor is given by the construction of Modules, Lemma 17.10.5. $\square$

Proof. Assume (2). Consider the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, \mathcal{O}_ X)$ and the composition map

To prove this is an isomorphism, we may work locally. Thus we may assume $\mathcal{O}_ X = \prod _{a \leq n \leq b} \mathcal{O}_ n$ and $M = \bigoplus _{a \leq n \leq b} \mathcal{H}^ n[-n]$. Then it suffices to show that

is zero if $n \not= m$ and equal to $\mathcal{O}_ n$ if $n = m$. The case $n \not= m$ follows from the fact that $\mathcal{O}_ n$ and $\mathcal{O}_ m$ are flat $\mathcal{O}_ X$-algebras with $\mathcal{O}_ n \otimes _{\mathcal{O}_ X} \mathcal{O}_ m = 0$. Using the local structure of invertible $\mathcal{O}_ X$-modules (Modules, Lemma 17.25.2) and working locally the isomorphism in case $n = m$ follows in a straightforward manner; we omit the details. Because $D(\mathcal{O}_ X)$ is symmetric monoidal, we conclude that $M$ is invertible.

Assume (1). The description in (2) shows that we have a candidate for $\mathcal{O}_ n$, namely, $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(H^ n(M), H^ n(M))$. If this is a locally finite family of sheaves of rings and if $\mathcal{O}_ X = \prod \mathcal{O}_ n$, then we immediately obtain the direct sum decomposition $M = \bigoplus H^ n(M)[-n]$ using the idempotents in $\mathcal{O}_ X$ coming from the product decomposition. This shows that in order to prove (2) we may work locally on $X$.

Choose an object $N$ of $D(\mathcal{O}_ X)$ and an isomorphism $M \otimes _{\mathcal{O}_ X}^\mathbf {L} N \cong \mathcal{O}_ X$. Let $x \in X$. Then $N$ is a left dual for $M$ in the monoidal category $D(\mathcal{O}_ X)$ and we conclude that $M$ is perfect by Lemma 20.50.8. By symmetry we see that $N$ is perfect. After replacing $X$ by an open neighbourhood of $x$, we may assume $M$ and $N$ are represented by a strictly perfect complexes $\mathcal{E}^\bullet$ and $\mathcal{F}^\bullet$. Then $M \otimes _{\mathcal{O}_ X}^\mathbf {L} N$ is represented by $\text{Tot}(\mathcal{E}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{F}^\bullet )$. After another shinking of $X$ we may assume the mutually inverse isomorphisms $\mathcal{O}_ X \to M \otimes _{\mathcal{O}_ X}^\mathbf {L} N$ and $M \otimes _{\mathcal{O}_ X}^\mathbf {L} N \to \mathcal{O}_ X$ are given by maps of complexes

See Lemma 20.46.8. Then $\beta \circ \alpha = 1$ as maps of complexes and $\alpha \circ \beta = 1$ as a morphism in $D(\mathcal{O}_ X)$. After shrinking $X$ we may assume the composition $\alpha \circ \beta$ is homotopic to $1$ by some homotopy $\theta$ with components

by the same lemma as before. Set $R = \Gamma (X, \mathcal{O}_ X)$. By Lemma 20.52.1 we find that we obtain

1. $M^\bullet = \Gamma (X, \mathcal{E}^\bullet )$ is a bounded complex of finite projective $R$-modules,

2. $N^\bullet = \Gamma (X, \mathcal{F}^\bullet )$ is a bounded complex of finite projective $R$-modules,

3. $\alpha$ and $\beta$ correspond to maps of complexes $a : R \to \text{Tot}(M^\bullet \otimes _ R N^\bullet )$ and $b : \text{Tot}(M^\bullet \otimes _ R N^\bullet ) \to R$,

4. $\theta ^ n$ corresponds to a map $h^ n : \text{Tot}^ n(M^\bullet \otimes _ R N^\bullet ) \to \text{Tot}^{n - 1}(M^\bullet \otimes _ R N^\bullet )$, and

5. $b \circ a = 1$ and $b \circ a - 1 = dh + hd$,

It follows that $M^\bullet$ and $N^\bullet$ define mutually inverse objects of $D(R)$. By More on Algebra, Lemma 15.126.4 we find a product decomposition $R = \prod _{a \leq n \leq b} R_ n$ and invertible $R_ n$-modules $H^ n$ such that $M^\bullet \cong \bigoplus _{a \leq n \leq b} H^ n[-n]$. This isomorphism in $D(R)$ can be lifted to an morphism

of complexes because each $H^ n$ is projective as an $R$-module. Correspondingly, using Lemma 20.52.1 again, we obtain an morphism

which is an isomorphism in $D(\mathcal{O}_ X)$. Setting $\mathcal{O}_ n = R_ n \otimes _ R \mathcal{O}_ X$ we conclude (2) is true.

If all stalks of $\mathcal{O}_ X$ are local, then it is straightforward to prove the equivalence of (2) and (3). We omit the details. $\square$