Lemma 28.21.2. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. The following are equivalent
$\mathcal{F}$ is locally projective, and
there exists an affine open covering $X = \bigcup U_ i$ such that the $\mathcal{O}_ X(U_ i)$-module $\mathcal{F}(U_ i)$ is projective for every $i$.
In particular, if $X = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{F} = \widetilde{M}$ then $\mathcal{F}$ is locally projective if and only if $M$ is a projective $A$-module.
Proof. First, note that if $M$ is a projective $A$-module and $A \to B$ is a ring map, then $M \otimes _ A B$ is a projective $B$-module, see Algebra, Lemma 10.94.1. Hence if $U$ is an affine open such that $\mathcal{F}(U)$ is a projective $\mathcal{O}_ X(U)$-module, then the standard open $D(f)$ is an affine open such that $\mathcal{F}(D(f))$ is a projective $\mathcal{O}_ X(D(f))$-module for all $f \in \mathcal{O}_ X(U)$. Assume (2) holds. Let $U \subset X$ be an arbitrary affine open. We can find an open covering $U = \bigcup _{j = 1, \ldots , m} D(f_ j)$ by finitely many standard opens $D(f_ j)$ such that for each $j$ the open $D(f_ j)$ is a standard open of some $U_ i$, see Schemes, Lemma 26.11.5. Hence, if we set $A = \mathcal{O}_ X(U)$ and if $M$ is an $A$-module such that $\mathcal{F}|_ U$ corresponds to $M$, then we see that $M_{f_ j}$ is a projective $A_{f_ j}$-module. It follows that $A \to B = \prod A_{f_ j}$ is a faithfully flat ring map such that $M \otimes _ A B$ is a projective $B$-module. Hence $M$ is projective by Algebra, Theorem 10.95.6. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: