Is it enough to consider only finitely generated projective modules having constant rank?
问题内容
It is known that every finitely generated projective module $M$ over a commutative ring $A$ has locally constant rank, i.e., for each $\mathfrak{p} \in \mathrm{Spec}(A)$, there are non-negative integer $r$ and an open neighborhood $U \subseteq \mathrm{Spec}(A)$ such that, for every $\mathfrak{q} \in U$, $M_{\mathfrak{q}}$ is free of rank $r$; see 00NX in Stacks Project. In fact, it is easy to see that each $U$ can be extended to be the connected component of $\mathrm{Spec}(A)$ having $\mathfrak{p}$.
Here is a stronger concept: a projective module $M$ is said to have constant rank if and only if there is a non-negative integer $r$ such that $M_{\mathfrak{p}}$ is free of rank $r$ for every prime $\mathfrak{p}$. With the above result, one can immediately solve one direction of Eisenbud, Exercise 20.12; every finitely generated projective module $M$ has constant rank if and only if $A$ has only trivial idempotents, which is equivalent to saying that $\mathrm{Spec}(A)$ is connected (by Atiyah, Exercise 1.22). The other direction of the exercise is also easy to show; if $e$ is a non-trivial idempotent of $A$, since $A = eA \oplus (1 - e)A$, $eA$ is a finitely generated projective $A$-module, and the localization of $eA$ at a prime ideal having $1 - e$ is nonzero, while the localization at a prime ideal having $e$ is $0$.
The facts I have so far tempt me to conclude that we need not be concerned with finitely generated projective modules that do not have constant rank. It seems reasonable based on our general experience; it might be possible to 'divide-and-conquer' things on each connected component and then 'merge' them. But is it enough to say this? What I'm concerned about is that I don't have enough devices to divide a given finitely generated projective module that does not have constant rank into those having constant rank (on each connected component) and merge them. If the module can always be written as a direct sum of two submodules, each of which is a (finitely generated) projective module over each connected component, then I don't need to worry about it. But I don't have it, unfortunately, and if it is not true in general, then I have no idea how to apply the 'divide-and-conquer' strategy. Even if there is no possible 'divide-and-conquer' strategy, then it will be hard to apply results on projective modules having constant rank to some finitely generated projective modules that do not have constant rank.
This is what I'm concerned about. Is there any study about this?
回答 (1)
If a scheme $X$ is Noetherian, then it has finitely many connected component, and they're both open and closed. Let $j_i:X_i \to X$ the open immersion (it is also closed). You have a natural map $\mathcal F \to \prod_i j_{i *}j_i^{*} \mathcal F$. As $j_i$ is both a closed and open immersion you have $$j_{i *}j_i^{*} \mathcal F_y = \mathcal F_y \text{ or } 0$$ depending on whether $y\in U_i$, and the aforementioned arrow is therefore an isomorphism. Thus your sheaf is a product/direct sum of sheaves that are supported on a connected component. Notice that the $j_{i *}j_i^{*} \mathcal F$ are also projective as they're direct factors of a projective quasi-coherent sheaf.
This should answer your question in the case where $A$ is noetherian.