Timeline for Connection: locally free - locally projective
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2010 at 17:00 | vote | accept | TonyS | ||
| May 22, 2010 at 15:27 | comment | added | TonyS | Ah okay. So there is some kind of global Morita equivalence for sheaves, i thought one has to use Morita equivalence over the local rings for every point $p$ of $X$. Thanks a lot to both of you for your help. | |
| May 22, 2010 at 14:47 | comment | added | Angelo | Yes, something like this. You don't need to localize. I would argue as follows. Suppose that $M$ is an $M_n(R)$-module which is projective as an $R$-module. Let $N$ the Morita-equivalent $R$-module; then $M$ is $N^n$, hence $N$ is projective. Since Morita equivalence preserves projectivity, $M$ is projective as an $M_n(R)$-module. | |
| May 22, 2010 at 14:35 | comment | added | TonyS | I am not so good with Morita equivalence, is the argument the following: If we look locally at a point $p \in X$ the $M_n(R)$-module $M_p$ is of the form $V^n$ for some $R$-module $V$. Since it is free over $R$, we have $M=R^(nk)$ for some k. Then the Morita equivalent $R$-module to $M_p$ is $R^k$, which is free, hence projective over $R$. Since Morita equivalence preserves projectivity we have that $M_p$ is projective. So $M$ is locally projective. | |
| May 22, 2010 at 14:09 | comment | added | Angelo | In the case of an Azumaya algebra, the statement is local in the étale topology, so it is enough to prove it for matrix algebras. In this case it follows immediately from Morita equivalence. In the case of hereditary orders, I am not sure, as I am really rusty on this stuff. | |
| May 22, 2010 at 14:01 | comment | added | Torsten Ekedahl | Yes, it is true for hereditary orders but unless I am mistaken if the order is hereditary then $X$ is necessarily a smooth curve. More interesting perhaps is the question for a maximal order (maximal orders over curves are hereditary). I am somewhat doubtful however if a maximal order in general has finite global dimension (which it must if your property is to hold). Artin has some results on the local form for maximal orders over surfaces which might give counterexamples (or support for that problem). | |
| May 22, 2010 at 13:43 | comment | added | TonyS | Yes i am especially/only interested in the case where the $O_X$-algebra $R$ lives in a given central simple $k(X)$-algebra $A$. This includes the Azumaya algebra case. I would really like to have a proof, an idea how to prove or a reference for this statement. According to the following paper, page 2, it could also hold more generally for "hereditary orders", although there $X$ is just a curve. web.maths.unsw.edu.au/~danielch/paper/moduli.pdf | |
| May 22, 2010 at 12:38 | history | answered | Angelo | CC BY-SA 2.5 |