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Let $(X,\mathcal O)$ be a ringed space. I am trying to prove that any $\mathcal O$-module that has an inverse with respect to the tensor product is locally free of rank $1$. I've found in Hartshorne a proof of the converse (i.e. any locally free sheaf of rank $1$ has an inverse), but not this.

Any hints?

Alexandru-Andrei Bosinta
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If you take a general ringed space, this can be wrong. Take $R$ a ring which has a module $M$, that is not free, but which has an inverse $N$ with respect to the tensor product.

Now take $X=\{x\}$ just a single point. Thus sheaves in sets/rings/modules are just equivalent to sets/rings/modules... Now take the structure sheaf given by $R$ and sheaf given by $M$. Then obviously $M$ is invertible (with inverse $N$). But $M$ is not free and since your space $X$ is just a point, it is also not locally free.

Notone
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    Sorry if this is a silly question, but do you have an example of such an $R$ and an $M$? – Hank Scorpio May 02 '22 at 21:12
  • Sure, see https://math.stackexchange.com/a/27213 i.e. take a Dedekind domain which is not a PID and an ideal generated by two elements is the module you are looking for – Notone May 02 '22 at 22:28
  • Oh silly me, of course that's the appropriate example. Should have thought of that! – Hank Scorpio May 02 '22 at 22:36
  • The term "invertible sheaf" is quite misleading then. Probably not all authors do that, but Hartshorne calls a an $\mathcal O$-module which is locally free of rank $1$ an invertible sheaf. Thanks! – Alexandru-Andrei Bosinta May 03 '22 at 08:33
  • As soon as you work with schemes, everything is fine and that is the set-up Hartshorne mostly takes place in, i.e. the interesting locally ringed spaces you care about are schemes (not only, but that is beyond the point right now) – Notone May 03 '22 at 08:36
  • I don't know that much about schemes. But is it really the case? Aren't all varieties schemes? Because a singleton set is always a variety, and over such a variety the sheaves with an inverse aren't necessarily locally free of rank $1$. – Alexandru-Andrei Bosinta May 03 '22 at 10:39
  • You have to be really careful here. A variety is always a topological space AND a structure sheaf that can not be chosen arbitrarily. My example is not a variety, for this $R$ would need to have only one prime ideal (and this would imply that you couldn't find such an $M$) – Notone May 03 '22 at 10:50
  • Ah, that makes sense! Thanks! – Alexandru-Andrei Bosinta May 03 '22 at 10:52