Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,O_X)$ be a ringed space, $E$ - finite locally free $O_X$-module. Let $E^*=Hom_{O_X}(E, O_X)$. How to show, that $E^{**} = E$? It's clear, that locally $E|_U = O_X^n|_U$, and then $E^*|_U = O_X^n|_U$, $E^{**}|_U = O_X^n|_U$. But the second isomorphism is canonical, then we can glue isomorphisms via intersections. It's obvious with canonicity, but how to show it more formally.

share|cite|improve this question
I guess you want to assume that $E$ is finite locally free as an $\mathscr{O}_X$-module? – Keenan Kidwell Nov 3 '12 at 17:14
Yes. I think, it's wrong in general case. – user46336 Nov 3 '12 at 17:18
up vote 1 down vote accepted

If $\mathscr{E}$ is a finite locally free $\mathscr{O}_X$-module, then in particular it is of finite presentation. This implies that for any $\mathscr{O}_X$-module $\mathscr{F}$, the natural map $\mathcal{H}om_{\mathscr{O}_X}(\mathscr{E},\mathscr{F})_x\rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{E}_x,\mathscr{F}_x)$ is an isomorphism.

With this in mind, the fact that the natural map $\mathscr{E}\rightarrow(\mathscr{E}^\vee)^\vee$ is an isomorphism should boil down to the fact that for a ring $R$ and a finite free $R$-module $M$, the canonical map from $M$ to its double $R$-linear dual is an isomorphism. Indeed, granted this fact, the stalks of the natural map from $\mathscr{E}$ to its double dual can be identified with the natural map from $\mathscr{E}_x$ to its double $\mathscr{O}_{X,x}$-linear dual. I'm using here that $\mathscr{E}^\vee$ is also finite locally free, hence finitely presented.

share|cite|improve this answer
I don't know this the fact about the isomorphism between stalk of inner $Hom$ and $Hom$ between stalks. What to read before Hartshorne to understand sheafs of modules better? – user46336 Nov 3 '12 at 17:24
This particular fact and many more can be found in the Stack Project's chapter Sheaves of Modules. Another thorough treatment of sheaves of modules on ringed spaces can be found in Goertz-Wedhorn's Algebraic Geometry Part I. – Keenan Kidwell Nov 3 '12 at 17:28
And what is the best to study algebraic geometry? – user46336 Nov 3 '12 at 17:29
My favorite textbook is Goertz-Wedhorn because it has EGA and Stacks Project generality but it is a bit more readable. But nothing is more comprehensive than EGA, although the Stacks Project is pretty great. I also love Liu's book, which goes much deeper into arithmetic stuff than GW, even if it doesn't have some foundational stuff that GW does (like relative spectrum and proj, finite locally free morphisms, etc.). The downside of GW is that it doesn't have cohomology. But they are writing a second volume entirely devoted to cohomology, so, you know, that's something to look forward to. – Keenan Kidwell Nov 3 '12 at 17:32
Where is GW available in the Internet? – user46336 Nov 3 '12 at 17:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.