Question on a (presumably) basic $O_X$-module isomorphism

Question

I'm working my way through The Rising Sea: Foundations of Algebraic Geometry and want to double-check my work and thinking (and more than likely my notation) on the following question it poses:

Exercise: If $\mathcal{F}$ is an $\mathcal{O}_X$-module, then show that $Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F}) \cong \mathcal{F}$.

The text provides the hint that 1 "generates" (in a reasonable sense) $\mathcal{O}_X$.

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My idea in broad strokes is to use the following:

1. Use the fact that we know there is a canonical isomorphism between $\mathcal{F}_x$ and $Hom_{\mathcal{O}_{X,x}}\left(\mathcal{O}_{X,x},\mathcal{F}_x\right)$ by virtue of the fact that $\mathcal{F}_x$ is a $\mathcal{O}_{X,x}$-module.

2. A morphism $\Psi:\mathcal{G}\to \mathcal{H}$ of $\mathcal{O}_X$ modules is an isomorphism if and only if it induces an isomorphism of the stalks $\Psi_x:\mathcal{G}_x \to \mathcal{H}_x$ for all $x\in X$.

Since the goal is an isomorphism between $Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})$ and $\mathcal{F}$, such an isomorphism would require that

$$ Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})_x \cong Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x) $$ as $Mod_{\mathcal{O}_{X,x}}-$modules. The right-hand side clearly has the structure of an $\mathcal{O}_{X,x}$ module, and the right-hand side can be seen to have such structure as each $$ Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})(U) $$ has the structure of a $\mathcal{O}_{X}(U)-$module, so taking direct limits gives the required structure.

To see that $$ Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})_x \cong Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x) $$ we consider an element $\varphi_x \in Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})_x$ and $U$ an open set containing $x$ such that $$ \varphi \in Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})(U) = Mor_{\mathcal{O}_U}\left(\mathcal{O}_U, \mathcal{F}|_U\right). $$ Then $\varphi$, being a morphism of $\mathcal{O}_U-$ modules, induces a morphism on the stalks, i.e an element of $Hom_{Mod_{\mathcal{O}_{U,x}}} (\mathcal{O}_{U,x}, \mathcal{F}_x)$.

Since for all $V\subset U$ we have $\mathcal{O}_U(V) = \mathcal{O}_X(V)$, and $\mathcal{F}|_U(V)=\mathcal{F}(V)$, it follows that the induces morphism on stalks is also an element of $Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x)$. Since this takes maps to themselves, it commutes with all required structures to be a morphism of $\mathcal{O}_{X}(U)-$modules.

To see this mapping has to be injective, assume that $\varphi_x \in Hom_{Mod_{\mathcal{O}_{X}}} (\mathcal{O}_{X}, \mathcal{F})_x$ induces the trivial map in $Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x)$ then it must be the case that $\varphi_x(1) = 0$, where $1$ is multiplicative identity of the ring $\mathcal{O}_{X,x}$ and $0$ is the identity of the abelian group $\mathcal{F}_x$.

As such, there must be some $V\subset X$ such that $1\in \ker\varphi(V)$. But this means that $\varphi(V)$ is the trivial map, thus the map $\varphi_x$ is itself the trivial map. Thus the only map that can induce the identity of $Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x)$ is the identity of $Hom_{Mod_{\mathcal{O}_{X}}} (\mathcal{O}_{X}, \mathcal{F})_x$, so our proposed mapping is injective.

To see the desired mapping is surjective, let $\psi \in Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x)$. Then we can uniquely identify $\psi$ with some element $f_x \in \mathcal{F}_x$ (picked as $\psi(1) = f_x \in \mathcal{F}_x$). Thus there is some open set $V$ containing $x$ and some morphism $\tilde{\psi}\in Mor(\mathcal{O}_{V},\mathcal{F}|_V)$ such that $\tilde{\psi}(V)(1) = f \in \mathcal{F}(V)$. But since $$ res_{W,V}\circ \tilde{\psi}(V)(1) = \tilde{\psi}(W)(1) = \tilde{\psi} \circ res_{W,V} (1) $$ we see that $\tilde{\psi}_x \in Hom_{Mod_{\mathcal{O}_{V,x}}} (\mathcal{O}_{V,x}, \mathcal{F}_x) = Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x)$ induces the $\psi$, hence proving surjectivity. Thus we indeed have $$ Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})_x \cong Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x) $$

This indicates that the isomorphism between $Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})$ and $\mathcal{F}$ ought to be of the form $$ \left(\Psi(U)\right)(\varphi) = \left(\varphi(U)\right)(1) $$ as taking direct limits on either side results in $$ \lim_{x\in U} \left(\Psi(U)\right)(\varphi) = \lim_{x\in U} \left(\varphi(U)\right)(1) $$ where the resulting lefthand side describes the proposed isomorphism $$ Hom_{Mod_{\mathcal{O}_X}} (\mathcal{O}_X, \mathcal{F})_x \cong Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x) $$ and the righthand side describes the canonical isomorphism $$ Hom_{Mod_{\mathcal{O}_{X,x}}} (\mathcal{O}_{X,x}, \mathcal{F}_x) \cong \mathcal{F}_x. $$ Thus the morphism $\Psi$ induces an isomorphism of stalks and thus must be an isomorphism of sheaves.

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My main question is: "Is there a more succinct way of proving the result?" as what I came up with feels very roundabout and vague at times.

Answer 1

Let me record my comments as an answer.

We can define two maps $\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})\to\mathcal{F}$ and $\mathcal{F}\to\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})$ and show they're mutually inverse. The first map takes a morphism $\varphi:\mathcal{O}_X|_U\to\mathcal{F}|_U$ (i.e. a section of the hom-sheaf $\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})$ over $U$) to the section $\varphi(U)(1)\in\mathcal{F}(U)$, while the second map takes a section $f\in\mathcal{F}(U)$ to the map $\varphi:\mathcal{O}_X|_U\to\mathcal{F}|_U$ defined by $\varphi(U')(s')=s'\cdot res_{U,U'}(f)$.

It is not so hard to show that these are really maps of sheaves which are $\mathcal{O}_X$-linear and mutually inverse - the key idea here is that for a map out of $\mathcal{O}_X$, everything is determined by where $1$ goes. Some advantages of this proof are that it works everywhere with not much in the way of technology. I would recommend working through this once yourself to get familiar with it, and then you are probably set to quote the result forever (or until you need to teach it to some new students :).

Answer 2

Here's an extremely succinct approach using topos theory. You can read more about this idea in Ingo Blechschmidt's excellent PhD thesis.

A topos is an alternate universe in which we can interpret mathematics, and proofs work exactly the same as proofs in our world (with the single caveat that our proofs must be constructive). This is useful because our interpretation lets us pretend that complicated objects (like sheaves) are actually simple objects (like sets). Understanding all the moving parts of how and why this machinery works is quite complicated, but thankfully using the machinery is much easier! In particular, if you're willing to believe that this "alternate universe" really does think that sheaves are sets, then @KReiser's comment becomes a very easy proof of your fact!

Inside the topos $\mathsf{Sh}(X)$ of sheaves on $X$, a sheaf of looks like an "ordinary" set, and a sheaf of algebraic objects (rings/modules/etc.) looks like a single "ordinary" ring/module/etc. Similarly all of our sheafy operations ($\mathcal{Hom}$, $\mathcal{Ext}$, $\otimes$, etc.) look like their "ordinary" counterparts¹. So inside $\mathsf{Sh}(X)$ the statement

If $\mathcal{F}$ is a a sheaf of $\mathcal{O}_X$ modules, then $\mathcal{Hom}(\mathcal{O}_X, \mathcal{F}) \cong \mathcal{F}$

becomes

If $\mathcal{F}$ is a $\mathcal{O}_X$-module, then $\text{Hom}(\mathcal{O}_X, \mathcal{F}) \cong \mathcal{F}$

where now everything in sight is just an "ordinary" ring, homset, etc.

Moreover, since we can prove things in $\mathsf{Sh}(X)$ just like we do in "ordinary math" (as long as our proofs are constructive), we can prove this theorem in the same way we prove it for "ordinary" rings!

$\ulcorner$ $\mathcal{O}_X$ is the free $\mathcal{O}_X$-module on one generator, so $\text{Hom}(\mathcal{O}_X, \mathcal{F})$ is in bijection with elements of $\mathcal{F}$ (choose where $1$ goes). But if $\varphi_x(1) = x$ and $\varphi_y(1) = y$ then by the definition of the module structure on $\text{Hom}(\mathcal{O}_X,\mathcal{F})$ we see $(\varphi_x + \varphi_y)(1) = x+y$ and $r \varphi_x(1) = \varphi_x(r) = rx$. But this means the map $\varphi \mapsto \varphi(1) : \text{Hom}(\mathcal{O}_X,\mathcal{F}) \to \mathcal{F}$ is more than just a bijection: It's a $\mathcal{O}_X$-module homomorhpism! Thus it's an isomorphism, as desired. $\lrcorner$

In fact, if you "externalize" this proof, it gives you the same proof as @KReiser's comment! It's worth taking a second to see why that's believable (even though I haven't told you the specifics of how internalization and externalization work). The nice thing about topos theory is that it makes these kinds of arguments obvious. Eventually, you become sensitive to what proofs are and aren't constructive, and proving a theorem like this is as quick as saying "it's true for rings and modules, and the proof is constructive. So it's true for sheaves."

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¹ In fact, this is a useful way of figuring out what a "sheafy" version of a construction should be! Never again do you have to wonder how someone could have come up with the definition of sheaf hom, etc. There's a completely algorithmic way to take a (constructive) classical definition, interpret it inside $\text{Sh}(X)$, and then see what that interpretation looks like on the level of sheaves!

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I hope this helps ^_^