Show that for an affine algebraic set $V$,$\operatorname{Shv}(V;\mathcal{C})\simeq\operatorname{Shv}(\operatorname{Spec}(\mathcal{O}_V);\mathcal{C})$ [duplicate]

Solution 1:

When $k$ is algebraically closed and $Z,Z'$ are of finite type over $k$:

• the data of $(Z(k),\mathcal{O}_Z|_{Z(k)})$ is enough to recover $Z$, and
• given a morphism $\varphi:(Z(k),\mathcal{O}_Z|_{Z(k)})\to (Z'(k),\mathcal{O}_{Z'}|_{Z'(k)})$, there is a unique morphism $\psi:Z\to Z'$ which has $\varphi$ as its restriction to the $k$-points. (Additionally, no two distinct morphisms $Z\to Z'$ restrict to the same morphism $Z(k)\to Z'(k)$.)

The recipe for recovering $Z$ from $Z(k)$ is straightforward: we need to freely adjoin generic points, which we can do by applying a nifty functor. Define a functor $t(-):Top\to Top$ which associates to any topological space $X$ the set of its irreducible closed subsets with the topology given by declaring that for every closed $Y\subset X$ we have that $t(Y)\subset t(X)$ is a closed set. This functor acts on a morphism $f:X\to X'$ by sending an irreducible closed subset $Y\subset X$ to $\overline{f(Y)}\subset X'$. For any topological space $X$, there is an induced morphism $\alpha:X\to t(X)$ by sending $x\mapsto \overline{\{x\}}$, and this map induces a bijection between open subsets of $X$ and $t(X)$.

To verify that $(t(Z(k)),\alpha_*(\mathcal{O}_Z|_{Z(k)}))\cong Z$, we'll need to make some observations about $Z(k)$. First, $Z(k)$ can be identified with the set of closed points of $Z$: by Zariski's lemma, the closed points of a scheme of finite type over a field $F$ are exactly those points which have as their residue field a finite extension of $F$. If $F$ is algebraically closed, it has no nontrivial finite extensions, so the closed points of $Z$ are exactly $Z(k)$, and they have the induced topology from $Z$. Next, for any scheme of finite type over a field, the closed points are very dense: they're dense in every nonempty closed subset.

Now we may define a map of topological spaces $t(Z(k))\to Z$ by sending any irreducible closed subset $Y\subset Z(k)$ to the generic point of its closure in $Z$. It's not difficult to see that this is a homeomorphism by our previous observations, and as $\alpha: Z(k)\to t(Z(k))$ gives a bijection on open sets, we have that $\alpha_*(\mathcal{O}_Z|_{Z(k)})\cong \mathcal{O}_Z$.

This rather quickly settles the second part, too: given $\varphi$, we can let $\psi$ be the morphism $(t(\varphi),\alpha_*\varphi^\sharp)$, and checking uniqueness is not difficult.

Essentially what we've proven here is that there's an equivalence of categories between schemes of finite type over an algebraically closed field $k$ and the $k$-points of schemes of finite type over an algebraically closed field equipped with the restriction of the structure sheaf. The upshot of this for you is that if you can construct the map in the statement you wish to prove in this restricted setting, you'll automatically get the map of schemes you want.

Both algebraically closed and finite type are necessary conditions: if either of these are violated, there are $k$-schemes which cannot be distinguished from the empty $k$-scheme by looking at their $k$-points, and our proof above breaks. In the non-algebraically closed case, consider the spectrum of the algebraic closure, and in the non-finite type case, consider the spectrum of the transcendental extension $k(x)$. In both cases, a $k$-point is equivalent to a series of $k$-algebra map $k\to F\to k$ which is the identity. But this can't happen: the map $F\to k$ must have nontrivial kernel by dimension reasons, which means that $F\to k$ is the zero map because $F$ is a field.