Let $X$ be a scheme. Does the forgetful functor

$$\mathrm{EtSch}/X \to \mathrm{Sch}/X$$

have a right adjoint $Z \mapsto \tilde{Z}$? One might call $\tilde{Z}$ the étalification of $Z$. So this is an étale $X$-scheme together with an $X$-morphism $\tilde{Z} \to Z$, which induces for every étale $X$-scheme $Y$ a bijection $\hom_X(Y,\tilde{Z}) \cong \hom_X(Y,Z)$.

If $X$ is the spectrum of a field $k$, the answer is yes, using 1) the equivalence of sites between étale $k$-schemes and continuous $G$-sets, where $G$ is the absolute Galois group of $k$, 2) the fact that every sheaf on $G$-sets is representable. Explicitly: If $Z$ is a $k$-scheme, let us denote by $Z_{\mathrm{sep}}$ the subset of all points $z \in Z$ such that $k(z)/k$ is finite and separable. Then $$\tilde{Z} = \coprod\limits_{z \in Z_{\mathrm{sep}}} \mathrm{Spec}(k(z)).$$ For example, $\widetilde{\mathbb{A}^1}=\coprod_{\alpha \in k} \mathrm{Spec}(k)$ and $\widetilde{\mathbb{A}^2}$ is the coproduct of spectra of the form $(k[x]/(p)[T])/(q)$, where $p \in k[x]$ is irreducible and $q$ is some irreducible separable polynomial over $k[x]/(p)$.

Perhaps this construction is well-known, therefore I've put the reference request tag.

  • $\begingroup$ This is reminding me of Stein factorization, but I can't put my finger on it. $\endgroup$ – Allen Knutson Jun 24 '12 at 13:46
  • $\begingroup$ Why do you care that it be a scheme and not just an algebraic space? $\endgroup$ – Ben Wieland Mar 10 '13 at 21:35
  • $\begingroup$ Any answer about the right adjoint from etale algebraic spaces over X to algebraic spaces over X is also appreciated. $\endgroup$ – Martin Brandenburg Mar 11 '13 at 1:44
  • $\begingroup$ You allude to the question: "Is the espace étalé of an étale sheaf a scheme?" Over a base of dimension 0, you answer yes. I believe it is also so in dimension 1. But in dimension 2, no: mathoverflow.net/questions/43542/… This leaves the question of whether the additional assumption of the sheaf coming from a scheme makes the espace étalé a scheme. $\endgroup$ – Ben Wieland Mar 11 '13 at 1:53

Given a scheme $Z$ over $X$, one can consider the sheaf of "sections $X\to Z$". More precisely, that sheaf associates to every étale open $U\to X$ the set of commutative diagrams $$ \begin{matrix} & & Z \\\ & \nearrow & \downarrow \\\ U & \to & X \end{matrix} $$ The scheme $\tilde Z$ you are looking for is the espace étalé of the above sheaf, a highly non-separated scheme.

  • $\begingroup$ Every sheaf on the small etale site is representable by an algebraic space etale over the base. Is it true that if the original sheaf is the sheaf of sections for a morphism of schemes that this algebraic space is a scheme? $\endgroup$ – David Carchedi Jun 24 '12 at 14:20
  • $\begingroup$ @André: This is basically just a reformulation of the question. Why is your sheaf represented by an étale scheme? $\endgroup$ – Martin Brandenburg Jun 24 '12 at 14:33
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    $\begingroup$ @David: Taking the espace etale of the Zariski sheaf certainly gives the wrong answer. If $X$ is a nontrivial etale cover of $Z$ then it has no Zariski sections, yet obviously a functor that takes it to the empty set is not adjoint. $\endgroup$ – Will Sawin Jun 24 '12 at 17:00
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    $\begingroup$ If Z is $\mathbb A^2$ with the obvious projection to $X=\mathbb A^1$ over, for simplicity, an algebraically closed field $k$, then a closed point of the espace etale is an $x$ coordinate, a $y$ coordinate, and a polynomial $f(x,y)$ with those roots whose $y$-derivative is nonzero at that point. A closed point of $Z$ is just an $x$ coordinate and a $y$ coordinate. How is that a monomorphism? If it is, why is it obvious that the other Zariski-local sections exist? $\endgroup$ – Will Sawin Jun 24 '12 at 22:17
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    $\begingroup$ @André: For me your answer is still just a reformulation of my question. You write down the sheaf represented by the scheme we are looking for, and then claim that it is a scheme. You don't offer any proof. Meanwhile I wonder about all the upvotes ... does this mean that I am too blind to see the answer? $\endgroup$ – Martin Brandenburg Mar 10 '13 at 18:33

Can we just do this the hard way?

First we construct the set of points of the etalification $\bar{Z}$. A point of the etalification is a point $P$ of $X$ plus a map from the spectrum $T$ of the etale local ring at $X$ to $Z$ that forms a commutative diagram with the natural map $T \to X$. (or, if $Z \to X$ is not finite type, we need to use a finitely-generated subring of the etale local ring.)

Then for $Y\to X$ is etale with a map $Y \to Z$. We get a set-theoretic map $Y \to \bar{Z}$. One easily checks that each point is in the image of such a map. We form the finest topology such that the image of each such map is open.

Now we need a sheaf of rings. The natural map $\bar{Z} \to Z$ given by taking the image of the special fiber is continuous. Pull back the structure sheaf of $Z$, mod out by the functions on a set $U$ that are $0$ in the images of the open balls $T$ corresponding to the points of $U$, and sheafify.

We will check that this is a scheme by finding affine neighborhoods of each point. Each point of $\bar{Z}$ is in the image of some etale $Y\to X$, and we can further choose $Y$ to be affome amd $T \to Z$ an immersion. We will check that $T \to \bar{Z}$ is an inclusion of locally-ringed spaces. Clearly it is injective. Next we check that it is continuous and open. Open subsets of $\bar{Z}$ wil come from other etale maps which agree on some etale local ring, but then these must agree on an open subset, so they will give open subsets of $Y$. Open subsets of $Y$ are just more schemes etale over $X$.

Then we check it gives an isomorphism of sheaves of rings. The sheaf of rings on $Y$ is the pullback of the sheaf of rings of $Z$, mod the functions that vanish on it. But the functions that vanish on $Y$ are exactly the functions that vanish on each etale local rings, so the two sheaves of rings are identical.

So in addition $\bar{Z}$ is etale over $X$. By construction of the sheaf of rings, $\bar{Z} \to Z$ is a morphism. Similarly the set-theoretic maps $Y \to \bar{Z}$ for all $Y\to Z$ etale over $X$ are clearly morphisms, and give the required bijections.

Obviously this is quite sketchy and there are a lot of technical details I'm missing but it seems doable.


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