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edit: Added smoothness condition.

The following may be rather trivial, but I cannot seem to find a reference. If $X$ is a scheme write $et/X$ for the category of étale schemes over $X$ and $Sm/X$ for the category of smooth schemes over $X$. Perhaps we may also wish to restrict to affine or relative affine objects. There is an obvious functor $i: et/X \to Sm/X$.

What I am wondering is: does this have a left adjoint?

If $X = Spec(k)$ is a field this is "known", but I cannot seem to find a reference. If $Y \to Spec(k)$ is a scheme let $A_Y$ be the normalisation of $k$ in $\mathcal{O}_X(X)$. Then $\pi_0(Y) := Spec(A_Y)$ is the sought-for left adjoint.

This construction seems sufficiently geometric that I am wondering if a relative version is possible. The obvious generalisation (using relative spec) does not work because $\pi_0$ must be a section of $i$ and not all étale schemes $Y \to X$ are (relative) affine.

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    $\begingroup$ You might find this question, the answer and references useful mathoverflow.net/questions/219637/… (Romagny considers both schemes of connected components and schemes of irreducible components in the cited paper). $\endgroup$
    – Daniel Loughran
    Commented Jul 5, 2016 at 13:23
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    $\begingroup$ You either need some kind of finite type hypothesis or you need to replace etale with pro-etale; otherwise already over a field you run into problems with e.g. Spec of an infinite product of copies of the field. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 5, 2016 at 18:47
  • $\begingroup$ @DanielLoughran: I think Theorem 2.5.2 and Remark 2.5.3 (1) of the paper show that the answer is affirmative if we replace schemes by algebraic stacks. Perhaps some form of smooth base change can be used to prove the required separatedness property for smooth schemes; I'm not sure. $\endgroup$
    – Tom Bachmann
    Commented Jul 6, 2016 at 13:00
  • $\begingroup$ In general these functors won't be representable by schemes; the best one could hope for is that they are represented by algebraic spaces. Do you have an explicit application in mind? Romagny works in great generality where he needs and works with stacks, but things can simplify in special cases. $\endgroup$
    – Daniel Loughran
    Commented Jul 6, 2016 at 13:32
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    $\begingroup$ If you are willing to focus on smooth and proper morphisms, then the answer is yes and provided by Stein factorisation. $\endgroup$
    – Daniel Loughran
    Commented Jul 7, 2016 at 8:33

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There definitely need to be some restrictions on what kind of schemes over $X$ you consider. Indeed, if $X=\mathbb{A}^1_k$ is the affine line (over a field $k$, say) and $Y \to X$ is the inclusion (closed immersion) of the origin $0$ in $X$, if we let $U\to X$ range over open immersions of $X$ (which are certainly étale), you are looking for an étale $F(Y)\to X$ which (a) factors through every $U\to X$ containing the $0$ but (b) not through any $U\to X$ that does not contain $0$. But (a) implies that the image of $F(Y)\to X$ is contained in every open set $U$ containing $0$; but étale morphisms are open, so the only possibility is that $F(Y)$ is empty, which then contradicts (b).

My guess is that "flat" would be a reasonable condition (or perhaps change the other category to schemes unramified over $X$ instead of étale?).

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  • $\begingroup$ As I said I am ok with using smooth schemes over $X$. I added an edit for clarity. $\endgroup$
    – Tom Bachmann
    Commented Jul 7, 2016 at 8:00
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    $\begingroup$ @TomBachmann But you had also written you did not think it made a difference, so I felt it was worth point out that it does. ☺ (I had to write this as an answer because it was too long for a comment.) $\endgroup$
    – Gro-Tsen
    Commented Jul 7, 2016 at 8:46
  • $\begingroup$ Yes, I'm not denying I learned something :). $\endgroup$
    – Tom Bachmann
    Commented Jul 7, 2016 at 9:12

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