What are all the etale morphisms from a scheme $X$ to $\Bbb A^1_k$?

Knowing that $X\to \Bbb A^1_k$ is etale means that $X$ is $1$-dimensional I think. Additionally, $X$ must admit a zariski open cover by affines, $X=\bigcup_{i\in I} Spec(A_i)$ where each $A_i$ is a $k[t]$-algebra.

So we have that $X$ is covered by $1$-dimensional affine $k[t]$-algebras. Additionally, these $U_i=Spec(A_i)$ are taken by open immersion $U_i\to X$ into $X$, and open immersions are etale, so each of these are etale over $\Bbb A^1_k$, so we can probably simplify our analysis first to affines over $\Bbb A^1_k$.

In which case we first want to consider $Spec(A_i)\to Spec(k[t])$ morphisms that are etale. I think $A_i$ should be finitely presented as a $k[t]$-algebra, so of the form $k[t][x_1,\dots,x_n]/(f_1,\dots,f_m)$ where being $1$-dimensional means that $(f_1,\dots,f_m)$ must cut out an $n$-dimensional subvariety of $\Bbb A^{n+1}_k$.

I'm not sure if I'm correct at this point, and I'm not sure how to find all of them. I think maybe one can argue like: 1) surjective finite etale morphisms to $\Bbb A^1_k$ are necessarily just isomorphisms $\Bbb A^1_k\to \Bbb A^1_k$, 2) any etale morphism $X\to \Bbb A^1_k$ can be covered by finite etale morphisms $U_i\to X\to \Bbb A^1_k$, and composites of etale morphisms are etale 3) ???, 4) profit

Bonus: I really would like to understand all etale coverings $\{U_i\to \Bbb A^1_k\}_{i\in I}$, where the question above was my first obstruction to working this out. So any ideas on that would also be helpful.

  • $\begingroup$ You can explicitly describe the etale site of any Dedekind scheme. If no one replies, I will write something later. $\endgroup$ Commented Apr 27, 2019 at 2:27
  • $\begingroup$ @AlexYoucis What is a dedekind scheme? Spectrum of a dedekind domain? $\endgroup$ Commented Apr 27, 2019 at 2:30
  • $\begingroup$ Integral normal scheme of dimension $1$ (locally spectrum of Dedekind domain( $\endgroup$ Commented Apr 27, 2019 at 2:32
  • $\begingroup$ @AlexYoucis In case noone answers, and you forget to come back here, did you have a reference for this explicit description (or some partial description)? $\endgroup$ Commented Apr 27, 2019 at 2:35
  • $\begingroup$ I'm in the process of writing it up (it's semi-long). I don't remember a reference off the top of my head. I think I read it in some work of Zink originally, but I don't recall. I'll finish writing it later when I have more time. $\endgroup$ Commented Apr 27, 2019 at 4:42

1 Answer 1



Disclaimer: Of course, none of the below is original. I don't remember where I first learned it (it was almost certainly something Brian Conrad wrote, but I can't find it--it might be something he posted on MO?).


Let $X$ be any Dedekind scheme. By defintion (for me) this means that $X$ is an integral normal Noetherian scheme of dimension $1$ (so locally the spectrum of a Dedekind domain). Let us set $K:=K(X)$. Note that for each point $x\in X$ we can define an inertia subgroup at $x$, denoted $I_x$, as follows. Let $\h_{X,\ov{x}}$ be the strict Henselization of $X$ at $\ov{x}:\Spec(k(x)^\sep)\to X$ (see this for more detail). Note that we can embed $\h_{X,\ov{x}}$ into $K^\sep$ essentially as follows. Choose a valuation $v'$ of $K^\sep$ lying over $v_x$. Then, take the union of the valuation rings $\{x\in F:v'(x)\geqslant 0\}$ as $L$ travels over the finite subextensions of $K^\sep/K$ such that $v'$ (restricted to that extension) is unramified over $K$. Let $L_x:=\mathrm{Frac}(\h_{X,\ov{x}})$. We then set $I_x:=\Gal(K^\sep/L_x)$. Note that $\Gal(K^\sep/K)/I_x\cong \Gal(k(x)^\sep/k(x)$.

So, in reality we won't explicitly parameterize all etale covers. Instead, we'll virtually parameterize all etale maps. Less cryptically, let us now suppose that $Y\to X$ is an etale morphism. Then, we know that $Y\to X$ is locally quasi-finite--there is an open cover $\{Y_i\}$ of $Y$ such that $Y_i\to X$ s quasi-finite. In particular, every etale map $Y\to X$ has a refinement (in the big Zariski site) by a cover of the form $\displaystyle \bigsqcup_i Y_i\to X$ with $Y_i\to X$ quasi-finite. Thus, for all intents and purposes it's really enough to describe the category $\mathscr{C}$ of all quasi-finite etale maps $U\to X$.

Let us make the further following reduction. Namely, note that if $Y\to X$ is quasi-finite then $Y_K\to\mathrm{Spec}(K)$ is a finite etale $K$-scheme. Indeed, we know that every etale scheme over $\mathrm{Spec}(K)$ is a disjoint union of spectra of finite separable extensions of $K$ (e.g. see this). Since $Y_K$ is quasi-finite it must be a finite disjoint union, and thus a finite etale cover.

By the 'spreading out principle' this implies that there exists some open subscheme $U$ of $X$ such that $Y_U\to U$ is finite etale. Let us set $\mathscr{C}_U$ to be the category of quasi-finite etale maps $Y\to X$ such that $Y_U\to U$ is finite. Then, what we will actually do here is give a fairly easy way to 'paramaterize' the category $\mathscr{C}_U$.

Description of $\mathscr{C}_U$

Let us take an object $Y\to X$ of $\mathscr{C}_U$. We shall essentially claim is that we can somehow capture $Y$ by the finite etale cover $Y_U\to U$ and the 'straggler fibers' over the points in $Z:=X-U$. moreover, we shall claim that both pieces of these data can be described in terms of Galois sets.

Let us begin by noting that if $Y\to X$ is in $\mathscr{C}_U$ the since $U$ is normal we know that $Y_U$ is normal. And, in fact, it's pretty easy to see that $Y_U$ is actually just the normalization of $X$ in $Y_K$. Thus, we see that $Y_U$ is actually determined from $Y_K$ which is determined by the finite (discrete) $\Gal(K^\sep/K)$-set $Y(K^\sep)$ which is unramified along $U$. Recall that a $\Gal(K^\sep/K)$-set $T$ is called unramified at $x$ if $I_x$ acts trivially on it, and that it's unramified along $U$ if its unramified at every $x\in U$.

In fact, what we have just described is the following well-known result:

Fact 1: The association $V\mapsto V(K^\sep)$ is an equivalence of categories from the category $\mathsf{Fet}(U)$ of finite etale covers of $U$ to the set of finite discrete $\Gal(K^\sep/K)$-sets unramified along $U$.

So, this accounts for $Y_U\to U$, but what about these straggler fibers $Y_x$ for $x\in Z:=X-U$?

Well, note that for each $x\in Z$ that we have a natural inclusion $Y(k(x)^\sep)\hookrightarrow Y(K^\sep)$. Indeed, since $Y_{\Spec(\h_{X,\ov{x}})}\to \Spec(\h_{X,\ov{x}})$ is etale, we can use Hensel's lemma to say that $Y(k(x)^\sep)=Y(\h_{X,\ov{x}})$. Since $\h_{X,\ov{x}}\hookrightarrow K^\sep$ (by the discussion at the beginning of the setup) this gives us a natural map $Y(k(x)^\sep)\to Y(K^\sep)$ which is easily seen to be injective. Moreover, since $I_x$ naturally acts trivially on $Y(k(x)^\sep)$ (by definition) we see that $Y(k(x)^\sep)$ naturally lands in $Y(K^\sep)^{I_x}$.

Note that since $I_x$ acts trivially on $Y(K^\sep)^{I_x}$ that $\Gal(k(x)^\sep/k(x)$ acts on $Y(K^\sep)^{I_x}$ and, in fact, the inclusion $Y(k(x)^\sep)\hookrightarrow Y(K^\sep)^{I_x}$ is $\Gal(k(x)^\sep/k(x)$-equivariant.

The main result is then the following:

Theorem: The association $Y\mapsto (Y(K^\sep),\{Y(k(x)^\sep)\}_{x\in Z})$ is an equivalence of categories from $\mathscr{C}_U$ to the category of tuples $(T,\{T_x\})$ where $T$ is a finite discrete $\Gal(K^\sep/K)$-set and $T_x$ is a $\Gal(k(x)^\sep/k(x))$-stable subset of $T^{I_x}$. Moreover, $T_x=T^{I_x}$ if and only if the associated quasi-finite etale scheme $Y\to X$ is finite etale in some open neigborhood of $X$.

So we see that $\mathscr{C}_U$ is parameterized by some concrete sets of Galois theoretic data.

Will add example with $\mathbb{A}^1_k$ later

  • $\begingroup$ when you say 'spreading out principle', do you mean this: stacks.math.columbia.edu/tag/02NW? $\endgroup$
    – user631975
    Commented Apr 27, 2019 at 14:04
  • $\begingroup$ @AknazarKazhymurat Right, (4) there is 'spreading out' in this situation. I'm just generally thinking of Theorem 3.2.1 of this: www-math.mit.edu/~poonen/papers/Qpoints.pdf Of course, you need to check finiteness is on the table, but it is. $\endgroup$ Commented Apr 27, 2019 at 17:28
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    $\begingroup$ Is it later yet :-)? $\endgroup$ Commented Aug 29, 2020 at 22:22
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    $\begingroup$ Based on the 4 upvotes of the comment above, it is safe to say that other people would be interested in you completing the answer, if you had time! $\endgroup$ Commented Oct 29, 2023 at 5:51
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    $\begingroup$ @EarthCracks Hey there. I hear you. I'm in the midst of applying for jobs and moving from Japan to the US, so I'm a bit busy right now. May I ask a favor though? I should be a lot less busy in like 3-4 weeks (I hope...). If you ping me then I will try to honestly finish this. $\endgroup$ Commented Oct 29, 2023 at 13:51

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