# All Etale morphisms $X\to\Bbb A^1$

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.

• You can explicitly describe the etale site of any Dedekind scheme. If no one replies, I will write something later. – Alex Youcis Apr 27 at 2:27
• @AlexYoucis What is a dedekind scheme? Spectrum of a dedekind domain? – Earth Cracks Apr 27 at 2:30
• Integral normal scheme of dimension $1$ (locally spectrum of Dedekind domain( – Alex Youcis Apr 27 at 2:32
• @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)? – Earth Cracks Apr 27 at 2:35
• 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. – Alex Youcis Apr 27 at 4:42


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?).

Setup:

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

• when you say 'spreading out principle', do you mean this: stacks.math.columbia.edu/tag/02NW? – user631975 Apr 27 at 14:04
• @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. – Alex Youcis Apr 27 at 17:28