# Extending section of étale morphism of adic spaces

This question is related to Lifting points via étale morphism of adic spaces.

Fix a complete non-archimedean field $$k$$. Let $$(A,A^+)$$ be a complete strongly noetherian Huber pair over $$(k,k^\circ)$$. Let $$f\colon A\to B$$ be a finite étale morphism of rings; endow $$B$$ with the natural $$A$$-module topology and let $$B^+$$ be the integral closure of $$A^+$$ in $$B$$. Then $$f$$ defines a morphism of complete Huber pairs $$f\colon (A,A^+)\to (B,B^+)$$ over $$(k,k^\circ)$$.

Let now $$x\in \operatorname{Spa}(A,A^+)$$ and suppose that $$\operatorname{Spa}(k(x),k(x)^+)\to X:=\operatorname{Spa}(A,A^+)$$ lifts to $$Y:=\operatorname{Spa}(B,B^+)$$. Denote the corresponding point in $$Y$$ by $$y$$. Then the inclusion $$k(x)\subset \widehat{k(x)}$$ factors through $$k(y)$$: $$k(x)\subset k(y)\subset \widehat{k(x)}.$$ Thus, if we assume that $$k(x)$$ is already complete, we obtain $$k(x)=k(y)$$. Now the morphism $$f_y\colon \mathcal{O}_{X,x}\to \mathcal{O}_{Y,y}$$ induced by $$f$$ is étale at the closed point of $$\mathcal{O}_{Y,y}$$. Since $$\mathcal{O}_{X,x}$$ is henselian, the equality $$k(x)=k(y)$$ implies that $$f_y$$ has a section $$s\colon \mathcal{O}_{Y,y}\to \mathcal{O}_{X,x}$$, so that we obtain a map $$s\colon B\to \mathcal{O}_{X,x}$$. The noetherian assumption implies that this lifts to a morphism $$s\colon B\to \mathcal{O}_X(U)$$, for $$U$$ a rational subset of $$X$$ containing $$x$$.
In other words, the morphism $$f\colon Y\to X$$ admits a section $$s\colon U\to Y$$ locally around $$x$$.

$$\textbf{Question}$$: How can I argue if the residue field $$k(x)$$ is not necessarily complete?

Help is highly appreciated.

The statement is correct and here is one way to prove it. The slogan is that "finite etale extensions of $$k(x)$$ and $$\widehat{k(x)}$$ are the same for "quasi-complete" fields $$k(x)$$" (this is a toy example of Gabber's Approximation Lemma).

I decided to write a proof in complete details mostly to be sure that there is no small mistake (there are always subtle points when trying to work with general (even strongly noetherian) adic spaces).

We start with a formulation of a very important Lemma (whose proof I will explain at the end of this answer)

Lemma 1: Let $$X:=\operatorname{Spa}(B, B^+)$$ and $$Y:=\operatorname{Spa}(A,A^+)$$ be strongly noetherian affinoid adic spaces, $$f: X\to Y$$ be a finite morphism. Choose any point $$y\in Y$$ and denote its preimage (as a finite set of points) by $$\{x_i\}$$. Then we have an isomorphism $$\mathcal O_{Y, y}\otimes_A B = \prod_{i} \mathcal O_{X,x_i}$$.

Let us now use this Lemma in the case of interest. Namely, we have finite etale morphism $$f:X \to Y$$ of strongly noetherian affinoid adic spaces. Then Lemma 1 implies that an $$A$$-homorphism $$\mathcal O_{Y,y} \to \prod_i \mathcal O_{X, x_i}$$ is a finite etale morphism! In particular, an extension of a maximal ideal $$\mathfrak{m}_y$$ to the ring $$\mathcal O_{X,x_i}$$ is equal to $$\mathfrak m_{x_i}$$ for all $$i$$ (i.e. $$\mathfrak{m}_y \mathcal O_{X,x_i}=\mathfrak{m}_{x_i}$$) and each residue field $$k(x_i)$$ is a finite separable extension of $$k(y)$$. In other words, $$k(y)=\prod_i k(x_i).$$ and each $$k(x_i)$$ is a finite separable extension of $$k(y)$$. Once we know that at least one of $$k(x_i)$$ is equal to $$k(y)$$ we can leverage the fact that $$\mathcal O_{Y,y}$$ is henselian to get a section analytically locally on $$Y$$ (this is well-explained in your question).

Proposition 2: Under notation as above, assume that $$f:X \to Y$$ has a section $$s: \operatorname{Spa}(k(y), k(y)^+) \to X,$$ then there is $$i$$ s.t. $$k(x_i)=k(y)$$.

Consider "a fiber" $$X_y$$, which is defined as a fiber product $$X\times_Y \operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})$$, this is a finite etale scheme over "a point" $$\operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})=\operatorname{Spa}(k(y), k(y)^+)$$ (this space is not necessary a one point space). Since finiteness is preserved by base change (at least for adic morphisms), we know that $$X_y$$ is a finite adic space over $$\operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})$$ and its algebra of functions is equal to $$\mathcal O_{X_y}(X_y)=(k(y)\otimes_{A} B)\hat{}= (k(y)\otimes_{\mathcal O_{Y,y}} \mathcal O_{Y,y} \otimes_A B)\hat{}=$$ $$=(k(y)\otimes_{\mathcal O_{Y,y}} \prod_i \mathcal O_{X,x_i})\hat{}=(\prod_{i} k(x_i))\hat{}=\prod_i \widehat{k(x_i)}.$$ Existence of a section $$s:\operatorname{Spa}(\widehat{k(y)},\widehat{k(y)^+}) \to X_y$$ implies that there is an integer $$i$$, s.t. $$\widehat{k(x_i)}=\widehat{k(y)}$$. Now we want to use some Approximation result to conclude that $$k(x_i)=k(y)$$. One way to show this is to prove that a natural functor $$\{\text{finite separable extensions of } k(y) \} \to \{\text{finite separable extensions of} \widehat{k(y) } \}$$ defined by $$l \mapsto \widehat{l}$$ is an equivalence of categories.

This would mean that an isomorphism $$\widehat{k(x_i)}\cong \widehat{k(y)}$$ implies an existence of an isomorphism $$k(x_i) \cong k(y)$$. It is possible to prove this statement using a (very) difficult theorem of Gabber, but there is a way to avoid it.

Indeed, we know that $$A$$ is an analytic ring, hence any valuation has a unique vertical generalization. This means that the field $$k(y)$$ has a rank-1 valuation subring $$k(y)^{\circ}$$. Since we want to relate finite separable extensions of $$k(y)$$ to those of $$\widehat{k(y)}$$, we can forget about $$k(y)^+$$ and work with its rk-1 valuation instead. Then there is a theorem of Berkovich (Thm 2.4.1 here) that shows the claim for quasicomplete fields and there is another Proposition in the same paper (Proposition 2.4.3) that says to prove quasi-completeness of a rk-1 valutation field $$k(y)$$ it is sufficient to show that $$k(y)^{\circ}$$ is henselian (with respect to an ideal generated by a pseudo-uniformizer). But this is a standard fact for analytic Huber pairs.

All in all, we have the desired equivalence of categories, thus we conclude that $$k(x_i)=k(y)$$ and this is exactly what we were looking for.

Proof of Lemma 1: The answer turned out to be bigger than I expected, so I only sketch a proof of this Lemma. If somebody is interested in a complete proof, I can add it later.

The Lemma basically follows from the following four facts.

• If two points $$x$$ and $$y$$ of a spectral space (such as $$\operatorname{Spa}(A, A^+)$$) don't have a common generalizing point, then there exist disjoint open neighborhoods containing $$x$$ and $$y$$.

• In a notation as above none of the points $$x_i$$ has a common generalizing point. Hence, there are disjoint open neighbourhoods $$U_i$$ of each $$x_i$$.

• Any finite morphism is closed, then a standard trick allows us to choose an open subset $$U \subset \cup U_i$$ of the form $$U=f^{-1}(V)$$ for some open $$V$$ containing $$y$$.

• The last bullet point implies (with a little work) that $$\mathcal O_{Y,y}\otimes_A B = \prod_i \mathcal O_{X, x_i}$$.

Here is a proof the second bullet point:

Assume that two points in the fiber over a point $$y$$ have a common generalization. Let us denote these two points by $$x_1$$ and $$x_2$$ and their common generalizing point by $$x$$. Then it is rather easy to see that $$z:=f(x)$$ is a generalization of $$y$$. This implies that $$z$$ lies in the image of $$\operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})$$ and, therefore, $$x\in \operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})\times_Y X$$. This means that we can base change the morphism along a map $$\operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+}) \to Y$$ to assume that $$Y$$ is an adic space over a (strongly noetherian) complete valuation field.

Under this assumption, we know that $$A:=\mathcal O_X(X)$$ should be a finite etale $$\widehat{k(y)}$$-algebra . In particular, it is semilocal. Let us denote all maximal ideals of $$A$$ by $$\mathfrak m_j$$. Then it is easy to see (by a classification of Artinian algebras over a field) that $$A=\prod_j A_{\mathfrak m_j}$$ and each $$A_{\mathfrak m_i}$$ is a field by etaleness of A (in the case of a general finite morphism just pass to $$A_{red}$$, it will not change an underlying topological space). Each $$x_i$$ has support on some $$\mathfrak m_j(i)$$ and it suffices to show that $$\mathfrak m_{j(1)}$$ is not equal to $$\mathfrak m_{j(2)}$$. The key is Theorem VI.8.3.1 and Corollary VI.8.2.2 in Bourbaki "Commutative Algebra". Together these two statements imply that there is exactly one extension of a valuation $$y$$ to each of $$A_{\mathfrak m_i}$$. Therefore $$x_1$$ and $$x_2$$ can't be supported on the same $$A_{\mathfrak m_i}$$.

• Dear gdb, thank you for the nice answer. But can't we avoid the use of Lemma 1 as follows: The morphism $f\colon X\to Y$ being étale already implies that the extension $k(y)\subset k(x)$ is finite separable; then as explained in my question (with your notation) one has $k(y)\subset k(x)\subset \widehat{k(y)}$ and hence $\widehat{k(x)}=\widehat{k(y)}$ and so one would obtain $k(y)=k(x)$ by the equivalence of categories you stated. Or am I missing something?
– Nib
Mar 22, 2019 at 5:58
• @Nib That's correct, it is sufficient to show that $k(y) \subset k(x)$ is finite separable. A priori, etaleness guarantees only that $\widehat{k(y)} \subset \widehat{k(x)}$ is finite and separable. Does it formally imply that $k(y) \subset k(x)$ is finite and separable? But I agree that this is a question how you set up general theory. I prefer to prove that around rk-1 points all quasi-finite maps are actually finite and then use the argument from my answer. Since all points have rk-1 vertical generalization it is sufficient for the claim. I guess you can argue somehow in a different way.
– gdb
Mar 22, 2019 at 8:52
• Ok, you can probably just require this property in the very definition of an etale map. I guess you will then face some other obstacles. But I am not sure.
– gdb
Mar 22, 2019 at 8:54
• If I see things correctly, then the second last bullet point reduces the last one to showing that for a rational subset $V=R(\frac{T}{s})$ of $Y$, one has $\mathcal{O}_Y(V)\otimes_A B=\mathcal{O}_X(f^{-1}(V))$. Now, $\mathcal{O}_X(f^{-1}(V))$ is a pushout of $A\to \mathcal{O}_Y(V)$, $A\to B$ (omitting the $+$-component). A pushout would also be the completion of $\mathcal{O}_Y(V)\otimes_A B$ wrt the topology defined by the open subring $\widehat{A_0[\frac{T}{s}]}\otimes_{A_0} B_0$ with ideal of definition $I^n\widehat{A_0[\frac{T}{s}]}\otimes_{A_0} B_0$, where $(A_0,I)$ is a...
– Nib
Mar 22, 2019 at 19:00
• pair of definition of $A$, $B_0$ is a ring of definition of $B$ such that $f(A_0)\subset B_0$. So it is enough to show that $\mathcal{O}_Y(V)\otimes_A B$ is complete for this topology. It would be enough to show that $(\ast)$ the topology on $\mathcal{O}_Y(V)\otimes_A B$ is the natural $\mathcal{O}_Y(V)$-module topology (or it would also follow from $B_0$ being of finite presentation over $A_0$). Why is the topology the natural topology? $(\ast)$ I only know that a finite module with the natural topology is complete if the underlying ring has a noetherian ring of definition.
– Nib
Mar 22, 2019 at 19:03