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Let $f:X\rightarrow Y$ be an etale morphism of schemes. By $f$ factoring locally as a composition of an open immersion and a finite etale map, I mean that for every $x \in X$ there exists an open $U$ of $x$ and an open $V \subset Y$ containing $f(U)$ such that the restriction $f:U \rightarrow V$ factors as $U \hookrightarrow W \rightarrow V$, where $U \hookrightarrow W$ is an open immersion and $W \rightarrow V$ is finite etale.

According to section 7.5 of the "Berkeley lectures" by Scholze and Weinstein, it is not true that every etale $f$ factors locally as a composition of an open immersion and a finite etale map in the category of schemes (though it is true in the category of analytic adic spaces). What is an example of this failure for schemes?

Thanks!

xlord
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  • Do you mean that $f$ factors or just factors locally? If the former, then you can take $Y$ the spectrum of $\mathbb{Z}$ and $X$ that of $\mathbb{Z}[1/2,i]$. – Aphelli Feb 17 '21 at 15:46
  • I meant locally. Thanks, I edited the question accordingly. – xlord Feb 17 '21 at 16:13

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Ariyan Javanpeykar's answer to a related question gives an example that works. Since you're asking for something a priori slightly stronger than in that question (no local factorization into open immersion + finite étale, not just no global factorization), I'll elaborate. The example is as follows: Let $X = \mathbb{A}^1 \setminus \{0, 2/3\}$, and define a morphism $f\colon X \to \mathbb{A}^1$ by $f(x) = x^2 (x - 1)$. Note that $f$ is étale.

Let $U \subseteq X$ be any open subset containing $1 \in X$, and let $V \subseteq \mathbb{A}^1$ be any open subset such that $f(U) \subseteq V$. Note that $0 = f(1) \in V$. Suppose we have a scheme $W$, an open immersion $j\colon U \to W$, and a finite morphism $h\colon W \to V$ such that $f\rvert_U = h \circ j$. Let $Z \subseteq W$ be the irreducible component containing $j(U)$ (with the reduced induced closed subscheme structure). Irreducible components are closed, and closed immersions are finite, so $g := h\rvert_Z\colon Z \to V$ is finite. Finite morphisms are affine, so $Z$ is affine, irreducible, reduced, 1-dimensional, and birational to $\mathbb{A}^1$, which means $Z$ is an open subscheme of $\mathbb{A}^1$. So $g$ is given by a rational function, but since $g$ agrees with $f$ on a dense open subset, this rational function must be $x^2 (x - 1)$.

But since $g$ is finite, all the fibers have the same degree, so $g^{-1}(0)$ consists of three points, counting multiplicity. Since $x^2 (x - 1)$ is ramified above $0$, this means $g$ (and hence $h$) cannot be étale.

Actually, now that I've written this up, I'm a little confused why this doesn't also give a counterexample for analytic adic spaces. I guess it has to do with the more flexible nature of open immersions in the analytic setting?

Daniel Hast
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  • Thanks! This makes sense. My intuition is that, in the adic setting, you can take $U$ to be a small disc containing 1, and $V$ to be its image via $f$ ($f$ is open since it is etale). Since 1 is not ramified, $f$ is locally 1-1 at 1 (which is not true in the scheme setting, since open sets are too large), so you can take $U$ and $V$ to be so small that already $U\rightarrow V$ is even an isomorphism. I think this picture at least works in the complex analytic setting. – xlord Feb 17 '21 at 17:54
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    @xlord You have to be careful. There are subtelties between maximal vs. non-maximal points. What you're describing is more accurate on the Berkovich space, but on the adic space you have to contend with the existence of things like Gauss points having non-trivial specialization. In particular, be careful that your intuition almost makes it seem as though for all $x\in X$ and $y=f(x)$ in $Y$ there exists opens $U$ and $V$ neighborhoods around both with $f(U)\subseteq V$ and $f:U\to V$ finite. This is true around maximal points, and in general if $f$ is partially proper, but not in general. – Alex Youcis Feb 17 '21 at 18:15
  • @AlexYoucis thanks, I have not thought of non maximal points. It was more like I was trying to explain what was happening only at the (maximal) evaluation at 1 point. Is this argument incorrect then? Do you know a nice formal way to prove in this case $f$ factors locally as a composition of an open immersion and a finite etale map? (I could in principle check this up in Huber's book, but perhaps it is simpler for this example?) – xlord Feb 17 '21 at 18:43
  • @xlord Are you asking about Lemma 2.2.8 of Huber's book? – Alex Youcis Feb 17 '21 at 18:45
  • Yes, this is Lemma 2.2.8 in his book. – xlord Feb 17 '21 at 18:47
  • @xlord What is your exact question then? How to prove the claim when you assume that $x$ (and therefore $y$) are maximal points? – Alex Youcis Feb 17 '21 at 18:47
  • Just for the point x=1. I was under the impression you were saying the sketch in my original comment was incorrect. How would you write it out formally, if not? – xlord Feb 17 '21 at 18:50
  • Let us continue this discussion in chat. – Alex Youcis Feb 17 '21 at 18:51