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Let $X\rightarrow Y$ be a finite etale map. Let $R$ be a strict henselian ring with residue field $k$. Say that we have a map $Spec(k)\rightarrow X$ and so also $Spec(k)\rightarrow Y$. Assume that the map $Spec(k)\rightarrow Y$ factors thusly: $Spec(k)\rightarrow Spec(R)\rightarrow Y$. Then the question is: would there be (a unique?) map $Spec(R)\rightarrow X$ that would make this commute?

This would be a little cleaner if I knew how to do commutative diagrams in mathoverflow, but hopefully you get the picture. Intuitively, this means that if you have a point on $Y$ and a point above it on $X$, and if you have a "path" (I use this word very loosely here) near that point on $Y$ then it extends to a "path" near the respective point on $X$. When trying to prove this, the first thing I thought about is that every etale map is formally etale, but in the definition of formally etale they only talk about lifting $1^{st}$ order deformations.

Do you know how to prove (or god forbid disprove) this?

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up vote 10 down vote accepted

I think this is true. Namely, we can lift each $\mathrm{Spec}(R/\mathfrak{m}^n)\to Y$ to $X$ (because formal etaleness allows the lift to exist for any nilpotent thickening) successively so as to be compatible with all the previous ones (in fact, it has to be, by uniqueness in the lifting property). This successive lifting gives a map $\mathrm{Spec} R \to X$ lifting $\mathrm{Spec} R \to Y$ (to see this, note that without loss of generality, $X, Y$ are affine, and then taking direct limits in the category of affine schemes is the same as taking inverse limits in the category of rings). Thus we get a map from the completion $\\mathrm{Spec} \hat{R}$, hence from $\mathrm{Spec} R$. I believe this is true even if we just assume $X \to Y$ to be smooth, since then we can still make successive liftings. Note that finiteness of the map $X \to Y$ is not necessary here. This argument seems to require modification because it is not obvious that the map will factor through the henselianization.

For another argument, note that we can reduce to the case where $Y$ is $\mathrm{Spec} R$ (by making a base-change via $\mathrm{Spec} R \to Y$). Then we have to show that there is a section $\mathrm{Spec} R \to X$: this is clear because $X$ is a product of strictly henselian rings finite and etale over $R$, and if one of them also has residue field $k$, it must be isomorphic to $\mathrm{Spec} R$. Here one uses the fact that a finite etale morphism of local rings with the same residue field must be an isomorphism, which follows from Nakayama's lemma and since flatness implies injectivity (for local rings).

(P.S. Just in case this was the question, recall the nilpotent lifting property for an etale morphism $X \to Y$: given any scheme $S$ and subscheme $S_0$ cut out by a nilpotent ideal (one can reduce to the square-zero case by induction), any diagram with $S_0 \to X$ and $S \to Y$ leads to a unique lift $S \to X$.)

(P. P. S. One doesn't need finiteness even in the second argument: if $X \to Y$ is an etale morphism with $Y$ the spectrum of a henselian ring $R$, then the local ring of a point of $x$ lying above the closed point of $Y$ is finite over $R$, by Zariski's Main Theorem: one characterization of henselian rings is that a quasi-finite local algebra over a henselian local ring is finite.)

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"I believe this is true even if we just assume X→Y to be smooth": but for uniqueness do we need étaleness? – Qfwfq Jul 17 '11 at 18:56
@unknowngoogle: Yes, I think so: smoothness guarantees a lift, but not uniqueness. – Akhil Mathew Jul 17 '11 at 20:07
This may very well be a dumb question, and I may be embarrassed for asking it later, but how does a morphism $\mathrm{Spec}(\hat{R})→X$ give a morphism $\mathrm{Spec}(R)→X$, when the natural arrow goes from $\mathrm{Spec}(\hat{R})$ to $\mathrm{Spec}(R)$? – Keenan Kidwell Jul 19 '11 at 0:26
@Keenan: Whoops! I don't see why it is obvious that the map should factor through the henselianization. – Akhil Mathew Jul 20 '11 at 4:30

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