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Let $X,Y$ be affine schemes over a strictly henselian base-ring $R$. Assume $X$ and $Y$ are $R$-smooth and consider a $R$-morphism $f$ between $X$ and $Y$.

I would like to use that $f$ is étale given that it is étale at every rational point of the special fiber.

So the questions is: is this true?

Thanks!

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    $\begingroup$ With these assumptions, the answer is trivially no: the special fiber of $X$ may be empty. $\endgroup$ Jun 22, 2018 at 6:21
  • $\begingroup$ Well, the special of X will be of finite type over an algebraically closed field. As it will be quasi-compact it has a closed point. Or am I getting something wrong? $\endgroup$
    – user125861
    Jun 22, 2018 at 6:25
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    $\begingroup$ The empty scheme is quasicompact and has no closed point. $\endgroup$ Jun 22, 2018 at 6:27

2 Answers 2

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As Piotr pointed out, the answer to your question is no as stated.

I just wanted to add a further comment (too long for an actual comment)

I guess what you're thinking is that it suffices to check etaleness at closed points, which is true.

However, it would be wrong to assume that all closed points of $X$ should lie on the special fiber. This is true if $X\rightarrow\text{Spec }R$ is proper, but false in general. This is the key point in Piotr's counterexample. The ramification locus of his map $f : X = \mathbb{A}^1_R\rightarrow\mathbb{A}^1_R = Y$ given by $f(x) = x+tx^n$ is a union of closed points which do not lie on the special fiber. If you consider his map, and extend it naturally to a map $\overline{X} = \mathbb{P}^1_R\rightarrow\mathbb{P}^1_R = \overline{Y}$, there you would be able to detect the ramification locus on the special fiber of $\overline{X}$. Indeed, the ramification locus is the divisor determined by the $(n-1)$th roots of $\frac{-1}{nt}$ in $\text{Frac}(R)$ (viewed as points on the generic fiber, the corresponding divisor being their closure in $X$). These elements of $\text{Frac}(R)$ are not integral over $R$, so the corresponding divisor intersects the special fiber of $\overline{X}$ at $\infty$. However, when you restrict back to $X = \mathbb{A}^1_R$, the divisor no longer intersects the special fiber. Indeed, since taking the closure of the $(n-1)$th roots of $\frac{-1}{nt}$ inside $\overline{X}$ only adds the $\infty$ of the special fiber, upon restricting to $X$, the ramification divisor becomes a disjoint union of closed points, all lying on the generic fiber.

This is also an interesting example of a nice (regular Noetherian) scheme of dimension 2 for which not all closed points have local rings of the same dimension. I believe this failure is stated as saying "the scheme is not equicodimensional" (it seems to be rather difficult to find references for this, but IIRC once upon a time I found a discussion of it in EGA...somewhere). For this, you can even take the scheme $\text{Spec }\mathbb{Z}_{(p)}[x]$.

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The answer is no, and the reason is that $X$ and $Y$ will not be henselian along their special fibers. If you replace $X$ and $Y$ by their henselizations (or completions) along the special fibers, the answer is positive.

For example, take $X = Y = \mathbb{A}^1_R$. Consider a map $f\colon X\to Y$ over $R$, which corresponds to a polynomial $f\in R[x]$. The non-smooth locus of $f$ is then given by the vanishing of the derivative $f' = df/dx$. If now $f'$ reduces to a non-zero constant in the residue field $k$ of $R$, then the reduction $f:X_k\to Y_k$ will be smooth, but $f$ itself is not smooth if $f'$ has nonzero degree.

For a concrete example, take $t\in R$ which is a nonzero element of the maximal ideal and $f = x + tx^n$, so $f' = 1 + tnx^{n-1}$.

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