# Desingularization of a fibered surface with smooth generic fiber - missing argument/assumption in Liu?

Theorem 8.3.50 in Liu's Algebraic Geometry and Arithmetic Curves states the following (this is only the relevant part of the statement):

Let $$S$$ be a Dedekind scheme of dimension $$1$$. Let $$\pi:X\to S$$ be a fibered surface with a regular generic fiber. Suppose that $$\operatorname{Sing}(X)$$ is contained in finitely many closed fibers $$X_{s_i}$$, and that $$X\times_S\operatorname{Spec Frac}(\widehat{\mathcal O_{S,s_i}})$$ is regular for each $$i$$. Then $$X$$ admits a desingularization.

This result is then used in proving Corollary 8.3.51:

Let $$S$$ be a Dedekind scheme of dimension $$1$$. Let $$\pi:X\to S$$ be a fibered surface with a smooth generic fiber. Then $$X$$ satisfies the above properties and hence admits a desingularization.

The proof of the Corollary uses an earlier result which almost immediately gives that $$\operatorname{Sing}(X)$$ is contained in finitely many closed fibers, and then states that this implies the conditions above hold. However, to me this argument seems to pretty blatantly ignore the condition on the formal fibers.

Does regularity of the formal fibers somehow follow automatically from smoothness of the generic fiber? Am I missing something obvious or is there really a missing argument here? Is the statement as written even true, or do we need extra assumptions?

Note: I had a thought there might be some running excellence assumption here which would imply it, but it seems to be quite the opposite - on page 363 Liu explicitly mentions he shall now treat the case where base is not necessarily excellent.

• Since $\eta_s:=\operatorname{Spec} {\rm Frac}(\widehat{\mathcal{O}}_{S, s_i})$ maps to the generic point $\eta$ of $S$, and $X_\eta$ is smooth over $\eta$, the base change $X\times_S \eta_s = X_\eta \times_\eta \eta_s$ is smooth over $\eta_s$. Smoothness over a field implies regularity, so you're good. Sep 5, 2020 at 15:58
• @PiotrAchinger Thank you, that makes sense. For some reason I was thinking of $\eta_s$ as lying over a closed point of $S$ instead of over the generic point. But it does of course lie over $\eta$, so your argument makes sense. If you'd like to post your reply as an answer I will gladly accept it. Sep 5, 2020 at 20:37

## 1 Answer

Posting my comment as an answer.

Since $$\eta_{s_i} := \operatorname{Spec} \operatorname{Frac}(\widehat{\mathcal{O}}_{S, s_i})$$ maps to the generic point $$\eta$$ of $$S$$, and $$X_\eta$$ is smooth over $$\eta$$, the base change $$X\times_S \eta_{s_i} = X_\eta \times_\eta \eta_{s_i}$$ is smooth over $$\eta_{s_i}$$. Smoothness over a field implies regularity, so the $$X\times_S \eta_{s_i}$$ are regular.