Are generically trivial finite unramified morphisms trivial

Question

Let $S$ be a smooth affine variety over $\mathbb C$ and let $f:X\to S$ be a finite unramified morphism.

Suppose that $X(K(S))$ is non-empty. (This means that $X\to S$ has a section generically. It does not imply $X\to S$ being generically trivial.)

Is $X(S)$ non-empty?

The answer is positive if $S$ is of dimension at most one.

$X(K(S))$ is always non-empty : just take some (strange) embedding of $K(S)$ into $\mathbb{C}$ ... — abx, Commented Nov 7, 2014 at 11:34
@JasonStarr How is ZMT relevant here? Am I missing something obvious? — Unit section, Commented Nov 7, 2014 at 11:37
@abx I see your point. I meant to assume that the generic fibre $X_{K(S)}$ of $X\to S$ has a $K(S)$-rational point. This is stronger than the set $X(K(S))$ being non-empty (unless we consider only elements of $X(K(S))$ that correspond to the generic point Spec $K(S) \to S$. Then $X(K(S)) = X_{K(S)}(K(S))$ which can be certainly be empty (take a curve with no rational points over $\mathbb C(t)$). — Unit section, Commented Nov 7, 2014 at 11:39
@Unitsection: Okay, let me be more clear. Look up the original formulation of "Zariski's Main Theorem", not necessarily the formulation given in some textbooks. There is a nice comparison of the different formulations in Mumford's "Red Book". Also the Wikipedia entry is pretty good. — Jason Starr, Commented Nov 7, 2014 at 12:54

Ariyan Javanpeykar · Accepted Answer · 2014-11-07 13:47:42Z

2

The answer is yes in far more generality; see Prop. 6.2 of Liu-Lorenzini-Gabber http://arxiv.org/abs/1404.5366

In your case you can take the closure of a generic section and use Zariski's main theorem as Jason Starr alluded to above.

answered Nov 7, 2014 at 13:47

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