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I have a very naive question concerning Hilbert's irreducilibity theorem over function fields.

Let $f: V \to \mathbb{A}^2_\mathbb{Q}$ be a finite morphism, with $V$ an irreducible variety over $\mathbb{Q}$.

It is known that the function field $\mathbb{Q}(t)$ is Hilbertian. Hence Hilbert's irreducibility theorem implies that there exists a point $x \in \mathbb{A}^2_\mathbb{Q}(\mathbb{Q}(t))$ such that $f^{-1}(x)$ is irreducible. However, it is very possible in the statement of the irreducibility theorem that actually $x \in \mathbb{A}^2_\mathbb{Q}(\mathbb{Q})$; I would like to know that one can choose an $x$ such that this is not the case.

Does there exist $x \in \mathbb{A}^2_\mathbb{Q}(\mathbb{Q}(t))$ with $x \notin \mathbb{A}^2_\mathbb{Q}(\mathbb{Q})$ such that $f^{-1}(x)$ is irreducible?

An equivalent way to phrase this is the following:

Is the set $\mathbb{A}^2_\mathbb{Q}(\mathbb{Q}(t)) \setminus \mathbb{A}^2_\mathbb{Q}(\mathbb{Q})$ not thin inside $\mathbb{A}^2_\mathbb{Q}(\mathbb{Q}(t))$?

Here I use the term thin set in the sense of Serre (https://en.wikipedia.org/wiki/Thin_set_(Serre))

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    $\begingroup$ It seems that it suffices to choose a $\mathbb{Q}(t)$ point which can be specialized to a $\mathbb{Q}$ point over which the fibre is irreducible (and reduced). $\endgroup$
    – naf
    Commented Feb 22, 2017 at 5:08
  • $\begingroup$ @Ulrich: I don't understand your comment. Could you please elaborate? $\endgroup$
    – Daniel Loughran
    Commented Feb 22, 2017 at 10:13
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    $\begingroup$ OK. Let $C$ be a rational curve in the plane defined over $\mathbb{Q}$ such that it has a smooth rational point $x$ over which the map $f$ is etale and the fibre is also irreducible. Let $y$ be the generic point of the curve $C$ and let $R$ be the local ring of $x$ on $C$. The fibre product $Z$ of $S := \mathrm{Spec}(R)$ with $V$ is finite etale over $S$ so it is the spec of a semilocal regular one dimensional ring $R'$. However, the inverse image of $x$ (which is the closed point of $S$) is a singleton, so $R'$ must be a dvr, so $f^{-1}(y)$ is also a singleton. $\endgroup$
    – naf
    Commented Feb 22, 2017 at 13:18
  • $\begingroup$ Great, thanks! Yes I understand now, very nice. If you would like to post this as an answer then I will accept it. $\endgroup$
    – Daniel Loughran
    Commented Feb 22, 2017 at 14:29

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