# Rational points and finite étale covers

This question is broadly about non-trivial examples where a map $f:Y\to X$ of smooth projective $k$-varieties ($k$ not algebraically closed) is such that existence of $k$-points (or in fact $k'$-points for $k'/k$ a finite extension) on $X$ ensures their existence on $Y$. For example, if $f$ is a birational $k$-map, the Lang-Nishimura Theorem guarantees an element of $Y(k)$ if $X(k)$ is not empty.

An interesting example is the case of $f:Y\to X$ being a finite étale cover. In general it will not be the case that a point in $X(k)$ lifts to one in $Y(k)$ (for example elliptic curves and an element in the Tate-Shafarevich group). Are there however examples or conditions that guarantee this happens?

We say that a variety $X/k$ is potentially dense if $X(k')$ is dense in $\bar{X}$ for some finite extension $k'/k$. Campana's classification program conjectures that $X$ is potentially dense if and only $X$ is special (i.e. has trivial core), so in this case, since it seems that a finite étale cover remains special if the base is special, we'd have that the arithmetic of special varieties is preserved under étale covers. But does potential density pull back under finite étale covers?

Related bibliography and any comments on the aforementioned or related topics mentioned are welcome!

## 2 Answers

Not a full answer but a collection of comments (Edit: Should be a full answer now, see edit at bottom):

Potential density does pull back under etale covers of projective varieties over number fields. The pull-backs of points of $X(K)$ are defined over $L$, with $L/K$ finite, by the Chevalley-Weil theorem.

If $X$ is a Fano variety (i.e. the negative of the canonical class is ample) then Colliot-Thelene conjectured that the Brauer-Manin obstruction is the only obstruction to the Hasse principle. If $f:Y \to X$ be an etale cover with $X$ Fano, then $Y$ is also Fano ($K_Y = f^*(K_X)$). So you might be able to get something that way by assuming this conjecture.

On the opposite end, if $X$ is general type, then so will be $Y$ and the Lang conjecture predicts that $X,Y$ will have few rational points and you probably can find some twist of $Y$ (if the cover is Galois) with no points.

In the intermediate case, your elliptic curve example generalizes to abelian varieties.

Edit: If the cover is etale and Galois then there should be a twist with no points, regardless of the geometry. Chevalley-Weil implies that the set of twists with points is finite, whereas the set of all twists is infinite. E.g. in the case of elliptic curves, never mind the Tate-Shafarevich group, choose an element of the Weil-Chatelet group with no local points.

• Is it not the case that every fano variety is simply connected so any étale cover is trivial? – Daniel Loughran Feb 2 '11 at 22:41
• @Daniel. You might be right, I don't know. – Felipe Voloch Feb 3 '11 at 0:06
• I was not aware of the Chevalley-Weil theorem, thank you! Though Fano varieties are rationally connected and so are simply connected, making étale covers uninteresting. As Campana's work suggests (as does your comment about Lang's conjecture) one would indeed expect interesting examples to lie in "middle" Kodaira dimensions. – Frank Feb 3 '11 at 0:06
• As a reference for the simple connectedness of rationally connected varieties, it goes back to Campana and probably Kollar-Miyaoka-Mori. It's proven for example in Debarre's book 'Higher dimensional algebraic geometry'. – Frank Feb 3 '11 at 0:09

For finite étale morphisms I found the following result in a survey article of Graber, Harris, Mazur and Starr:

Call a flat morphism $\pi:X\to B$ of schemes over a number field $F$ with $B$ smooth and irreducible arithmetically surjective if for every finite degree extension $L/F$, the induced mapping of rational points $X(L)\to B(L)$ is surjective.

$\textbf{Proposition:}$ Let $B$ be an integral, integrally closed scheme over a number field $F$. A finite flat morphism $\pi:X\to B$ defined over $F$ is arithmetically surjective if and only if it admits a section over $F$.