Suppose we have an étale morphism between smooth quasi-projective (complex) varieties $X \rightarrow Y$ and assume that $Y$ is unirational. I am wondering whether we can somehow deduce that $X$ is unirational, or rationally connected, or uniruled, or has negative Kodaira dimension.
And what if $Y$ is just uniruled but non-necessarily unirational?
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$\begingroup$ An étale cover of a rational curve is a disjoint union of rational curves... $\endgroup$– Francesco PolizziMay 27, 2019 at 14:11
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1$\begingroup$ @FrancescoPolizzi What you write is correct for an etale morphism that is also finite, e.g., proper. However, the OP only indicates that the morphism is etale, not etale and finite. $\endgroup$– Jason StarrMay 27, 2019 at 14:17
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$\begingroup$ @JasonStarr: I guessed that finite was implict, but of course you are right... $\endgroup$– Francesco PolizziMay 27, 2019 at 14:18
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7$\begingroup$ Without properness assumptions you cannot deduce anything. For instance take any projective variety $X$ of dimension $n$; a general projection to $Y=\mathbb{P^n}$ is finite. Just take out of $X$ the ramification locus, you get an étale map to $\mathbb{P}^n$. $\endgroup$– abxMay 27, 2019 at 15:34
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1$\begingroup$ "assuming all the other hypotheses": I am lost with which hypotheses you want to make. Anyway there is no reasonable statement of the type you ask for. $\endgroup$– abxMay 28, 2019 at 16:13
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1 Answer
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In the case where $Y$ is unirational and projective, there exists no non-trivial étale cover of $Y$. In fact, the fundamental group of a complex, projective, smooth unirational variety is trivial, see
J. P. Serre: On the fundamental group of a unirational variety, J. Lond. Math. Soc. 34, 481-484 (1959). ZBL0097.36301.