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Suppose that $p:Y\rightarrow X$ is a surjective etale morphism of affine varieties (both over an algebraically closed field of characteristic zero), and let $x \in X$.

Then is there a subvariety $V$ of $Y$ containing $y \in p^{-1}(x)$, such that $V$ isomorphic to $X$?

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  • $\begingroup$ I suppose you mean to either have $p$ surjective, or "locally isomorphic" instead of "isomorphic". $\endgroup$
    – Zhen Lin
    May 21, 2013 at 17:49
  • $\begingroup$ @Zhen Lin: yes, thanks, I've edited the question. $\endgroup$
    – Bob G
    May 21, 2013 at 18:11

1 Answer 1

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Let $C \subset \mathbb{A}^2$ be the elliptic curve $\{ y^2 = x^3 - x \}$, let $Y = C \setminus \{ (-1, 0), (0, 0), (1, 0) \}$, and let $X = \mathbb{A}^1 \setminus \{ -1, 0, 1 \}$. It is not hard to see that $X$ and $Y$ are affine varieties: in fact, $X$ corresponds to the $k$-algebra $A = k [t, (t-1)^{-1}, t^{-1}, (t+1)^{-1}]$, and $Y$ corresponds to the $k$-algebra $B = k [x, y, y^{-1}] / (y^2 - x^3 + x)$. Consider the morphism $p : Y \to X$ defined by $(x, y) \mapsto x$. This corresponds to the unique $k$-algebra homomorphism $p^* : A \to B$ where: \begin{align} p^* (t) & = x & p^* ((t-1)^{-1}) & = \frac{x^2 + x}{y^2} \\ p^* (t^{-1}) & = \frac{x^2 - 1}{y^2} & p^* ((t+1)^{-1}) & = \frac{x^2 - x}{y^2} \end{align} It is also easy to check that $p : Y \to X$ is surjective, so $p^* : A \to B$ is injective. By considering what is happening in the projective closure, we see that $p : Y \to X$ is a finite étale morphism. But there is no open subset of $Y$ that is isomorphic to $X$, because an elliptic curve has genus 1 and $\mathbb{P}^1$ has genus 0, and the genus of a smooth projective curve can be computed from any non-empty open subset (via the function field).

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  • $\begingroup$ Well, the closed subvarieties are either the whole variety or finite sets of points, so those won't do either. And I don't think you really want to consider subvarieties that are neither open nor closed... $\endgroup$
    – Zhen Lin
    May 21, 2013 at 20:20

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