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Let $f: X \to Y$ be a finite, surjective morphism of smooth, projective, irreducible varieties over $\mathbb{C}$ and let $y \in Y$.

  1. Can I find a smooth curve $C \subseteq Y$ with $y \in C$ such that $f^{-1}(C)$ is again a smooth curve?

  2. If not, can I find a curve $C \subseteq Y$ such that $y$ is a smooth point of $C$ and such that every $x \in f^{-1}(\{y\})$ is a smooth point of $f^{-1}(C)$?

  3. Can I find a curve $C \subseteq Y$ with the property of 1. or 2. with the additional property that $f$ is unramified along a Zariski dense subset of $f^{-1}(C)$?

Edit: As Yusuf recommended, I'd like to point out that I am especially interested in the case where $y$ lies in the branch locus. Also $y$ is not necessarily a smooth point of the underlying reduced scheme of the branch divisor.

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  • $\begingroup$ Is $y \in Y$ allowed to be in the branch divisor $B$ of $f$? If not, 1 and 3 can be settled by taking $C$ to be a complete intersection of sufficiently ample divisors on $Y$ such that $C$ contains $y$ and intersects $B$ transversally (this follows from Bertini). $\endgroup$
    – Yusuf Mustopa
    Commented Aug 13, 2014 at 11:37
  • $\begingroup$ Thank you. But in fact, the case where $y$ is in the branch locus is the case where I am interested in. $\endgroup$
    – Hans
    Commented Aug 13, 2014 at 11:43
  • $\begingroup$ I figured that this might be the case. You are still okay if $y$ is in the smooth locus of the underlying reduced scheme $B_{red}.$ However, you may want to edit the question to emphasize your interest in the branch point case. $\endgroup$
    – Yusuf Mustopa
    Commented Aug 13, 2014 at 11:45
  • $\begingroup$ Okay thanks again. At the moment, I can't see whether my problem reduces to such a situation. $\endgroup$
    – Hans
    Commented Aug 13, 2014 at 12:11

1 Answer 1

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Consider the map $\mathbb P^2 \to \mathbb P^2$ with equations $(xy,x^2-y^2,z^2)$. Any smooth curve passing through the point $(0,0,1)$ has an equation that looks locally linear, and its inverse image has an equation that looks locally quadratic.

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    $\begingroup$ This is correct. Jouanolou's theorem that I cited [Theoreme 6.3 (2) in Theoremes de Bertini et Applications] implies the smoothness of the inverse images of a Zariski-dense set of hyperplane sections, but the condition of passing through a fixed point is Zariski closed. So this theorem does not apply here. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 13, 2014 at 12:37
  • $\begingroup$ Thank you for this counterexample. Now, I have another question which is related to that. $\endgroup$
    – Hans
    Commented Aug 13, 2014 at 14:57

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