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All varieties are over algebraically closed field of characteristic zero. Let $S$ be a smooth projective surface. Let $D$ be an irreducible divisor on $S$, $H$ be another divisor and $Z\subset S$ be a closed subset.

Assume I know that $D\cap H=\varnothing$. Does exist $D'$ rationally equivalent to $D$, such that $D'$ intersects $Z$ properly and $D'\cap H=\varnothing$?

If possible the proof should be elementary and use something like Bertini theorem.

If $H$ is an exceptional divisor of some blow-up than this statement follows from the classical Chow moving lemma.

In the case when $S$ is fibered over some curve and $D,H$ are two fibers, the statement holds as well.

But I don't know what to do in general case.

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  • $\begingroup$ That is not true. The divisor $D$ could be rigid. Also, even when $S$ is fibered over a curve, fibers need not be rationally equivalent (it depends on whether or not the target of the morphism is a rational curve). $\endgroup$
    – Jason Starr
    Commented yesterday
  • $\begingroup$ Can you explain what is a counterexample? Or where can I read about it? $\endgroup$
    – Galois group
    Commented yesterday
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    $\begingroup$ Let $S$ be the blowing up of the projective plane at one point. Let $D$ be the exceptional divisor. Let $H$ be the inverse image of a line that does not contain the point. Let $Z$ be $D$. $\endgroup$
    – Jason Starr
    Commented yesterday
  • $\begingroup$ But if I know that $H$ is contracted (there is a map $\pi$ from $S$ to may be singular $S'$, such that $\pi^{-1}(p)=H$ and $\pi$ induces an isomorphism $S\backslash H\to S'\backslash p$) than this statement is true, yes? $\endgroup$
    – Galois group
    Commented 9 hours ago

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