Let $X$ be a smooth projective surface. Let $C\subset X$ is a smooth and irreducible curve. By the Kawamata covering lemma, for any natural $m$ there exists a smooth projective surface $Y$ and a finite map $p:Y\rightarrow X$ together with a smooth divisor $D\subset Y$ such that $p^*C= mD$. Now will $D$ be irreducible. Will it be a disjoint union of smooth curves, if so are they isomorphic? Can choose $Y$ such that $D$ is irreducible?

  • $\begingroup$ Where did you learn the Kawamata covering lemma? For example, Prop. 4.1.12 in Lazarsfeld's book Positivity in Algebraic Geometry I says that $D$ will be smooth and irreducible. $\endgroup$ – Takumi Murayama May 17 '17 at 17:58
  • $\begingroup$ @TakumiMurayama, I studied it from the same source. It says $D$ is just smooth right. In principle, it could mean $D$ can have disjoint irreducible components, isn't it? $\endgroup$ – user349424 May 17 '17 at 18:36
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    $\begingroup$ The proof goes by taking first a Bloch–Gieseker cover $X_1 \to X$ to get a new divisor $D^+$ which is smooth and irreducible if $C$ is (Thm. 4.1.10). Then, you take a cyclic covering $Y \to X_1$, which maps $D$ isomorphically onto $D^+$ (Prop. 4.1.6). So I think $D$ ends up being smooth and irreducible. $\endgroup$ – Takumi Murayama May 17 '17 at 21:24
  • $\begingroup$ @TakumiMurayama, thanks for the clarification. Are there other versions of this lemma, where the divisor $D$ is reducible? $\endgroup$ – user349424 May 20 '17 at 21:10
  • $\begingroup$ @TakumiMurayama, thanks for the help with lazarsfeld's proof. I have one question, since $D$ is Smooth and irreducible, consider the map $D\rightarrow C$. What will the degree of the map be? Is it an isomorphism?? $\endgroup$ – user349424 Jun 10 '17 at 7:03

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