1
$\begingroup$

Let $X$ be a non-singular projective variety over the field of complex numbers, of dimension $\geq 2$. Suppose $D$ is a non-singular and irreducible divisor of $X$.

The Kawamata covering lemma (Proposition 4.1.12, Positivity in Algebraic Geometry) says that - for any $m>0$, there is a smooth variety $Y$ and a finite flat morphism $f:Y\rightarrow X$ such that $f^*D=mD'$ for some smooth divisor $D'$ in $Y$.

(a) Looking at the proof, it seems to me that the divisor $D'$ is irreducible as well. Is this right?

(b) Is $f$ a Galois morphism? I think it is, since it is constructed from the Bloch-Gieseker covering and cyclic covering.

(c) Consider the morphism between non-singular irreducible varieties $f|_{D'}:D'\rightarrow D$. Can we say what the degree of this map is? Is it possible to obtain a covering $f$ so that $f|_{D'}$ is of some prescribed degree?

$\endgroup$
10
  • $\begingroup$ If you take ramification locus in $X$ is nc , then there is such $f$ I think $\endgroup$
    – user21574
    Jun 10, 2017 at 16:54
  • $\begingroup$ @HassanJolany, if $D$ is irreducible and if $f^*D =mD'$, then $D'$ is smooth by the statement of Kawamata covering lemma. Hence $D'$ cannot be a normal crossing divisor right. It is either irreducible and smooth, or it is a disjoint union of smooth irreducible components right? $\endgroup$
    – user349424
    Jun 10, 2017 at 17:13
  • $\begingroup$ Ahh, I had assumed $D$ is snc $\endgroup$
    – user21574
    Jun 10, 2017 at 17:22
  • $\begingroup$ @HassanJolany, so in the case when $D$ is irreducible, $f^*D_{red}$ will be irreducible? $\endgroup$
    – user349424
    Jun 10, 2017 at 17:31
  • $\begingroup$ I think it can be reduced $\endgroup$
    – user21574
    Jun 10, 2017 at 18:24

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.