0
$\begingroup$

Let $X$ and $Y$ be normal projective varieties over $\mathbb{C}$ of dimension $n$. Let $f: X \rightarrow Y$ be a finite morphism. Also, let $H$ be a very ample divisor on $Y$.

Is there a constant $C=C(\mathrm{deg}(f),n)$ depending just on degree and dimension such that $Cf^*H$ is very ample? If the answer is in general no, are there natural conditions to impose?

$\endgroup$
7
  • 4
    $\begingroup$ What do you fix?? If you want a universal $C$ depending only on $\deg(f)$, this doesn't exist already for $\deg(f)=2$. $\endgroup$
    – abx
    Commented Jul 16, 2018 at 16:09
  • $\begingroup$ @abx I am fixing the degree and the dimension. I forgot to include the dimension in the statement. Is it still not enough? $\endgroup$
    – Stefano
    Commented Jul 16, 2018 at 16:12
  • 4
    $\begingroup$ No, it is not enough — already with degree 2 and dimension 1. $\endgroup$
    – abx
    Commented Jul 16, 2018 at 16:13
  • 3
    $\begingroup$ Well, some kind of bound on the ramification would eliminate the counter-examples that I have in mind. But I have no idea of what would be a reasonable formulation. $\endgroup$
    – abx
    Commented Jul 16, 2018 at 16:24
  • 2
    $\begingroup$ Since $f^*H$ is ample and gg, then $K_X+(dim X+1)f^*H $ is gg (and something similar will be very ample)...But to answer your question you would need to control when $cf^*H-K_X$ is nef (in abx's example, if $f_*O_X=O_Y\oplus (-tH)$, and say $K_Y=0$, then $c\geq t$ can be arbitrarily big). Maybe one can show something similar with $K_{X/Y}$ instead of $K_X$, but I doubt you can do any better. $\endgroup$
    – Hacon
    Commented Jul 16, 2018 at 19:17

0

You must log in to answer this question.