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Let $X$, $Y$ be proper schemes (can assume over a field) and let $L$ be a line bundle on $X$. Assume I have a finite morphism $q:Y\rightarrow X$ and that I know that $q^{*}L$ is very ample. Is it also true that $L$ is very ample? if not do you have a counterexample

user65187
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    It is not true with very ample. If $f:C\to E$ is a morphism where $C$ is a genus 2 curve and $E$ is an elliptic curve, then $\deg f^(0)=2\deg f$, and so if $\deg f\geq3$, we have that $f^(0)$ is very ample but $0$ is not ample on $E$. Perhaps you could ask the question for ample instead of very ample? – rfauffar Aug 20 '13 at 12:39
  • Dear @Robert, I think you mean in your nice comment that $\mathcal O(1.O)$ is not very ample instead of "is not ample". – Georges Elencwajg Aug 20 '13 at 14:44
  • Yes you are right! Sorry about that! Unfortunately I can't edit it! – rfauffar Aug 20 '13 at 15:01
  • Dear Robert @RobertAuffarth, I think you meant $\deg f^*(0)=\deg f$. So we need $\deg f\ge 5$. Anyway, using an isogeny of degree $\ge 3$ on $E$ gives a similar counterexample. – Cantlog Sep 01 '13 at 21:15
  • Yes of course! Sorry about all the mistakes, below I will post what I really meant. – rfauffar Sep 01 '13 at 22:47
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    It is not true with very ample. If $f:C\to E$ is a morphism where $C$ is a genus 2 curve and $E$ is an elliptic curve, then $\deg f^(0)=\deg f$, and so if $\deg f\geq 5$ we have that $f^(0)$ is very ample but 0 is not very ample on $E$. – rfauffar Sep 01 '13 at 22:49

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