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First, let me emphasize that for $X$ a not-necessarily proper variety, we say that a line bundle $L$ on $X$ is ample, if for some positive integer $n$, $L^{\otimes n}$ arises as $j^*O(1)$ for some (not-necessarily closed) immersion $j:X\rightarrow \mathbb{P}^n$.

Now, let $X$ be a variety and $L$ a line bundle on $X$, and let
$f:X^{nor}\rightarrow X$ be the normalization map. Suppose that $f^*L$ is ample. Is it true that $L$ is ample?

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  • $\begingroup$ With your definition, a line bundle is ample if and only if it is when you pull it back under a finite map. $\endgroup$
    – Mohan
    Commented Sep 14, 2018 at 13:19
  • $\begingroup$ Could you please explain why? $\endgroup$
    – jacob
    Commented Sep 14, 2018 at 13:20
  • $\begingroup$ The best proof (not necessarily the easiest) would be to show that with your definition, a line bundle $L$ is ample if and only for any coherent sheaf $F$, $H^i(F\otimes L^n)=0$ for all large $n$ and all $i>0$. This property transfers nicely under finite maps. $\endgroup$
    – Mohan
    Commented Sep 14, 2018 at 13:24
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    $\begingroup$ @Mohan: your criterion is not true for non-proper varieties. For example, consider $\mathscr L = \mathcal O_X$ on $X = \mathbb A^2 \setminus \{0\}$, which is ample by Tag 01QE (and take $\mathscr F = \mathcal O_X$ as well). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 14, 2018 at 14:21
  • $\begingroup$ @R.vanDobbendeBruyn You are right, it is a mistake. $\endgroup$
    – Mohan
    Commented Sep 14, 2018 at 15:25

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There is a non-quasi-affine variety $X$ with quasi-affine normalization. See Tag 0271. Then $\mathcal{O}_X$ is a counter example.

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