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?

  • With your definition, a line bundle is ample if and only if it is when you pull it back under a finite map. – Mohan Sep 14 at 13:19
  • Could you please explain why? – jacob Sep 14 at 13:20
  • 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. – Mohan Sep 14 at 13:24
  • 1
    @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). – R. van Dobben de Bruyn Sep 14 at 14:21
  • @R.vanDobbendeBruyn You are right, it is a mistake. – Mohan Sep 14 at 15:25
up vote 6 down vote accepted

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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  • Thanks! Thats a beautiful example. – jacob Sep 16 at 5:50

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