Let $f:X \rightarrow Y$ be a finite surjective map between quasi-projective varieties. Let L be a line bundle on Y. Suppose $f^*L$ is ample on X. Is it true that L is ample on Y?
How about converse?
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2$\begingroup$ For projective varieties, it's true that $L$ is ample if and only if $f^*L$ is ample (Hartshorne Exercise III.5.7). So is you question really about what happens for quasi-projective varieties? $\endgroup$– Joe SilvermanCommented Aug 2, 2017 at 0:29
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$\begingroup$ does that follow from Hartshorne? $\endgroup$– user111251Commented Aug 2, 2017 at 6:21
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$\begingroup$ The converse holds (since $O_X$ is $Y$-ample: see EGA II 4.6.13(ii) and 5.1.6(a)(c')). If $f$ is flat or if $Y$ is normal then the equivalence is EGA II 6.6.3 (where "$g:X' \to X$" corresponds to your $f$ and where "$X\to Y$" corresponds to your map "$Y\to {\rm{Spec}}(k)$" for the implicit ground field $k$; see the end of 6.5.1 for meaning of "(II bis)" there). The point is making useful sense of "norm" of line bundles through $f$ (possible in such cases), and relating "norm" of $f^*(L)$ to a power of $L$. Passing to reduced schemes is harmless (EGA II, 4.5.14), but $f$ the normalization...? $\endgroup$– nfdc23Commented Aug 2, 2017 at 10:05
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$\begingroup$ yes f is normalisation $\endgroup$– user111251Commented Aug 2, 2017 at 12:15
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1$\begingroup$ If you're mainly interested in the case of a normalization map then please say so in the question! $\endgroup$– nfdc23Commented Aug 3, 2017 at 1:37
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