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Jun 7, 2022 at 11:40 vote accept Maqui
Jun 7, 2022 at 10:02 comment added Ennio Mori cone You can also see it because if $\phi_L$ maps to something lower-dimensional, then $0=f^*(O(1))^n$, so $L^n=0$. However, since $L$ is globally generated (hence nef), it is big if and only if $L^n>0$.
Jun 6, 2022 at 19:26 answer added Karl Schwede timeline score: 1
Jun 5, 2022 at 20:37 comment added Maqui @KarlSchwede Thanks! if you post this as an answer I'd be happy to accept it.
Jun 5, 2022 at 3:38 comment added Karl Schwede I don't see it in Positivity, but one way to see it is this, let $Z = \phi_L(X)$, let $\eta$ be the generic point of $Z$ and let $f : X_{\eta} \to \eta$ be the base change. Note $L|_{X_{\eta}}$ is still big and globally generated. Furthermore $X_{\eta}$ is a projective variety over the field $k(\eta)$ and $L|_{X_{\eta}} = \phi_L^* O(1)|_{X_{\eta}} = f^* k(\eta) = O_{X_{\eta}}$. Now the only time the structure sheaf of a projective variety can be big is if the variety is 0-dimensional, so $X_{\eta}$ is zero dimensional, thus $X \to Z$ is generically finite.
Jun 5, 2022 at 1:28 comment added Maqui @JasonStarr Thanks for the prompt reply! Do you happen to know a reference?
Jun 5, 2022 at 1:25 comment added Jason Starr Welcome new contributor. Yes, the morphism $\phi_L$ is generically finite to its image.
S Jun 5, 2022 at 1:11 review First questions
Jun 5, 2022 at 2:03
S Jun 5, 2022 at 1:11 history asked Maqui CC BY-SA 4.0