Mori cones and projective morphisms

Question

Let $f:X\rightarrow Y$ be a morphism of smooth projective varieties, and $NE(X),NE(Y)$ the Mori cones of curves of $X$ and $Y$. Assume that $NE(X)$ is finitely generated. Then is $NE(Y)$ finitely generated as well?

If $f$ is birational I think the answer is positive. Let $C\subset Y$ be an irreducible curve and $\Gamma$ its strict transform in $X$. Then $\Gamma\sim a_1\Gamma_1+\dots + a_r\Gamma_r$ with $a_1,...,a_r\geq 0$ and where $\Gamma_1,\dots,\Gamma_r$ are the generators of $NE(X)$. So $C\sim a_1f_{*}\Gamma_1 + \dots + a_rf_{*}\Gamma_r$ and hence $NE(Y)$ is generated by $f_{*}\Gamma_1,\dots,f_{*}\Gamma_r$. Is this argument correct?

If $f$ is not birational could it happen that $NE(Y)$ is not finitely generated even if $NE(X)$ is?

Thank you very much.

It suffices to assume that $f$ is dominant. Then given an irreducible curve $C \subset Y$ there always exists an irreducible curve $D \subset X$ such that $f(D) = C$ so $f_*$ induces a surjection $NE(X) \to NE(Y)$. — naf
– naf, Commented Apr 30, 2023 at 1:37

naf · Accepted Answer · 2023-05-01 04:09:42Z

It suffices to assume that $f$ is surjective (equivalently, dominant). Then for $C \subset Y$ any irreducible curve there exists an irreducible curve $D \subset X$ such that $f(D) = C$. (A schemy proof: let $D$ be the closure in $X$ of any closed point in $f^{-1}(c)$, where $c$ is the generic point of $C$.) It follows that $f$ induces a surjection from $NE(X)$ to $NE(Y)$, so if $NE(X)$ is finitely generated then so is $NE(Y)$.

The same argument also implies that if $\overline{NE(X)}$ is finitely generated then so is $\overline{NE(Y)}$.

R. van Dobben de Bruyn · Accepted Answer · 2023-04-29 23:42:15Z

The argument is correct, and in fact can be easily extended to the case where $f$ is generically finite. Indeed, we have $$f_*([X]) = (\deg f) [Y],$$ so the projection formula [Fulton, Prop. 8.3(c)] or [Stacks, Tag 0B2X(3)] gives $$f_*f^*([Z]) = (\deg f)[Z] \in A^*(Y)$$ for any subvariety $Z \subseteq Y$. Moreover, $f_*$ preserves numerical equivalence [Fulton, Ex. 19.1.6], so if $f^*C \equiv a_1\Gamma_1 + \ldots + a_n\Gamma_n$ with $a_1,\ldots,a_n \in \mathbf R_{>0}$, then $$C = \tfrac{1}{\deg f}f_*f^*C \equiv \tfrac{a_1}{\deg f}f_*\Gamma_1 + \ldots + \tfrac{a_n}{\deg f}f_*\Gamma_n.$$ So indeed we see that $f_*\Gamma_1,\ldots,f_*\Gamma_n$ generate $\operatorname{NS}(Y)$.

Note that it doesn't matter that we use the full pullback $f^*C$ instead of the strict transform, since their difference is contracted by $f$ (hence vanishes under $f_*$).

Remark. The result is false if you don't make any assumption on $f$. For instance, let $Y$ be any variety for which $\operatorname{NE}(Y)$ is not finitely generated, and let $X \subseteq Y$ be any smooth curve. If you think this example is too trivial (as $\operatorname{NE}(X)$ is not very interesting when $X$ is a curve), you can look instead at $X \times \mathbf P^1 \to Y \times \mathbf P^1$.

References.

[Fulton] W. Fulton, Intersection theory, second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 2. Springer, 1998. ZBL0885.14002.

[Stacks] The stacks project.

Thanks a lot. Yes, I am assuming the morphism to be dominant. I think that @naf is right, it is enough to assume that $f$ is dominant. — Puzzled
– Puzzled, Commented Apr 30, 2023 at 14:48

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