# Mori cones and projective morphisms

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
Commented Apr 30, 2023 at 1:37
• @naf why not post it as an answer? Commented Apr 30, 2023 at 14:58

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 $$\overline{NE(Y)}$$.

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. Commented Apr 30, 2023 at 14:48