Lemma 29.37.10. Let $g : Y \to S$ and $f : X \to Y$ be morphisms of schemes. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. If $\mathcal{L}$ is $g \circ f$-ample and $f$ is quasi-compact1 then $\mathcal{L}$ is $f$-ample.

Proof. Assume $f$ is quasi-compact and $\mathcal{L}$ is $g \circ f$-ample. Let $U \subset S$ be an affine open and let $V \subset Y$ be an affine open with $g(V) \subset U$. Then $\mathcal{L}|_{(g \circ f)^{-1}(U)}$ is ample on $(g \circ f)^{-1}(U)$ by assumption. Since $f^{-1}(V) \subset (g \circ f)^{-1}(U)$ we see that $\mathcal{L}|_{f^{-1}(V)}$ is ample on $f^{-1}(V)$ by Properties, Lemma 28.26.14. Namely, $f^{-1}(V) \to (g \circ f)^{-1}(U)$ is a quasi-compact open immersion by Schemes, Lemma 26.21.14 as $(g \circ f)^{-1}(U)$ is separated (Properties, Lemma 28.26.8) and $f^{-1}(V)$ is quasi-compact (as $f$ is quasi-compact). Thus we conclude that $\mathcal{L}$ is $f$-ample by Lemma 29.37.4. $\square$

[1] This follows if $g$ is quasi-separated by Schemes, Lemma 26.21.14.

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