The Stacks project

Proof. Assume $\mathcal{L}$ is $f$-ample and $\mathcal{M}$ ample. By assumption $Y$ and $f$ are quasi-compact (see Definition 29.37.1 and Properties, Definition 28.26.1). Hence $X$ is quasi-compact. By Properties, Lemma 28.26.8 the scheme $Y$ is separated and by Lemma 29.37.3 the morphism $f$ is separated. Hence $X$ is separated by Schemes, Lemma 26.21.12. Pick $x \in X$. We can choose $m \geq 1$ and $t \in \Gamma (Y, \mathcal{M}^{\otimes m})$ such that $Y_ t$ is affine and $f(x) \in Y_ t$. Since $\mathcal{L}$ restricts to an ample invertible sheaf on $f^{-1}(Y_ t) = X_{f^*t}$ we can choose $n \geq 1$ and $s \in \Gamma (X_{f^*t}, \mathcal{L}^{\otimes n})$ with $x \in (X_{f^*t})_ s$ with $(X_{f^*t})_ s$ affine. By Properties, Lemma 28.17.2 part (2) whose assumptions are satisfied by the above, there exists an integer $e \geq 1$ and a section $s' \in \Gamma (X, \mathcal{L}^{\otimes n} \otimes f^*\mathcal{M}^{\otimes em})$ which restricts to $s(f^*t)^ e$ on $X_{f^*t}$. For any $b > 0$ consider the section $s'' = s'(f^*t)^ b$ of $\mathcal{L}^{\otimes n} \otimes f^*\mathcal{M}^{\otimes (e + b)m}$. Then $X_{s''} = (X_{f^*t})_ s$ is an affine open of $X$ containing $x$. Picking $b$ such that $n$ divides $e + b$ we see $\mathcal{L}^{\otimes n} \otimes f^*\mathcal{M}^{\otimes (e + b)m}$ is the $n$th power of $\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}$ for some $a$ and we can get any $a$ divisible by $m$ and big enough. Since $X$ is quasi-compact a finite number of these affine opens cover $X$. We conclude that for some $a$ sufficiently divisible and large enough the invertible sheaf $\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}$ is ample on $X$. On the other hand, we know that $\mathcal{M}^{\otimes c}$ (and hence its pullback to $X$) is globally generated for all $c \gg 0$ by Properties, Proposition 28.26.13. Thus $\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a + c}$ is ample (Properties, Lemma 28.26.5) for $c \gg 0$ and (1) is proved.

Part (2) follows from Lemma 29.37.6, Properties, Lemma 28.26.2, and part (1). $\square$