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Suppose $f: Y \to Z$ is a projective morphism of smooth varieties with connected fibers. If an effective divisor $H$ on $Y$ is relatively ample over $Z$, and $\dim Y >\dim Z$, is $h^0(Y, mH)>1$ for some $m\in \mathbb N$?

I'm interested in the following special case:

Suppose $g: X \to Z$ is a projective morphism, and $D$ is an effective divisor on $X$ such that $h^0(X, mD)=1$ for any $m\in \mathbb N$. Suppose we can run a $D$-MMP over $Z$ and get $X' \to Z$ such that the push-forward of $D$, $D'$, is semi-ample for $X' \to Z$. Then $D'$ defines a fibration $h: X' \to Y$ over $Z$. Let $H$ on $Y$ such that $h^*H=D'$, then $H$ is relatively ample over $Z$. The question is equivalent to ask can $\dim Y > \dim Z$?

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    $\begingroup$ Unfortunately that is not true. Let $Z$ be a smooth, projective, genus $0$ curve. Let $f:Y\to Z$ be a Hirzebruch surface fibered over $Z$ whose "directrix" $H$ has negative self-intersection, i.e., any Hirzebruch surface other than $\mathbb{P}^1\times \mathbb{P}^1$. The divisor class of $H$ is $f$-relatively ample, yet the only global sections of $\mathcal{O}_Y(m\underline{H})$ are scalar multiples of the section whose zero scheme is $m\underline{H}$. $\endgroup$
    – Jason Starr
    Commented Sep 5, 2017 at 3:21
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    $\begingroup$ My comment add almost nothing to Jason's one, but indeed more generally, take any vector bundle $E\to Z$ or rank greater than one such that its symmetric powers (including the vector bundle itself) have no global sections. Now take $Y=\mathbb P(E)$ its projectivization. Then, the Serre (anti)tautological line bundle $\mathcal O_E(1)\to Y$ is relatively ample by construction. But $H^0(\mathbb P(E),\mathcal O_E(m))\simeq H^0(Z,S^m E)=\{0\}$. $\endgroup$
    – diverietti
    Commented Sep 5, 2017 at 13:05

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