Let $X$ be a smooth projective variety and $A$ an ample $\mathbb{Q}$-divisor on $X$.
Is the round up $\ulcorner A\urcorner$ of $A$ ample?
I think it's true. But I do not know how to arrange an argument.
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$\begingroup$ It's definitely not true, but if the difference between $\langle A \rangle$ and $A$ has nice singularities you can still get some vanishing theorems. See for instance things like Lazarsfeld's positivity in algebraic geometry. $\endgroup$– Karl SchwedeCommented Apr 5, 2015 at 18:31
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This doesn't seem right. Let $\pi : X \to \mathbb P^2$ be the blow-up of $\mathbb P^2$ at a point, and let $A = L + 2/3 E$, where $E$ is the exceptional divisor, and $L$ is the strict transform of a line through the point. Take $H$ to be the pullback of a line in $\mathbb P^2$. Then $A$ is linearly equivalent to $(H-E)+2/3 E = H - 1/3 E$, which is ample. But the round-up of $A$ is $L+E$, linearly equivalent to $H$, which isn't ample.
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$\begingroup$ Hi, sorry can you explain why $H-1/3E$ is ample but $H$ is not? Thanks! $\endgroup$– fwgCommented Apr 25, 2016 at 14:09