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## (Very) Ampleness And Numerical Equivalence

Algebraic Geometry
Tsakanikas Nickos
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### (Very) Ampleness And Numerical Equivalence

Let $X$ be a non-singular projective surface over an algebraically closed field $\mathbb{K}$. If $D$ is an ample divisor on $X$ and $D \equiv D^{\prime}$ (numerically equivalent divisors), show that $D^{\prime}$ is ample too. Furthermore, give an example to show that the same statement does not hold if "ample" is replaced by "very ample".