(Very) Ampleness And Numerical Equivalence

Algebraic Geometry
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Tsakanikas Nickos
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(Very) Ampleness And Numerical Equivalence

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Post by Tsakanikas Nickos »

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".
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