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Given a semiample divisor $D$ (that is, the induced morphism $\phi_{|mD|}$ for $m\gg 0$ is base point free) on a smooth projective variety $X$, it is easy to prove (using projection formula) that $D$ is nef, that is $D\cdot C\geq 0$ for any irreducible curve $C\subset X$.

Therefore, I asked my self why the converse does not hold, that is I'd like to find an example of nef divisor which is not semiample. Unfortunately since I'm quite at the beginning I don't know how to proceed, and any hint or reference would be much appreciated, thanks in advance!

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An example is given in Example 2.3.1 of "Positivity in Algebraic Geometry, I" by Robert Lazarsfeld.

It is available free on line at https://cims.nyu.edu/~rodion/lib/R.%20K.%20Lazarsfeld.%20Positivity%20in%20Algebraic%20Geometry,%20I.%20Classical%20Setting:%20Line%20Bundles%20and%20Linear%20Series%20-%202003.pdf.

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    $\begingroup$ It's a good idea to include some information past just the link - links can and do disappear. $\endgroup$
    – KReiser
    May 12, 2021 at 22:03
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Nick L's answer links to one kind of example, where the divisor is nef and big, but not semiample.

If you just ask for nef but not semiample, examples are maybe easier to come by.

One kind of behaviour is given in Example 1.5.1 of the Lazarsfeld book that Nick L linked to. This example is a ruled surface $S$ with a line bundle $L$ which is nef, but such that for every natural number $m$, the bundle $L^m$ has no global sections. So $L$ is not semiample.

If you want an example where the divisor is nef and effective but not semiample, fix a smooth cubic curve $C'$ in $\mathbf P^2$, take $S$ to be the blowup of $\mathbf P^2$ in 9 very general points, and take $C$ to be the proper transform of $C'$. Then $C$ is an irreducible curve with selfintersection $C^2=0$, so it is nef. But by choosing the points appropriately, you can arrange that $O_C(C)$ is any given line bundle $L$ of degree $0$ on $C$. In particular if we choose $L$ to be a non-torsion line bundle of degree 0, this implies that $O_C(mC)$ is nontrivial for all natural numbers $m$, and therefore that $mC$ is not basepoint-free for any $m$.

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