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Is it possible to find an example of an $\mathbb{R}$-Cartier divisor $D$ on an irreducible variety $X$ that is non-trivial, nef, effective and numerically rigid?

By "numerically rigid" I mean that if $E$ is another $\mathbb{R}$-Cartier effective divisor such that $E$ is numerically equivalent to $D$ then $D=E$.

For curves this clearly cannot be the case, since an effective non-trivial divisor is always ample.

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Blow up $P^2$ at nine very general points and choose $E$ to be the anticanonical divisor. – damiano Nov 16 '10 at 11:32
Is there a difference between 'general' points and 'very general' points? – Ketil Tveiten Nov 16 '10 at 14:07
"General" refers to the complement of a proper closed subset; "very general" refers to the complement of a countable union of proper closed sets. General would suffice for the purpose of the question; using very general points, you can also make sure that every positive multiple of the anticanonical divisor has the required property. – damiano Nov 16 '10 at 15:38
up vote 2 down vote accepted

Take a minimal surface $S$ of general type with $p_g=1$, $q=0$ and zero torsion.

Then $S$ contains a unique effective canonical curve $K$, which is nef and numerically rigid.

In fact, since $q=0$ and there is no torsion, we have $\textrm{Pic}^0(S)=0$, the Neron - Severi group $\textrm{NS}(S)$ coincides with the Picard group $\textrm{Pic}(S)$ and any two numerically equivalent divisors on $S$ are linearly equivalent.

Examples of these surfaces, with $K^2=2$, are given in the paper of Debarre and Catanese

"Surfaces with $K^2=2$, $p_g=1$, $q=0$",

J. reine angew. Math. 395 (1989), 1-55.

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