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Let $X$ be a projective variety.

Is it true that if the canonical divisor $K_X$ is not $\mathbb Q$-Cartier then it is not nef?

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    $\begingroup$ "Nef" is a property that is only defined for $\mathbf Q$-Cartier divisors. $\endgroup$ – user64687 Oct 21 '14 at 14:56
  • $\begingroup$ Can you explain more, please. Because I don't see it in the definition of Q-Cartier $\endgroup$ – user61135 Oct 21 '14 at 15:01
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    $\begingroup$ Sure. It is in the definition of "nef", not of "$\mathbf Q$-Cartier". A Cartier or $\mathbf Q$-Cartier divisor $D$ on a projective variety has well-defined interesection numbers $D \cdot C$ with curves $C$. If all those intersection numbers are $\geq 0$, we say $D$ is nef. But if $D$ is not $\mathbf Q$-Cartier, there is no way in general to define $D \cdot C$, and so it is not meaningful to ask if $D \cdot C \geq 0$ for all curves $C$. $\endgroup$ – user64687 Oct 21 '14 at 15:03
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A Cartier or $\mathbf Q$-Cartier divisor $D$ on a projective variety has well-defined interesection numbers $D \cdot C$ with curves $C$. If all those intersection numbers are $\geq 0$, we say $D$ is nef. But if $D$ is not $\mathbf Q$-Cartier, there is no way in general to define $D⋅C$, and so it is not meaningful to ask if $D⋅C \geq 0$ for all curves C.

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