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In the paper http://arxiv.org/pdf/1207.5011.pdf of Chi Li and Song Sun, they say that "$D$ is a smooth divisor which is $\mathbb{Q}$-linearly equivalent to $−\lambda K_X$ for some $λ \in \mathbb{Q}$", and $X$ is a Fano manifold. I do not understand what is the meaning of "$\mathbb{Q}$-linearly equivalent".

In my opinion, if $L$ is linear equivalent to $-\frac{m_1}{m_2}K_X$, then we mean that $m_2 L$ is linear equivalent to $-m_1K_X$, for $m_1,m_2\in\mathbb{Z}^+$ and $m_1,m_2$ are prime with each other. Another question is whether $m_2$ here can be taken along all $\mathbb{Z}^+$? I wonder if $m_2$ is too large, whether there will be such line bundle $L$.

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    $\begingroup$ Your first "opinion" is correct. I do not understand the question that follows : you can always multiply both $m_1$ and $m_2$ by the same integer. $\endgroup$
    – abx
    Commented Nov 21, 2014 at 7:17
  • $\begingroup$ I have added that $\gcd (m_1,m_2)=1$. $\endgroup$
    – Daniel
    Commented Nov 21, 2014 at 12:50
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    $\begingroup$ Then $m_2$ is bounded by $\dim(X)+1$, and any value between $1$ and $\dim(X)+1$ is possible. $\endgroup$
    – abx
    Commented Nov 21, 2014 at 13:56
  • $\begingroup$ Why require that $m_1$ and $m_2$ are relatively prime, let me not limit myself to $-K_X$ also. You can have a Cartier divisor that is torsion in the Picard group but not actually linearly equivalent to zero. Then it is $\mathbb{Q}$-linearly equivalent to zero but not itself linearly equivalent to zero. $$\text{ }$$ Definition: on a normal variety, two $\mathbb{Q}$-divisors $F,G$ are $\mathbb{Q}$-linearly equivalent if $mF \sim mG$ for some integer $m > 0$ so that $mF$ and $mG$ are honest Weil divisors. $\endgroup$
    – Karl Schwede
    Commented Nov 22, 2014 at 1:07
  • $\begingroup$ I voted to leave this open because while the phrasing is confusing, this is exactly the sort of question a student might ask when they are starting research. In particular, it isn't really a question that would be asked outside of a research environment. I could also imagine some nice answers with nice examples. $\endgroup$
    – Karl Schwede
    Commented Nov 22, 2014 at 1:11

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