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Suppose $X$ a normal project variety over complex numbers, $D$ is an integral Cartier divisor on $X$, and $L$ is a line bundle on $X$. Suppose one know the natural map $H^0(X, L) \to H^0(D, L|_D)$ is surjective, I want to know if the following statement holds:

For any effective divisor $G_D \in |L|_D|$, is there an effective divisor $G \in |L|$, such that $G|_D = G_D$? (Of course $G, D$ should not have common components in order to make sense of $G|_D$.)

Actually, my feeling is that this holds when there exists a divisor $R \in |L|$ such that $D$ is not a component of $R$, but may not hold in general. Hence I try to find a counterexample by looking at the case when $|L|$ has fixed component, and $D$ is one such fixed component. However, I don't have too much such example on hand.

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Any effective divisor in $|L\vert_D|$ is the zero locus of a global section of $L\vert_D$. If $s_D$ is such a section, by the surjectivity assumption there is a global section $s$ of $L$ on $X$ that restricts to $s_D$. If $G$ is its zero locus then $G\vert_D = G_D$. So, what you ask is a tautology.

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  • $\begingroup$ I don't think this is tautology: suppose $D$ is a fixed component of $|L|$, and $s \in K(X)$ is one section whose image is $s_D$, then simply restrict $s$ to $D$ may not be in $K(S)$ (for example, $s$ has poles containing $D$). That is the case I am worry about. $\endgroup$
    – Li Yutong
    Commented Dec 15, 2016 at 9:09
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    $\begingroup$ If $D$ is a fixed component, then the restriction map is zero. $\endgroup$
    – Sasha
    Commented Dec 15, 2016 at 9:18
  • $\begingroup$ Sorry for keep asking: suppose $L=2D>0$, and $D$ is a fixed component of $|L|$, and then suppose $s\in K(X)$ is a section of $L$, such that $(s) = H - D$ (hence $(s)_0 = H-D + 2D = H+D$). Then how to make sense of $s$ restrict to $D$? I cannot see why it is zero on $D$, actually it has poles on $D$. $\endgroup$
    – Li Yutong
    Commented Dec 15, 2016 at 9:32
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    $\begingroup$ Your $s$ is not a section of a line bundle, it is the ratio of two sections (with the section defining $2D$ in the denominator). It makes no sense to restrict this ratio to $D$. $\endgroup$
    – Sasha
    Commented Dec 15, 2016 at 10:39

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