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Can anyone help me with the following question? Let $X$ be a smooth, projective algebraic variety over a field $k$ of characteristic zero. Let $D$ be an effective divisor on $X$ and $m \geq 2$ an integer.

Under which conditions there exists a line bundle $L$ such that $O_X(D)=L^m$?

There is of course the obvious one: $m$ should divide de degree of $D$. Is that sufficient?

Thanks

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1 Answer 1

up vote 3 down vote accepted

There is no standard way to deduce that a divisor $D$ on a projective variety $X$ is $m$-divisible in the Picard group, and even for $m=2$ this can be a difficult problem. Geometrically, this condition is equivalent to the existence of a simple cyclic cover $Y \to X$ of degree $m$ branched precisely on $D$.

The condition you give is not sufficient, and the following is a counterexample. Take a quartic surface $X \subset \mathbb{P}^3$ and assume that $X$ contains a conic $D$. Then the degree of $D$ with respect to the hyperplane section of $X$ is $2$. On the other hand, there is no divisor $L$ on $X$ such that $2L=D$. In fact, since $X$ is a $K3$ surface, by adjunction formula one finds $D \cdot D=-2$, that would imply $L \cdot L=-1/2$, a contradiction.

However, the condition is true for curves (see my comment below) and also in the following situation. Assume that $\textrm{Pic}(X)=\mathbb{Z}$, generated by an effective divisor $H$. If $\deg_H D = m \deg_H H$, then $D$ is linearly equivalent to $mH$. This happens for instance when $X=\mathbb{P}^n$, when $X \subset \mathbb{P}^3$ is a very general smooth surface of degree at least $4$ and when $X \subset \mathbb{P}^n$ is a smooth complete intersection of dimension at least $3$.

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Thanks for your answer, Francesco! What about curves? Is it sufficient in that case? –  div Dec 28 '12 at 21:36
    
Yes, because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group. In fact, assume that the degree of $D$ is divisible by $m$ and take any effective divisor $F$ such that $\deg F = (\deg D)/m$. Then $D-mF \in \textrm{Pic}^0(X)$, so there exists a degree $0$ divisor $\psi$ such that $D-mF=m \psi$, that is $D=m(F+ \psi)$. –  Francesco Polizzi Dec 28 '12 at 21:44
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There was no assumption on the ground field in the original question. –  Felipe Voloch Dec 28 '12 at 22:59
    
Right. I was tacitly assuming $k=\mathbb{C}$. –  Francesco Polizzi Dec 29 '12 at 9:42

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