Let $\pi \colon X \rightarrow C$ be a smooth minimal elliptic surface over an algebraically closed ground field $k$. Furthermore assume that $\pi$ has a section and that the fundamental line bundle $R^1\pi_*(\mathcal{O}_X) \in \text{Pic}(C)$ has positive degree. A standard result, found e.g. as Theorem 6.5 in Schütt's survey on elliptic surfaces, states that the Neron-Severi group $\text{NS}(X) = \text{Pic}(X)/ \text{Pic}^0(X)$ and the numerical group $\text{Num}(X) = \text{Pic}(X) / \text{Pic}^{\tau}(X)$ agree. Since the numerical group is the torsion-free part of the Neron-Severi group, this can be reformulated as the Neron-Severi group of such an elliptic surface being torsion-free. I would like to understand why that statement is true.

If we denote by $\mathcal{L}$ a representative of a non-trivial torsion class in $\text{NS}(X)$, then I can manage to prove that the class of $\mathcal{L}$ is trivial based on the assumption that $h^0(\mathcal{L}) = 0$, i.e. that $\mathcal{L}$ has no non-trivial global sections. This assumption seems very plausible to me:

$\bullet$ If $X$ were a curve, so that algebraically trivial line bundles are exactly the line bundles with degree $0$, then it is standard to show that such line bundles have no non-trivial global sections, unless they agree with the structure sheaf.

$\bullet$ If $X$ were an abelian variety, I also have an argument. It is quite simple to show that a tensor power $\mathcal{L}^{\otimes n}$ of a line bundle is the pullback of $\mathcal{L}$ along the $n$-th power map of the abelian variety (by induction). Therefore, if $\mathcal{L}$ has a section, so does its dual $\mathcal{L}^{\otimes -1}$. Hence $\mathcal{L} = \mathcal{O}_X$.

$\bullet$ Furthermore, algebraically trivial line bundles are in particular numerically trivial and hence have degree $0$ when restricted to any curve in $X$. Therefore we have no non-trivial global sections on curves in $X$. Maybe one can pull this back?

Most likely there is a simple generalization of the argument for curves to general algebraically trivial line bundles, but I cannot seem to figure it out or to find it online. Therefore:

Question: Why do non-trivial algebraically trivial line bundles (on a smooth minimal elliptic surface if necessary) have no non-trivial global sections?


1 Answer 1


I believe I finally understand:

Let $X$ be an integral projective scheme over a field $k$ and let $\mathcal{L}$ be a numerically trivial line bundle on $X$ (for example an algebraically trivial line bundle). Let me explain why if $\mathcal{L}$ has a non-zero global section, $\mathcal{L}$ must be isomorphic to $\mathcal{O}_X$:

Assume that $\mathcal{L}$ is not isomorphic to the structure sheaf of $X$ and let $D$ be an associated effective Cartier divisor corresponding to a non-zero global section $s$ of $\mathcal{L}$. There always exists an integral curve $C \subset X$ not lying in $D$ with non-trivial intersection $C \cap D \neq \emptyset$ (this is not obvious, but can be shown by induction on the dimension of $X$ (choose $C = X$ for dimension $1$) via prime avoidance. Then we get

$$0 = \text{deg}(\mathcal{L}_C) = h^0(\mathcal{O}_{C \cap D}) \geq 1,$$ a clear contradiction to our assumption. Hence such $\mathcal{L}$ must be isomorphic to $\mathcal{O}_X$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .