What is a non-effective holomorphic line bundle $L$ of degree 0 such that $L \otimes L$ is effective?

Question

I am trying to solve an exercise that asks for a non-effective holomorphic line bundle $L$ of degree 0 such that $L \otimes L$ is effective.

I think I have mostly solved it but since I am a bit shaky on the details and would appreciate if someone could see if my approach is valid.

We know that every line bundle has a meromorphic section thus we can find a divisor $D$ such that $L$ is nothing but the line bundle $[D]$ obtained by the map $[\cdot]: Div(X) \to Pic(X)$. Hence in terms of divisors we want to find some $D \in Div(X)$ of degree 0 such that $h^0(X, D) = 0$ and $h^0(X, 2D) = 1$, by an application of Riemann-Roch this is equivalent to $2D \sim 0$ and $D \nsim 0$.

If we have any hope of finding something like this we should look for it in some surface $X$ with non-zero genus.

According to this top answer the Picard group of the torus has several $2$-torsion elements that should do the trick, however, I have not been able to define the line bundle explicitly.

Answer 1

From your post, it seems like you're working on curves over $\Bbb C$, but everything I'm about to say will work over any algebraically closed field.

Let $X$ be an elliptic curve with identity point $P_0$, and let $P_1$ be a nontrivial $2$-torsion point. Then $\mathcal{O}_X(P_1-P_0)$ is degree $0$ and not the trivial bundle: if it were, $P_1$ and $P_0$ would be linearly equivalent divisors, but two distinct points are linearly equivalent only on a curve of genus $0$. On the other hand, $\mathcal{O}_X(P_1-P_0)^{\otimes 2} \cong \mathcal{O}_X(2P_1-2P_0)\cong\mathcal{O}_X$, which is principal.

Here are two ways to see that this divisor is principal, depending on how much background with elliptic curves you have:

• Suppose $X$ is embedded in $\Bbb P^2$ as $y^2z=x(x-z)(x-\lambda z)$ with $P_0=[0:1:0]$ and $P_1=[0:0:1]$ (this can always be done if we're outside characteristic two). The rational function $x/z$ has valuation $2$ at $P_1$, valuation $-2$ at $P_0$, and valuation $0$ everywhere else, which we may compute from looking at the intersection of $V(x)$ and $V(z)$ with $X$.
• There's a homomorphism $X\to\operatorname{Pic} X$ given by sending $P\to \mathcal{O}_X(P-P_0)$. Since $P_1$ is $2$-torsion in $X$, its image must be 2-torsion in $\operatorname{Pic} X$. On the other hand, we showed above that $\mathcal{O}_X(P_1-P_0)$ was not trivial. (In fact, this homomorphism is an isomorphism on to $\operatorname{Pic}^0(X)$, the subgroup consisting of line bundles of degree zero.)