Two copied of $\mathbb{P}^2_k$ glued by a non-torsion point on an elliptic curve - how to prove all invertible sheaves are trivial?

Question

This is 19.11.11 in Vakil's Foundations of Algebraic Geometry, July 31 2023 version.

Let $E$ be an elliptic curve, $p$ be a non-torsion point, and $\infty$ be the point at infinity. Consider the usual closed immersion $E \rightarrow \mathbb{P}^2_k$, and also the closed immersion given by translation by $p$ then the usual one. Call these $i$ and $j$ respectively.

Then, consider the fiber coproduct $X = \mathbb{P}^2_k \coprod_{E, i, j} \mathbb{P}^2_k$. The claim is that all invertible sheaves on $X$ are trivial.

Let $L$ be an invertible sheaf. Then, its pullbacks to the copies of $ \mathbb{P}^2_k$ are $O(d_1)$ and $O(d_2)$ respectively. $i$ pulls back $O(d_1)$ to $O(3 d_i \infty)$, and $j$ pulls it back to $O(3 d_2 p)$. By $p$ being non-torsion, this is only possible if $d_1 = d_2 = 0$.

Thus $L$ pulls back to the structure sheaf on both copies of $ \mathbb{P}^2_k$. But how do we conclude that $L$ itself is the structure sheaf?

The problem I'm facing is that the two copies of $ \mathbb{P}^2_k$ are not necessarily open in $X$, so I don't know how to use information on $ \mathbb{P}^2_k$ to conclude anything about $X$.

Answer 1

Let $X_1$ and $X_2$ be the two irreducible components of $X$ and let $X_{12} = X_1 \cap X_2$ be their intersection (the elliptic curve). Then there is an exact sequence $$ 0 \to \mathcal{O}_X \to \mathcal{O}_{X_1} \oplus \mathcal{O}_{X_2} \to \mathcal{O}_{X_{12}} \to 0 \tag{*} $$ where all maps are given by restriction of functions. Now, if $L$ is a line bundle on $X$, tensoring $(*)$ by $L$ one obtains $$ 0 \to L \to L\vert_{X_1} \oplus L\vert_{X_2} \to L\vert_{X_{12}} \to 0. $$ If $L\vert_{X_1} \cong \mathcal{O}_{X_1}$ and $L\vert_{X_2} \cong \mathcal{O}_{X_2}$, the sequence takes the form $$ 0 \to L \to \mathcal{O}_{X_1} \oplus \mathcal{O}_{X_2} \to \mathcal{O}_{X_{12}} \to 0. $$ Now, since for $i = 1,2$ one has $$ \mathrm{Hom}(\mathcal{O}_{X_i}, \mathcal{O}_{X_{12}}) \cong k, $$ it follows that this sequence is isomorphic to $(*)$, hence $L \cong \mathcal{O}_X$.