# 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?

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 $$X$$.

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 $$L \cong \mathcal{O}_X$$.