ag.algebraic geometry - Line bundles trivial after extension of the base-field

This is not true. Let $E$ be an elliptic curve over $mathbb{R}$, such that $E(mathbb{R})$ has one connected component. Let $u$ be a point of $E(mathbb{C}) setminus E(mathbb{R})$. Writing $sigma$ for complex conjugation; $u + sigma(u)$ is in $E(mathbb{R})$. Two of the solutions to $2v=u + sigma(u)$ are real and the other two are complex conjugate to each other; let $v$ and $sigma(v)$ be the complex conjugate pair. So

$$2v=2sigma(v) = u + sigma(u)$$

in the group law of $E$.

Now, let $X = E setminus { v, sigma(v) }$ and consider the line bundle $L:=mathcal{O}(u + sigma(u))$. Since this is a $sigma$ invariant divisor, the line bundle $L$ is defined over $mathbb{R}$. I claim that $L$ is nontrivial, but becomes trivial after base extending to $mathbb{C}$.

Proof that $L$ is nontrivial:
If not, there would be a real function $f$ on $X$, with zero divisor precisely $u + sigma(u)$. Extending to a meromorphic function on $E$, we would have
$$u + sigma(u) = k v + (2-k) sigma(v).$$
But, since $f$ is $sigma$ invariant, it has poles of the same order at $v$ and $sigma(v)$, so $u + sigma(u) = v + sigma(v)$. Using the linear relation $u + sigma(u) = 2v$, we deduce that $v = sigma(v)$, contradicting that $v$ was chosen not to be real.

Proof that $L times_{mathbb{R}} mathbb{C}$ is trivial:
The relation $u + sigma(u) = 2v$ means there is a meromorphic function on $E$ with zeroes at $u$ and $sigma(u)$, and a double pole at $v$. Restricting this function $X$, we get a trivialization of $L$.