Qing Liu's "Algebraic Geometry and Arithmetic Curves" page 299 COrollary 7.4.41 gives the following result.

Let $X$ be a smooth, connected, projective curve over an algebraically closed field $k$, of genus $g$. Let $Pic^0(X)$ denote the subgroup of $Pic(X)$ consisting of divisors of degree $0$. Let $n\in \mathbb{Z}$ be non-zero and $Pic^0(X)[n]$ denote the kernel of the multiplication by $n$ map.

- If $(n,\text{char} (k))=1$, then $Pic^0(X)[n]\cong (\mathbb{Z}/n\mathbb{Z})^{2g}$;
- If $p=\text{char} (k)>0$, then there exists an $0\leq h\leq g$ such that for any $n=p^m$, we have $Pic^0(X)[n]=(\mathbb{Z}/n\mathbb{Z})^h$.

From this it is easy to deduce that if $X$ is a smooth, connected, projective curve over an algebraically closed field $k$, of genus $g\geq 1$, then $Pic^0(X)$ is not a finitely generated abelian group. (This is Exercise 4.9 (d) in page 301 of Qing Liu's book.)

$\textbf{My question}$ is: if the base field $k$ is not algebraically closed, is the above statement still true? I.e. if $X$ is a smooth, geometrically connected, projective curve over a field $k$, of genus $g\geq 1$, then is $Pic^0(X)$ finitely generated?