Is $Pic^0(X)$ of a curve of genus $\geq 1$ over a non-algebraically closed field still non-finitely generated?

Question

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.

1. If $(n,\text{char} (k))=1$, then $Pic^0(X)[n]\cong (\mathbb{Z}/n\mathbb{Z})^{2g}$;
2. 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?

Answer 1

As Will Sawin says, the key phrase here is Mordell-Weil theorem. Also, this isn't really a theorem about Picard groups of curves, it's a theorem about abelian varieties. Here is a fairly general statement:

Theorem (Mordell-Weil-Lang-Neron) Let $K$ be a field that is of finite type over its prime field (where the prime field is either $\mathbb Q$ or $\mathbb F_p$), and let $A/K$ be an abelian variety. Then $A(K)$ is finitely generated.

More generally, let $k$ be any field and let $K/k$ with $K$ of finite type over $k$. Then for any abelian variety $A/K$ there is an abelian variety $B/k$ (possibly trivial) called the $K/k$-trace of $A$ and an inclusion $i:B\times_kK\hookrightarrow A$. Roughly speaking, $B$ is the largest piece of $A$ that comes from an abelian variety defined over $k$. Then $A(K)/i(B(k))$ is finitely generated.

I believe that this is all proven in Lang's Fundamentals of Diophantine Geometry.

Conversely, if $K$ is not finitely generated over its prime field, then I suspect that there always exists an abelian variety such that $A(K)$ is not finitely generated. (But you probably won't be able to prove it using torsion points.)

Answer 2

Yes for number fields by the Mordell-Weil theorem.

For arbitrary fields it may depend on the curve. For instance for a function field of an algebraically closed field, elliptic curves defined over the base field have infinitely generated Picard groups but other elliptic curves have finitely generated ones.

Answer 3

The answer is 'no'. If $k = \mathbb{R}$, then $Pic^0(X)(\mathbb{R})$ is a commutative real Lie group of dimension $g$, isomorphic to $(\mathbb{R}/\mathbb{Z})^g \times (\mathbb{Z}/2\mathbb{Z})^c$ for some $0 \le c \le g$, so is not finitely generated as an abelian group.