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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?

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    $\begingroup$ One thing for sure is that the answer to the last question in the text is the opposite of the answer to the question in the title. $\endgroup$ – Fan Zheng Dec 19 '15 at 18:09
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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.)

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    $\begingroup$ The kernel of the canonical map $\tau:{\rm{Tr}}_{K/k}(A)_K \rightarrow A$ is $K$-finite with infinitesimal dual, but it can be etale and nontrivial. (Much deeper is that it is infinitesimal when $K/k$ is "regular": separable with $k$ algebraically closed in $K$.) The existence of ${\rm{Tr}}_{K/k}(A)$ and structure of $\ker(\tau)$ are discussed in Lang's book on abelian varieties, but not in his book on Diophantine geometry (which incorrectly says "$\tau$ is injective", but gives no reference for that assertion). $\endgroup$ – nfdc23 Dec 20 '15 at 6:33
  • $\begingroup$ @nfdc23 Thanks for the clarification. $\endgroup$ – Joe Silverman Dec 20 '15 at 13:06
  • $\begingroup$ I should have given a reference for an example in which $\ker \tau$ is nontrivial and etale; see Example 6.3 in the 2006 paper "Chow's K/k-trace and K/k-image and the Lang-Neron theorem" in L'enseignement Math. 52(1). $\endgroup$ – nfdc23 Dec 20 '15 at 16:02
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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.

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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.

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    $\begingroup$ There can be more than one component of the real group, right? So you're really describing the identity component of the real points of Pic^0... $\endgroup$ – Daniel Litt Dec 19 '15 at 18:19
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    $\begingroup$ E.g. Take $y^2=f(x)$ where $f$ is a separable cubic with 3 real roots... $\endgroup$ – Daniel Litt Dec 19 '15 at 18:21
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    $\begingroup$ @DanielLitt: Thanks, you're right. I was for some reason thinking that $Pic^0$ already takes care of that, but I overlooked that connectedness may nevertheless be lost on real points. $\endgroup$ – Kestutis Cesnavicius Dec 19 '15 at 18:24

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