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

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?

• 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. – Fan Zheng Dec 19 '15 at 18:09

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.)

• 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). – nfdc23 Dec 20 '15 at 6:33
• @nfdc23 Thanks for the clarification. – Joe Silverman Dec 20 '15 at 13:06
• 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). – nfdc23 Dec 20 '15 at 16:02

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.

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.

• 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... – Daniel Litt Dec 19 '15 at 18:19
• E.g. Take $y^2=f(x)$ where $f$ is a separable cubic with 3 real roots... – Daniel Litt Dec 19 '15 at 18:21
• @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. – Kestutis Cesnavicius Dec 19 '15 at 18:24