# Points of abelian varieties over purely transcendental extensions

I heard about the result in the theory of abelian varieties which says the following: given an abelian variety $X$ defined over a field $k$ and a purely transcendental extension $k\subset L\subset L'$ we have that the group $X(L)$ of $L$-points is the same as the group $X(L')$. Could you give me (the reference for) the proof of this result?

This follows from the following well-known lemma.

Lemma. Let $A$ be an abelian variety over $k$. Then any map $f \colon \mathbb P^1 \to A$ is constant.

Proof 1. The map $f$ induces a map on the Albanese $f_* \colon \operatorname{Alb}_{\mathbb P^1} \to \operatorname{Alb}_A$ sitting in a commutative diagram $$\begin{array}{ccc}\mathbb P^1 & \stackrel{f}\longrightarrow & A \\ \downarrow & & \downarrow \\ \operatorname{Alb}_{\mathbb P^1} & \stackrel{f_*}\longrightarrow &\ \operatorname{Alb}_A. \end{array}$$ But $\operatorname{Alb}_{\mathbb P^1}$ is trivial, and the map $A \to \operatorname{Alb}_A$ is an isomorphism. $\square$

Proof 2. After translating on $A$, we may assume $f(0) = 0$. The point $0\in \mathbb P^1$ is the identity element for two different group structures: the open subscheme $\mathbb P^1 - \infty$ is isomorphic to $\mathbb G_a$, and $\mathbb P^1 - \{\infty, 1\}$ is isomorphic to $\mathbb G_m$. Then $f|_{\mathbb G_a}$ and $f|_{\mathbb G_m}$ are both group homomorphisms, which is impossible unless $f$ is constant. $\square$

Proof 3. There is a third proof using triviality of $\Omega_A$ and the fact that $\Omega_{\mathbb P^1}$ has no global sections. See e.g. Bhatt's notes, Cor. 3.6. $\square$

Corollary. Let $A$ be an abelian variety over $k$, and let $C$ be a rational curve (not necessarily smooth or proper). Then any map $f \colon C \to A$ is constant.

Proof. By assumption, there is an open $U \subseteq C$ that is isomorphic to an open in $\mathbb P^1$. Moreover, $f|_U$ extends uniquely to a morphism $\mathbb P^1 \to A$ because $A$ is proper. Hence, $f|_U$ is constant. Because $U$ is dense, this forces $f$ to be constant. $\square$

Corollary. Let $A$ be an abelian variety over $k$, and let $Y$ be a rational variety (not necessarily smooth or proper). Then any morphism $f \colon Y \to A$ is constant.

Proof. It suffices to prove that $f$ is constant on a big open, so we may replace $Y$ by an open $U \subseteq Y$ that is isomorphic to an open in $\mathbb P^n_k$. Then $Y$ is covered by (not necessarily proper) rational curves $C \subseteq Y$. By the corollary above, we conclude that $f$ is constant on all of those, hence $f$ is constant. $\square$

The answer to the OP now follows: $L$-points of $X$ are the same as the $L$-points of $X_L = X \times_k L$, so we may assume that $L = k$. Then a morphism $\operatorname{Spec} L' \to A$ spreads out to a rational variety $Y$ with function field $L'$, hence is constant (i.e. comes from a $k$-point) by the arguments above.

• I think the simplest proof of the lemma uses that there are no nontrivial finite, etale covers of $\mathbb{P}^1$. Thus, for every odd prime $\ell$ different from the characteristic, every morphism from $\mathbb{P}^1$ to $A$ factors through the "multiplication-by-$\ell$" morphism. For any ample invertible sheaf $L$ on $A$, this implies that the degree of the pullback of $L$ on $\mathbb{P}^1$ is divisible by $\ell$ (by the Theorem of the Cube). Thus, the degree is zero, and the morphism is constant. – Jason Starr Feb 20 '18 at 17:44
• @JasonStarr: ah, that's pretty neat. I'm a little agnostic as to which proof is simpler, but the one you gave is certainly pretty nice. – R. van Dobben de Bruyn Feb 20 '18 at 19:39
• mathoverflow.net/questions/9066/… – Felipe Voloch Feb 20 '18 at 21:41
• @AnnaAbasheva: if $L'$ is the function field of $\mathbb A^n$ (or $\mathbb P^n$), then any morphism $\operatorname{Spec} L' \to X$ to any finite type $k$-scheme $X$ spreads out to some open neighbourhood of the generic point in $\mathbb A^n$. If you don't know this, you should do this as an exercise. (Hint: first think about the case that $X$ is affine. Use that $X$ is of finite type.) – R. van Dobben de Bruyn Feb 21 '18 at 18:52
• @BenLim: any $k(x_1,\ldots)$-point of a finite type $k$-scheme factors through the finitely generated extension $k(x_1,\ldots,x_n)$ for some $n$. – R. van Dobben de Bruyn Feb 21 '18 at 18:54