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?



Points of abelian varieties over purely transcendental extensions
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](http://www-personal.umich.edu/~stevmatt/abelian_varieties.pdf), **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.


