Let $X$ be an algebraic variety defined over a field $k$ of characteristic zero. Suppose $X(K)$ is non-empty for some extension $K/k$. Is it true that $X(L)$ is non-empty where $L$ is the algebraic closure of $k$ in $K$? An equivalent formulation is: suppose $X(k)$ is empty; does that force $X(K)$ to be empty for every purely transcendental extension $K/k$?


In my particular application $k$ is a number field with a real place and $K=\mathbb{R}$ and then claim is true since the theory of real closed fields eliminates quantifiers, but I'm wondering about the general case. Relatedly Chevalley's Theorem (elimination of quantifiers for algebraically closed fields) does produce $\bar{k}$-points, but I don't see why they have to be $\bar{k}\cap K$-points.



Can a variety acquire points in a purely transcendental extension?
The following answers your second question.



> 
> 
> > 
> > Let $X$ be a finite type separated scheme over an infinite field $k$.
> > Let $K/k$ be a purely transcendental extension.
> > If $X(k)$ is empty, then $X(K)$ is empty.
> > 
> > 
> > 
> 
> 
> 


*Proof.* We may assume that $K$ has finite transcendence degree over $k$. Then, any element of $X(K)$ extends to a morphism $U\to X$, where $U$ is a dense open of some $\mathbb{P}^n$. Since $k$ is infinite, the set $U(k)$ is dense (hence non-empty). This contradicts $X(k)$ being empty.


Your first question is answered by Laurent Moret-Bailly.



> 
> 
> > 
> > Let $X$ be a geometrically connected finite type scheme over $k$ (and thus non-empty). Let $K=K(X)$ be the function field of $X$. Then $X(K)$ is non-empty. However, there is no reason for $X(k)$ to be non-empty. Take $X$ to be a smooth proper conic over $\mathbb{Q}$ with no $\mathbb{Q}$-points, for example.
> > 
> > 
> > 
> 
> 
> 


