By a *curve* I mean an integral one-dimensional scheme of finite type over a spectrum of a field.


Let $C$ be a curve over an arbitrary field $k$. It's probably a very well known fact, that $C$ is rational over $k$, if and only if $C$ is rational over any field extension $L/k$. I'm wondering, if there an elementary proof of this property.


The easiest argument I can think about at this moment, is to choose a smooth projective curve $\overline{C}\subset\mathbb{P}\_k^n$ birational to $C$ and observe, that it's Hilbert polynomial does not depend on base extension $L/k$. This is a consequence of the fact that $L$ is a flat $k$-module. Therefore, the arithmetic genus of $C$ does not depend on base extension. Finally, by Bézout theorem and the genus formula, one can verify that arithmetic genus equal to zero implies that there exist a $k$-parametrization $\mathbb{P}\_k^1\longrightarrow\overline{C}$.


All this said, I believe that there should exist a shorter and more algebraic argument. However, I can't find one. I will be grateful for any piece of advice.


**EDIT** As @Abhinav pointed out in his answer below, the statement is not even true in general. Luckily for me, it holds true as long as the field extension $L/k$ is pure transcendental (see the answer of @Pete L. Clark below), which is good enough for the applications I was thinking about. Thank you! :)



Rationality of curve does not depend on base change
Well, why not turn my comment on Abhinav's answer into an answer?
$\newcommand{\ra}{\rightarrow}$
$\newcommand{\PP}{\mathbb{P}}$
$\newcommand{\F}{\mathbb{F}}$


This is a cut-and-paste from a passage in a paper I wrote earlier today.



> 
> Lemma: 
>  Let $L/K$ be a purely transcendental field extension.  
> 
>  a) Let $V\_{/K}$ be an algebraic variety. Suppose *either* that $K$ is infinite or $V$ is complete. Then $V(L) \neq \varnothing \implies V(K) \ \neq \varnothing$.  
> 
>  b) For every abelian variety $A\_{/K}$, we have $H^1(L/K,A) = 0$. 
> 
> 
> 


Proof: a) 
Step 1: Let $\{t\_i\}\_{i \in I}$ be a transcendence basis for $L/K$. If $P \in V(L)$, there is a finite subset $J \subset I$ such that $P \in V( K(\{t\_i\}\_{i \in J})$. Thus we are 
reduced to the case in which $L/K$ has finite transcendence degree. Induction reduces us to the case $L = K(t)$.  

Step 2: A point $P \in V(K(t))$ corresponds to a rational map $\varphi: \mathbb{P}^1 \ra V$. The locus on which $\varphi$ is not defined is a finite set of closed points of $\mathbb{P}^1$. If $K$ is infinite, so is $\mathbb{P}^1(K)$, and thus there is $P \in \mathbb{P}^1(K)$ at which $\varphi$ 
is defined, and then $\varphi(P) \in V(K)$. On the other hand, any rational map from a regular curve to a complete variety is a morphism, so if $V$ is complete then 
e.g. $\varphi(0) \in V(K)$.  

b) Since $\eta \in H^1(K,A)$ corresponds to a torsor $V$ under $A$ and thus a 
projective variety, this follows immediately from part a). 



> 
> Remark: 
>  a) If in the statement of the Lemmma we strengthen "complete" to "projective", a more elementary proof can be given: let $\varphi: V \ra \mathbb{P}^N$ be a $K$-embedding. Since $K(t)$ is the fraction field of 
>  the UFD $K[t]$, if $P \in V(L)$, we can write $\varphi(P) = [f\_0(t):\ldots:f\_N(t)]$ with $\operatorname{gcd}(f\_0,\ldots,f\_N) = 1$. In particular, some $f\_i(t)$ is not divisible by $t$ and thus $(f\_0(0):\ldots:f\_N(0)) \in V(K)$.  
> 
>  b) Let $K = \F\_q$ be a finite field. Then the affine curve $V = \PP^1\_{\F\_q} \setminus \PP^1(\F\_q)$ has $K(t)$-rational points but no $K$-rational points.
> 
> 
> 


There follows (in the manuscript) a result that if you start with $V\_{/K}$ and replace $K$ by the function field $K(X)$ of a variety $X\_{/K}$ with a rational zero-cycle of degree $1$, the least degree of a rational zero-cycle on $V$ does not change. Even this result is rather well-known, I think, but it can be hard to find these types of things written down in "proper generality".


Also the OP mentions something about his genus zero curve $C\_{/K}$ having enough rational points to be rational. For this: it is not hard to show that a geometrically integral curve of arithmetic genus $0$ over an arbitrary field is birational to $\mathbb{P}^1$ if and only if it has at least one nonsingular $K$-rational point. By Riemann-Roch the curve is birational to a conic, and having a nonsingular rational point is a birational invariant (think in terms of valuations on the function field).


