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! :)

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    $\begingroup$ The argument fails at the last step: you need a $k$-rational point on $\overline{C}$ to carry this step out (if I understand you correctly) and such a point may not exist. $\endgroup$ – Qiaochu Yuan Nov 1 '13 at 1:59
  • $\begingroup$ That's what I was afraid of :) Hopefully, in my particular use-case I can prove that $C$ has enough $k$-rational points. $\endgroup$ – Tomasz Lenarcik Nov 1 '13 at 3:03
  • $\begingroup$ Actually, I only need this lemma in case when $L$ is a pure transcendental extension of $k$. $\endgroup$ – Tomasz Lenarcik Nov 1 '13 at 3:05

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).

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    $\begingroup$ The $H^1(L|K,A)$ in part $b)$ is mysterious: the extension $L|K$ is not galois ! $\endgroup$ – Chandan Singh Dalawat Nov 1 '13 at 7:36
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    $\begingroup$ @Chandan: It means the kernel of the restriction map from $H^1(K,A)$ to $H^1(L,A)$, which is defined for any field extension. (Anyway, it is has little to do with the OP's question, so should perhaps best be ignored or deleted.) $\endgroup$ – Pete L. Clark Nov 1 '13 at 16:17

If you're taking the definition of rational to be: birational to $\mathbb{P}^1$ over the field $k$, then the stated property is not even true. There are conics which have no rational points, and so are not rational, but are rational over a quadratic extension. For example, the affine conic $x^2 + y^2 + 1 = 0$ over the field $\mathbb{Q}$.

Added: since you only need the result when $L/k$ is a pure trancendental extension - there the result is indeed true. The idea is that you can represent any genus $0$ curve as a conic (take the embedding corresponding to $-K_C$), and so you need to show that if the conic has a rational point over a pure transcendental extension of $k$, then it has a rational point over $k$ itself. There are many ways to see this (essentially you specialize the indeterminates suitably so that no denominators vanish) - one way is by induction on the transcendence degree. For a reference see Lam's "Quadratic forms over fields", Lemma 1.1 of Chapter IX. The result holds more generally for quadratic forms in any number of variables.

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    $\begingroup$ That's a funny coincidence. Earlier today I wrote up a proof of the following Lemma: Let $K$ be a field and $L/K$ a purely transcendental field extension. Let $V_{/K}$ be a variety. If either $K$ is infinite or $V$ is complete, then $V(L) \neq \varnothing \implies V(K) \neq \varnothing$. The proof is not at all hard, but it becomes easier if you strengthen "complete" to "projective", as you note. Note that some hypothesis is needed to rule out things like $\mathbb{P}^1_{/\mathbb{F}_p} \setminus \mathbb{P}^1(\mathbb{F}_p)$! $\endgroup$ – Pete L. Clark Nov 1 '13 at 6:05

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