Rationality of curve does not depend on base change

Question

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! :)

Answer 1

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).

Answer 2

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.