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 peace of advice.

EDIT As @Abhinav and pointed out in his answer below, the result is not true in general. On the other hand, 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! :)

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! :)

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 belive that there should exist a shorter and more algebraic argument. However, I can't find one. I will be grateful for any peace of advice.

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 peace of advice.

EDIT As @Abhinav and pointed out in his answer below, the result is not true in general. On the other hand, 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! :)