Is the circle a rational curve and what is its function field? It does seem like the circle ($S^1=\{X^2+Y^2=1\}\subseteq k^2$ for $k$ a field) is a rational curve: it has parameterization $X=2T/(T^2+1)$ and $Y=(T^2-1)/(T^2+1)$.
On the other hand, we have a theorem that a variety is rational iff its function field is pure transcendental (eg Miles Reid, Ugrad Alg. Geo. 5.9). I think the function field of the circle is $k(X,\sqrt{1-X^2}),$ which is an algebraic extension of $k(X)$, so not purely transcendental. What is wrong with this picture?
 A: I think that there is some confusion about purely transcendental extensions. Note that $k(\sqrt{x})/k$ is purely transcendental, despite $k(\sqrt{x})/k(x)$ being algebraic. Similarly, if we can show that if $k(x, \sqrt{1 - x^2})$ is generated by a single transcendental element over $k$, then $k(S^1)$ will be a purely transcendental extension of $k$.
Consider the purely transcendental extension $k(x + \sqrt{1-x^2})/k$. For simplicity we assume that the characteristic of $k$ is not $2$. Clearly $k(x + \sqrt{1-x^2}) \subseteq k(x, \sqrt{1-x^2})$, so it suffices to prove the reverse inclusion. Notice that
$$
(x+\sqrt{1-x^2})^2 = x^2 + 2 \sqrt{1 - x^2} + (1 - x^2) = 2 \sqrt{1-x^2} + 1.
$$
Hence $\sqrt{1-x^2} \in k(x+ \sqrt{1-x^2})$, which also implies that $x \in k(x + \sqrt{1-x^2})$. Thus $k(x+\sqrt{1-x^2}) = k(x, \sqrt{1-x^2})$ and so the function field of the circle is, in fact, a purely transcendental extension of $k$, hence $S^1$ is rational.
Edit: There is a mistake in this proof that is corrected in my answer here.
It is important to note that the above argument does fail for other curves. For instance, the function field of the elliptic curve $y^2 = x^3 - x$ is $k(x, \sqrt{x^3 - x})$, which is not a purely transcendental extension of $k$.
A: Too long for a comment:
First, I think Reid's theorem talks of the function field over $\,k\,$ , not over another function field, e.g. $\,k(X)\,$, unless, of course, you "begin" with the last field.
Second, I don't think you have the easiest expression for the function field of the circle, and I think you may want to read this, the first 2/3 pages. Don't get confused by the title of the paper.
