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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?

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good question! I wish I know the answer.... –  Bombyx mori Jan 23 '13 at 4:47

2 Answers 2

up vote 6 down vote accepted

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 $k$.

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Thank you! That pretty much clears it up. But what is the difference between $k(x,\sqrt{1-x^2})$ and $k(x,\sqrt{x^3-x})$? Is there a way to determine whether extension fields are transcendental? I guess in the case of function fields, one way (explained in the paper that DonAntonio linked) is to find a parametrization of the variety, which realizes the function field as a field of rational functions in the parameters. That will demonstrate that $k(x,\sqrt{1-x^2})$ actually is transcendental. But to show that $k(x,\sqrt{x^3-x})$ isn't? –  Joe Hannon Jan 23 '13 at 16:35
@ziggurism: You can see Robin Chapman's answer here for a proof that $k(x, \sqrt{x^3 - x})$ is not transcendental. One way to check whether function fields of curves are transcendental without going explicitly through algebra is to use other geometric invariants of the curves. For instance, genus is a birational invariant so any projective curve of positive genus cannot admit a rational parametrization (as $\mathbf{P}^1$ has genus $0$). –  Brandon Carter Jan 23 '13 at 16:44
In fact, this is one way to see that $k(x, \sqrt{x^3-x})$ is not purely transcendental. This is because the projective closure of $y^2 = x^3 - x$ is a genus $1$ curve. –  Brandon Carter Jan 23 '13 at 16:47

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.

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