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Consider the function field $k(x,\sqrt{1-x^2})$ of the circle over an algebraically closed field $k$. Is $k(x,\sqrt{1-x^2})$ a purely transcendental extension over $k$?

I'm curious because I was reading the answer here. The proof shows $k(x+\sqrt{1-x^2})=k(x,\sqrt{1-x^2})$. However, there is line which says, $$ (x+\sqrt{1-x^2})^2 = x^2 + 2 \sqrt{1 - x^2} + (1 - x^2) = 2 \sqrt{1-x^2} + 1. $$ But I think $$ (x+\sqrt{1-x^2})^2 = x^2 + 2x\sqrt{1 - x^2} + (1 - x^2)=2x\sqrt{1-x^2}+1 $$ so, assuming $\operatorname{char}(k)\neq 2$, I believe at most one can conclude is $x\sqrt{1-x^2}\in k(x+\sqrt{1-x^2})$. I've been struggling to salvage this.

How can you show $k(x,\sqrt{1-x^2})$ is a purely transcendental extension over $k$? Is the extension still purely transcendental when $\operatorname{char}(k)=2$?

user26857
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Yong Pan
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1 Answers1

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You are correct that there was a mistake in my original proof, but the result is still true. Instead, in the case that $\mathrm{char}(k) \neq 2$, consider the purely transcendental extension $k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$. This field contains $$ \frac{1 - \left(\frac{\sqrt{1-x^2}}{1+x}\right)^2}{1 + \left(\frac{\sqrt{1-x^2}}{1+x}\right)^2} = \frac{(1+x)^2 - (1-x^2)}{(1+x)^2 + (1-x^2)} = x $$ From this it is easy to see that $x, \sqrt{1-x^2} \in k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$, hence $k(x, \sqrt{1-x^2}) = k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$.

Really this all just comes from the rational parametrization of the circle $t \mapsto \left( \frac{1 - t^2}{1+t^2}, \frac{2t}{1+t^2} \right)$.

In the case that $\mathrm{char}(k) = 2$, the extension $k(x)$ works, as $$(1+x)^2 = 1 + x^2 = 1- x^2,$$ hence $\sqrt{1-x^2} \in k(x)$. This case is much easier because the circle $x^2 + y^2 + 1 = 0$ is really just a degenerate double line $(x+y+1)^2 = 0$ in characteristic 2.

bzc
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