Isomorphism of Varieties

Question

Let $V=V(x^2+y^2-1) \subset \mathbb{R}^2$ be an affine variety. Show that $V$ is rational, but isn't isomorphic to $\mathbb{R}^1$.

I could show that $V$ is rational, by parametrization $$x=\frac{1-t^2}{1+t^2} , y=\frac{2t}{1+t^2}, t \in \mathbb{R}$$ but I do not know how to show that $V$ isn't isomorphic to $\mathbb{R}^1$.can anyone help me?

Well, it's a circle, so you don't expect it to be isomorphic to a line. An isomorphism $V \simeq \mathbb R^1$ would give you an isomorphism $\mathbb R^1 \backslash {p} \simeq \mathbb R^1$ by composing with your parametrization. Can you finish the argument? (This is an actual question, I'm pretty sure it works but I'm not convinced ; didn't think about it enough.) — Patrick Da Silva, Jun 23 '15 at 18:32
I converted my comment into an answer, feel free to upvote it. — Patrick Da Silva, Jun 23 '15 at 19:39

score 3 · Accepted Answer · answered Jun 23 '15 at 19:39

Well, it's a circle, so you don't expect it to be isomorphic to a line. An isomorphism $V \simeq \mathbb R^1$ would give you an isomorphism $\mathbb R \backslash \{p\} \simeq \mathbb R$ by composing with your parametrization. Can you finish the argument?

Hope that helps,

score 2 · Answer 2 · answered Jun 23 '15 at 19:53

It's enough to show that $\mathbb{R}[x,y]/(x^2 + y^2 - 1) \not \cong \mathbb{R}[t]$. Suppose there exists such an isomorphism which sends $x$ to $f(t)\in \mathbb{R}[t]$ and $y$ to $g(t)\in \mathbb{R}[t]$. This implies that $f^2(t) + g^2(t) - 1=0$. This is impossible if $f$ and $g$ are both non trivial polynomials with coefficients in $\mathbb{R}$. Then ...

Isomorphism of Varieties

2 Answers2