Prove $S = \{ (a,b) \in \mathbb{A}^2 \ | \ a \overline{a} + b \overline{b} = 1\}$ is not Zariski closed

Question

I'm working on a problem where I have to prove that

Consider the subset $S$ of $\mathbb{A}^2$, the affine space over $\mathbb{C}$, defined by $$ S = \{ (a,b) \in \mathbb{A}^2 \mid a \overline{a} + b \overline{b} = 1\}. $$ Prove that $S \subset \mathbb{A}^2$ is not a Zariski closed set.

Here $\overline{a}$ is the complex conjugate of $a$. I know that in the Zariski topology, the opens are complements of affine algebraic varieties. So I have to prove $S$ is not an affine algebraic variety. This means I have to prove I can not write $S$ as $$ S = \mathbb{V}(\{ F_i \}_{i \in I})$$ where $F_i$ are complex polynomials on $\mathbb{C}^n$. I don't see where I can start, I thought about proving it by assuming I can write it like this and then getting a contradiction, but I don't see how.

Could anyone help me please or give me a hint? Thanks!

Answer 1

If your set were Zariski-closed, then it's intersection with any copy of $\Bbb A^1\subset \Bbb A^2$ would be a Zariski-closed subset of $\Bbb A^1$. Do you know what the Zariski-closed subsets of $\Bbb A^1$ are? Can you think of a good candidate $\Bbb A^1$ to use?

Here's the answer to the first bit:

The Zariski-closed subsets of $\Bbb A^1$ are exactly the finite sets of points: any ideal of $k[t]$ is principal. So if you can find a copy of $\Bbb A^1\subset\Bbb A^2$ so that $S\cap \Bbb A^1$ is not finite, you win.

Here's how to do the second bit:

If we fix $a=0$, then we're looking at $\{b\in\Bbb C\mid b\overline{b}=1\}$. This is infinite, as it's the unit circle in $\Bbb C$.