# Prove a set $V$ is not algebraic

I need help showing that the set $$V = \{(a,b) \in \mathbb{A}^2(\mathbb{C})\ \vert \ \vert a\vert^2 + \vert b\vert^2 = 1\}$$ is not an algebraic subset of complex affine 2-space.

I believe that I can just choose two complex numbers $$a$$ and $$b$$ with moduli that sum to 1 and just show that $$(a,b) \neq 0$$. For example, if I choose $$a = i$$ and $$b=0$$, then $$\vert a\vert^2 + \vert b\vert^2 = \vert i\vert^2 + \vert 0\vert^2 = 1^2 = 1,$$ but $$(a,b) = (i,0) \neq (0,0)$$, and we are done.

Is this approach correct? Thanks in advance.

• But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument. – Matt Samuel Dec 11 '18 at 4:56
• No, the approach is not correct. What do you think the definition of "algebraic set" is? – Alex Kruckman Dec 11 '18 at 4:56
• @mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one. – André 3000 Dec 11 '18 at 8:24

## 2 Answers

If this set were algebraic, then its intersection with the set $$\{b=0\}$$ would be an algebraic subset of $$\Bbb A^1(\Bbb C)$$. But that intersection is $$\{a\in\Bbb C:|a|^2=1\}$$. But apart from $$\Bbb A^1(\Bbb C)$$ itself, the algebraic subsets of $$\Bbb A^1(\Bbb C)$$ are all finite.

• Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation. – Lord Shark the Unknown Dec 11 '18 at 7:24

Whether or not $$(a,b) = (0,0)$$ has nothing to do with anything. For example $$\{xy = 1\}$$ is algebraic and doesn't contain $$(0,0)$$. Recall that a set $$V$$ is algebraic if it is the vanishing set of some collection of polynomials.

There are a couple strategies to show that a set is not algebraic. First, you can take a polynomial that vanishes on $$V$$ and show that it must always vanish at some other point not in $$V$$ (where "must always" means this is true for any polynomial that vanishes on $$V$$, not just a specific one). Think about why this is. Suppose $$V = V(f_1,\dots,f_r)$$ and each $$f_i$$ vanishes at some other point not in $$V$$.

Another strategy is to consider dimensions and degrees and stuff like that. A common example for this is the set $$\{(t,\sin(t)) : t \in \mathbf{R}\} \subseteq \mathbf{R}^2$$. Then when you intersect this with the line $$\{y = 0\}$$ you get infinitely many points. But what sort of polynomial has infinitely many zeroes?

So, what sort of polynomials vanish on your set? That is the first question you want to ask.

• @mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V \cap \{b = 0\} = \{a \in \mathbf C : |a| = 1\}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw. – Trevor Gunn Dec 11 '18 at 5:23