Consider the Zariski topology on $\mathbb{C}^2$. How can I prove that a set is not an algebraic variety? For example let $Y=\{(z_1,z_2)\in \mathbb{C}^2| \text{Im}(z_1) \cdot \text{Im}(z_2)>0\}$ be a subset of $\mathbb{C}^2$. I think that it is "too large" to be an algebraic set, but I don't know how to prove it. I think that I can prove it with the theorem:
An open, not empty, set in the Zariski topology is dense.
But I have this exercise before the proof of this theorem. So there is a way to solve the exercise without the theorem but I can't find it. And after that I have to say if it is irreducible or not, but I have the same problem.
As one of the commenters pointed out, if $Y\subseteq \mathbb{C}^2$ is algebraic, then for each $z_2\in \mathbb{C}$, the set $\{z\in \mathbb{C}:(z,z_2)\in Y\}$ is also algebraic. In particular, this is the fibre of the projection map $\pi_2:\mathbb{C}^2\to \mathbb{C}^1$ i.e. the set I just wrote is $\pi_2^{-1}(z_2)$. Now suppose the fibre $\pi_2^{-1}(z_2)$ is an algebraic subse $\mathbb{C}^1$. For ease, pick $z_2$ such that $\operatorname{Im}(z_2)>0$.
It is known that the algebraic subsets of $\mathbb{C}^1$ are either finite sets of points, the emptyset, or the whole of $\mathbb{C}^1$. So one just needs to check that for fixed $z_2$, there are infinitely many $z\in \mathbb{C}$ s.t. $(z,z_2)\in Y$ and that not every $z\in \mathbb{C}$ has $(z,z_2)\in Y$. On fixing $\operatorname{Im}(z_2)$, we can consider $z=ai$ for $a$ a positive real number which are in the fibre. However, $z=0$ is not in the fibre. So the fibre cannot be algebraic.