Chapter 1, Exercise 1.4: Product of topologies on $\mathbb A^1\times\mathbb A^1$

If we identify $\mathbb A^2$ with $\mathbb A^1\times\mathbb A^1$ in the natural way, show that the Zariski topology on $\mathbb A^2$ is not the product topology of the Zariski topologies on the two copies of $\mathbb A^1$.

This is clearly much more a problem in topology than in algebra, and in particular we can reduce it to a fairly well-known statement in general topology. We note that the diagonal $Z(y-x)$ in $\mathbb A^2$ is an algebraic set and thus closed. Recall that a topological space is Hausdorff if and only if its diagonal $\Delta =\{(x,x):x\in X\}$ is closed in the product $\mathbb A^1\times\mathbb A^1$. Thus, if $\mathbb A^1$ is not a Hausdorff space, then we have found a distinction.

To do this, we will use the fact that the Zariski topology on $\mathbb A^1$ is just the finite complement topology. More generally, we will show that the finite complement topology on any infinite set is not Hausdorff. Consider any such infinite space $X$ endowed with the finite complement topology and suppose $x,y\in X$ are separated by neighborhoods $x\in U,y\in V$. Then $(U\cap V)^c = X$ since $U\cap V=\emptyset$. By De Morgan, $U^c\cup V^c = X$, but $U^c,V^c$ are both finite and $X$ is infinite so this is a contradiction. Hence this topology is not Hausdorff as desired.