Let $X \subset k^m , Y \subset k^n$ be algebraic sets ($k$ an algebraically closed field). Then $X\times Y \subset k^{m+n}$ is an algebraic set whose Zariski topology is finer than the product topology of the Zariski topologies of the factors. Could someone give me an example that demonstrates that it can be strictly finer?
[Math] Open in Zariski but not product topology
algebraic-geometry
Best Answer
(I will take $n=m=1$, you will adapt in higher dimensions.) Let $P = X-Y$ and $Z = V(P)$, so that $Z$ is closed. Now $Z = \{(x,x)\;|\; x\in \mathbf{A}_k^1\}$. Now (easy) $Z$ is closed in the product topology if and only if $\mathbf{A}_k^1$ is Hausdorff, and this is not the case as $k$ is infinite, being algebraically closed, and as (independently of $k$ finiteness) the closed sets of the affine line are the whole affine line and finite sets.