The Zariski topology on an affine product is generally not the product topology

By [F1], the affine product has coordinate ring $\mathbb{C}[x,y]$, so the diagonal $D=\{(c,c):c\in \mathbb{C}\}$ is the algebraic set $V(x-y)$ and is Zariski closed.

Each factor has the cofinite Zariski topology by [F2]. For a point $(a,b)$ off the diagonal, every basic product neighbourhood $U\times V$ of it has $U\cap V\mathrm{\ne }\mathrm{\varnothing }$, because both $U$ and $V$ are cofinite in the infinite field $\mathbb{C}$. Choosing $c\in U\cap V$ gives $(c,c)\in (U\times V)\cap D$. Thus no product-topology neighbourhood of $(a,b)$ is contained in the complement of the diagonal.

Hence the diagonal is not product-topology closed although it is Zariski closed. For arbitrary affine varieties $X,Y$, the product property in [F1], applied to a one-point affine variety, identifies points of $X\times Y$ with pairs $(x,y)$; the fibre of the first projection over $x$ is therefore set-theoretically $\{x\}\times Y$. In this example its coordinate ring is $\mathbb{C}[x,y]\mathrm{/}(x-a)\cong \mathbb{C}[y]$. Thus the expected fibres coexist with the topology difference.