Identify \(\AA^2\) with \(\AA^1 \times \AA^1\) in the natural way. Show that the Zariski topology on \(\AA^2\) is not the product topology of the Zariski topologies on the two copies of \(\AA^1\).
Erratum: an alternative argument sometimes given uses the hyperbola \(V(xy-1)\), whose projection \(\pi_x(V(xy-1)) = \AA^1 \sm \ts{0}\) is not closed, and claims that projections would be closed maps if \(\AA^2\) had the product topology. Projections from a product topology are open maps but need not be closed, so that argument does not work.
Solution
The diagonal \(\Delta = V(x-y)\) is Zariski closed in \(\AA^2\), and it is not closed in the product topology.
Suppose it were. Then \(\AA^2 \sm \Delta\) is open in the product, so it contains a basic box \(U \times V\) around each of its points. On \(\AA^1\) the Zariski topology is cofinite, so \(U = \AA^1 \sm \ts{p_1,\ldots,p_m}\) and \(V = \AA^1 \sm \ts{q_1,\ldots,q_n}\). Over an infinite field pick \(z\) distinct from every \(p_i\) and \(q_j\); then \((z,z)\) lies in \(U \times V\) and on \(\Delta\), contradicting \(U \times V \subseteq \AA^2 \sm \Delta\).
The product topology sees only finite unions of horizontal and vertical lines, and no curve that is not one; the Zariski topology on \(\AA^2\) has a closed set for every plane curve.