The Zariski topology on $\AA^2$ is not the product topology

The diagonal $\Delta = V(x-y) \subseteq \AA^2$ is Zariski closed. It is not closed in the product topology.

Suppose it were. Then $\AA^2 \sm \Delta$ would be open in the product, so it would contain a basic open box $U \times V$ around any of its points. Each factor carries the cofinite topology, 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 choose $z$ distinct from every $p_i$ and $q_j$; then $(z,z) \in (U \times V) \intersect \Delta$, contradicting $U \times V \subseteq \AA^2 \sm \Delta$.

The product topology is strictly coarser: it sees only finite unions of horizontal and vertical lines, and no curve that is not one. The same phenomenon recurs for schemes: $\Spec$ does not take products to products, so the product in the category of varieties is not the topological product.

Is the Zariski topology on $\AA^2$ the product of the Zariski topologies on the factors?

Dependencies

D-BIVAUThe Zariski topologyuses

Backlinks

P-AGHPRODTOPThe Zariski topology on $\AA^2$ against the product topologyrelated-to

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The dictionary