the Zariski topology
/ zuh-RISS-kee /
A topology is just a rule for which sets count as closed. The natural rule in algebraic geometry is: a set is closed precisely when it is the common zero locus of some polynomials. This is the Zariski topology. It is deliberately coarse — it has very few closed sets — because the only shapes it can see are the ones that polynomials can cut out. On a line, the only proper closed sets are finite collections of points, since a one-variable polynomial has finitely many roots.
Formally, on affine space A^n the closed sets are exactly the algebraic sets V(S), and one checks the axioms: A^n and the empty set are closed (V(0) and V(1)), arbitrary intersections of algebraic sets are algebraic, and finite unions are algebraic too (V(I) union V(J) = V(IJ)). The open sets are the complements. A basis of opens is given by the distinguished opens D(f) = { p : f(p) is not 0 }, where a single polynomial is nonzero. This same recipe defines a Zariski topology on any variety (the subspace topology) and, more abstractly, on the spectrum of any commutative ring.
The Zariski topology is strange if you expect it to behave like the metric topology. It is almost never Hausdorff: on an irreducible variety any two nonempty open sets meet, so distinct points cannot be separated by disjoint opens. Nonempty open sets are huge and dense; closed sets are thin. This coarseness is a feature, not a bug — it makes irreducibility, generic points, and dimension into clean topological notions, and it is the only topology that is purely algebraic, defined without any reference to the field's metric or analytic structure. Caveat: continuity in the Zariski sense is far weaker than ordinary continuity, and compactness here (quasi-compactness) does not imply Hausdorffness.
On the affine line A^1 over an infinite field, the closed sets are exactly the finite subsets together with the whole line. So the open set 'complement of {0}' is everything except the origin — enormous, and it meets every other nonempty open set, which is why A^1 is not Hausdorff in this topology.
Closed = cut out by polynomials. Because polynomials have few zeros, closed sets are thin and open sets are dense.
The Zariski topology on A^2 is not the product of the Zariski topologies on the two factors A^1: the diagonal x = y is Zariski-closed in A^2 but is not closed in the product topology, a standard trap.