An <affine variety> is irreducible when it is nonempty and cannot be expressed as the union of two proper <Zariski-closed sets>. If a nonempty open subset $U$ were not dense, $X=\overline U\cup(X\setminus U)$ would be such a union: the first set is proper by assumption, and the second is proper because $U$ is nonempty. Hence \b[every nonempty open subset of an irreducible variety is dense].

Solution

Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 24H / i / Solution

 Solution (source code)