Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 24H (Algebraic Geometry)

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 every nonempty open subset of an irreducible variety is dense.

The coefficients of $\det (TI-A)$ are polynomial functions of the matrix entries. Its polynomial discriminant $\Delta (A)$ is therefore a polynomial in those entries, and it is nonzero exactly when all roots of the characteristic polynomial are distinct. Thus the required set is the principal open subset $D(\Delta )$ of affine space. It is nonempty: an algebraically closed field is infinite, so choose a diagonal matrix with $n$ pairwise distinct entries.

It is also affine as a variety. The map $A\mapsto (A,1/\Delta (A))$ identifies it with the closed hypersurfaceProjection to $A$ is its regular inverse. This proves the requested affineness while retaining the open embedding into matrix space.

Each entry of ${\mathrm{char}}_A(A)$ is a polynomial in the entries of $A$, so their simultaneous zero set $Z$ is Zariski closed. A matrix in $D(\Delta )$ has distinct eigenvalues and is diagonalizable. If $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$, thenThus $D(\Delta )\subseteq Z$. Affine space is irreducible because its coordinate ring is an integral domain; by part (i), its nonempty open subset $D(\Delta )$ is dense. The closed set $Z$ containing it must be all of affine space. This proves the Cayley-Hamilton theorem: