Inspired by $$ $$ http://mathoverflow.net/questions/65738/when-matrix-multiplication-commutes $$ $$ and $$ $$ http://www.imdb.com/title/tt0293702/ $$ $$ is it true that, when $$ A \in SL_n(\mathbf Z),$$ then all integral matrices that commute with $A$ are an integral (or at least rational) polynomial in $A$? I dimly recall proving this for a specific 3 by 3 $A$ that was all 0's and 1's, so calculations were easy. The use of the unit determinant is that $A^{-1}$ is an integral polynomial in $A$ by Cayley-Hamilton. The degree of the polynomial need be no larger than $n-1,$ also by Cayley-Hamilton.

With or without commutativity, I once made a "multiplicative" function out of $$ f(x_0, x_1, \ldots, x_{n-1}) = \det (x_0 I + x_1 A + x_2 A^2 + \cdots + x_{n-1} A^{n-1}), $$ amounting to a kind of fake norm form. The guy I asked about it laughed at me but said that's what I had.