Is it that easy to be Zariski dense? I've just heard about diagonizable matrices being Zariski dense as a consequence of Cayley-Hamilton, and this is the proof I came up with (in $\mathbb{C}$):
Let $O_{\cal F} = \{M\in{\cal M}_n(\mathbb C);\ \exists f \in {\cal F}, f(M)\neq 0\}$ be a non-empty open set, and $f\in {\cal F}$.
Let $\lambda$ be such that $P_\lambda = X - \lambda$ does not divide $f$. The matrix $\lambda I_n$ has $P_\lambda$ as its minimal polynomial, which does not divide  $f$, so $f(\lambda I_n) \neq 0$ and it belongs to  $O_{\cal F}$.
We showed that there are diagonizable matrices in every open set, Q.E.D.
This proof seems strange to me, we proved it quite easily, without directly involving Cayley-Hamilton, and in fact proved something much stronger that is to say that $\mathbb C I_n$, a line, is Zariski dense. Moreover every other proof I've found in 5 minutes of googling looked much more complicated and involved more advanced concepts. Basically my proofs generalizes to any set that contains matrices with minimal polynomials with arbitrary roots.
Did I understand Zariski's topology wrong, made a mistake in my proof, or is it really that easy to be Zariski dense ?
 A: You are confusing two quite different notions of what it means for a polynomial to not vanish on a matrix. The Zariski topology concerns polynomial functions in the entries of a matrix; for $n \times n$ matrices this is a polynomial algebra in $n^2$ variables $x_{ij}, 1 \le i, j \le n$. These polynomial functions take as input (the entries of) a matrix and return as output a scalar; you are thinking of polynomials (in one variable) in a matrix, which take as input a matrix and return as output a matrix.
$\mathbb{C} I_n$ is certainly not Zariski dense since the polynomials $x_{ij}, i \neq j$ all vanish on it. More generally, no proper vector (or affine) subspace can be Zariski dense.
A correct proof that the diagonalizable matrices are Zariski dense is the following: it suffices to show, and it is a little easier to show, that the set of matrices with distinct eigenvalues is Zariski dense. This is because this set of matrices is the open subset determined by the condition that the discriminant of the characteristic polynomial does not vanish, and because $\mathbb{A}^n$ is irreducible, every Zariski open subset is dense (this boils down to $\mathbb{C}[x_1, \dots x_n]$ being an integral domain, and it is worth carefully working through the definition of the Zariski topology as an exercise to see this).
