The following is my proof of the assertion that the set of diagonalizable matrices is Zariski-dense in $M_n(\mathbb F)$. Is this right ?
Let $\mathbb F$ be an infinite field (not necessarily algebraically closed)and $M_n(\mathbb F)$ the set of all $n \times n$ matrices with entries in $\mathbb F$.
We denote by $D_n(\mathbb F)$ the set of $n \times n$ diagonalizable matrices with entries in $\mathbb F$.
For each $A \in M_n(\mathbb F)$, we denote by $d(A)$ the discriminant of the characteristic polynomial of $A$.
Since $d(A)$ is a polynomial in the entries of $A$ with coefficients in $\mathbb F$,
the set $U := \{ X \in M_n(\mathbb F) : d(X) \not = 0 \}$ is Zariski-open.
(Here, we are identifying $M_n(\mathbb F)$ with ${\mathbb A}^{n^2}$.)
It follows from the fact that ${\mathbb A}^{n^2}$ is irreducible that $U$ is Zariski-dense in ${\mathbb A}^{n^2}$. As $U$ is contained in $D_n(\mathbb F)$, $D_n(\mathbb F)$ is also Zariski-dense in ${\mathbb A}^{n^2}$.
Thanks in advance.
Best Answer
Your proof is only correct if by "diagonalizable" you mean "diagonalizable" over an extension field of $\mathbb F$.
However, in my experience this is not the most usual interpretation of diagonalizable.
The rotation $\begin {pmatrix} 0&-1\\1&0\end {pmatrix}$ by $\pi/2$ in the plane over $\mathbb R$ for example is not diagonalizable over $\mathbb R$, even though its characteristic polynomial is $X^2+1$ has nonzero discriminant.
In your proof however it counts as diagonalizable, and that is the controversial point.
Edit
I have just checked that Hoffman-Kunze explicitly write on page 185 of their Linear Algebra that the above matrix is not diagonalizable.