Let $k$ be an algebraically closed field, $n$ a positive integer, and consider the action of $\mathrm{GL}_n(k)$ on $M_n(k)$ by conjugation.
My professor tells me that semisimple conjugacy classes are closed, and his argument is as follows. Fix $f$ a monic polynomial of degree $n$, and consider the set of matrices $A_f$ whose characteristic polynomial is $f$. Then $A_f$ is closed, a finite union of orbits (by considering Jordan forms), and the orbit of minimal dimension is the unique semisimple conjugacy class with characteristic polynomial $f$. Since orbits of minimal dimension are closed, this semisimple orbit is closed.
A few questions I have to understand this proof:
- Why do we need to note that $A_f$ is closed? It is true, but doesn't the argument work without this fact?
- How can I see that the semisimple orbit in $A_f$ has minimal dimension?
- Consider the semisimple class $S_f$ of $2\times 2$ matrices defined by the characteristic polynomial $f(x)=(x-1)(x-2)$. Since $I_2$ satisfies this polynomial, wouldn't it be in the closure of $S_f$? That is, doesn't $I_2$ satisfy any polynomial equation that the elements of $S_f$ satisfy, and hence $I_2$ is in the Zariski closure? However, this can't be right because then $S_f$ is not closed (because $I_2\notin S_f$). What is wrong with my reasoning here?
I read only 3.
$S_f=\{\begin{pmatrix}a&b\\c&3-a\end{pmatrix}|a(3-a)-bc=2\}$. Then $S_f\subset k^3$ is defined by the relation $a(3-a)-bc=2$. Thus $S_f$ is Zariski-closed. The concept of annulator polynomial is different:
Let $T_f=\{A|f(A)=(A-I)(A-2I)=0\}$. Then $T_f=S_f\cup\{ I\}\cup \{2I\}$, that is $S_f$ $\cup$ two isolated points.