Consider the $n\times n$ matrix $$ M=\begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & 0 & -c_0\\ 1 & 0 & 0 & \cdots & 0 & 0 & -c_1\\ 0 & 1 & 0 & \cdots & 0 & 0 & -c_2\\ \vdots & \vdots & \vdots & \ddots & \vdots &\vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & 0 & -c_{n-2}\\ 0 & 0 & 0 & \cdots & 0 & 1 & -c_{n-1}\\ \end{pmatrix} $$ with rational entries. It is called the companion matrix of its charateristic polynomial $p_M$ which equals its minimal polynomial. Assume that $p_M$ is irreducible over $\mathbb{Q}$. By considering $p_M$ over the extension $\mathbb{C}\supset\mathbb{Q}$, it splits as $$ p_m(x)=(x-a_1)^{e_1}\cdots (x-a_k)^{e_k}, $$ but according to Wikipedia, $M$ does not have to be diagonalizable in general.
What is an example of an $M$ (such that $p_M$ is irreducible over $\mathbb{Q}$) that is not diagonalizable over the complex numbers?
As $p_M$ is irreducible over $\mathbb{Q}$, the matrix $M$ defines an algebraic field extension $L=\mathbb{Q}[x]/(p_M)$ over $\mathbb{Q}$ of degree $n$. How is the property that $M$ is diagonalizable over $\mathbb{C}$ reflected in properties of the field extension?
Edit Sorry, I want to assume that $p_M$ is irreducible over $\mathbb{Q}$, of course.
We have
$$A=\Bbb C[M]\cong\frac{\Bbb C[x]}{(p(x))}\cong\prod_{i=1}^k \frac{\Bbb C[x]}{(x-a_i)^{e_i}}.$$
Thus the Jordan blocks $J(\lambda,n)$ of $M$ are $J(a_i,e_i)$ for $i=1,\cdots,k$. A matrix $M$ is diagonalizable over $\Bbb C\Leftrightarrow$ all its Jordan blocks have size $1$, so $M$ is diagonalizable $\Leftrightarrow p(x)$ is separable.
There was (edit: originally) no hypothesis that $p$ was irreducible, so $\Bbb Q[x]/(p(x))$ needn't be a field.
What we can say is that the maximal ideals of $A$ are in bijection with $p$'s roots, and the $\Bbb C$-algebra $A$ is reduced $\Leftrightarrow M$ is diagonalizable. This relates to the rupture ring $R=\Bbb Q[x]/(p(x))$ via the complexification $A\cong R\otimes_{\Bbb Q}\Bbb C$.