Is the linear operator given by multiplication by an element of an algebraic field extension always diagonalizable?

Question

I noticed this to be true for nontrivial extensions given by adjoining a square root. (By diagonalizable i mean over an algebraic closure). Given a nontrivial field extension of form $k\hookrightarrow k[\sqrt{\sigma}]$, and an element $\alpha := a+b\sqrt{\sigma}$ of it, the linear operator multiplication by $\alpha$ can be represented, in the basis $(1, \sqrt{\sigma})$, by the matrix $\begin{bmatrix} a & \sigma b\\ b & a\end{bmatrix}$. And, over an algebraic closure of $k$, this matrix can be easily seen to be diagonalizable*.

Is this always the case? Is there any intuitive reason for this to be the case here and/or in general? I'm particularly interested in finite extensions of $\mathbb Q$, but an answer for finite extensions of other fields would be interesting too.

*Actually this can fail in characteristic 2

Answer 1

A linear map is diagonalizable (over the algebraic closure) if and only if its minimal polynomial is separable. The minimal polynomial of $x\mapsto \alpha x$ coincides with that of $\alpha$, so the map is diagonalizable (over the algebraic closure) iff $\alpha$ is separable.

Original answer: Let $L/K$ be finite and $\alpha\in L$. We may assume that $L=K(\alpha)$ because the matrix for $x\mapsto \alpha x$ in $L/K$ is block diagonal with the blocks coming from the matrix of $x\mapsto \alpha x$ on $K(\alpha)/K$. Then $L=K[t]/(p)$ where $p$ is the minimal polynomial of $\alpha$. So $L\otimes K^{\mathrm{alg}}=K^{\mathrm{alg}}[t]/(p)$ and multiplication by $\alpha$ corresponds to multiplication by $t$. This is diagonalizable iff $p$ has only simple roots, so in conclusion the map is diagonalizable over the algebraic closure (normal hull is sufficient) iff $\alpha$ is separable.