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

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

• Re: intuition, its eigenvalues are the conjugates! Commented Aug 7 at 17:39

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.
• Did you mean to say iff $\alpha$ is separable in the end? Commented Aug 7 at 16:06
• Also, doesn't this contradict the particular case in my answer where $K = \mathbb F_2(t)$ and $L = K[\sqrt{t}]$? It seems like multiplication by $\sqrt{t}$ is still diagonalizable here, despite $\sqrt{t}$ not being separable. Commented Aug 7 at 16:52
• Just realized, this is unnecessarily complicated. A linear map is diagonalizable (over the algebraic closure) iff its minimal polynomial is separable, and the minimal polynomial of the map $x\mapsto \alpha x$ is is the same as that of $\alpha$ Commented Aug 7 at 16:55
• @Carla_ No, in that case it won't be diagonalizable. In the basis $1,\sqrt t$ the matrix is $A=\begin{pmatrix}0&t\\1&0\end{pmatrix}$ with only eigenvalue $\sqrt t$ (of algebraic multiplicity $2$), so if it were diagonalizable, we would have $A = \sqrt tI$ which is not the case Commented Aug 7 at 16:58