Let $k>0$ be an integer, let $R$ be a ring (commutative, unital), which contains $\mathbb{Q}$ (i.e. with a ring homomorphism $\mathbb{Q}\to R$) and all $k$-roots of unity. The examples I have in mind are polynomial rings over $\mathbb{C}$.


Is every element of $\mathrm{SL}(n,R)$ of order $k$ diagonalizable?



Is every element of $\mathrm{SL}(n,R)$ of finite order diagonalizable?
Yes. One way to see this is to decompose the group ring $R[G],$ where $G$ is a cyclic group of order $k$, into the product of simple rings. It follows from your assumptions that this is possible and since $G$ is cyclic, every simple component is isomorphic to $R.$ An $n\times n$ matrix $g$ of order $k$ with entries in $R$ gives rise to an $n$-dimensional representation of $G$ over $R$, or equivalently, an $R[G]$-module that is free of rank $n$ as an $R$-module. By the above decomposition, this module is a direct sum of one-dimensional representations. Translating this into the language of matrices, we get an eigenbasis for $g$ and hence $g$ is diagonalizable over $R.$ 


