The title has it all. I'm looking for a proof/disproof of the fact that an algebraically closed field, say $\mathbb K$, has characteristic zero iff the following property (R) holds: For all $n,k \in \mathbb N^+$, every invertible $n$-by-$n$ matrix with entries in $\mathbb K$ has at least one $k$-th root. The question is certainly well-known, and boils down to the case of Jordan blocks. I myself have a sense, but not a proof, that it *must* have an answer in the positive. I'm not interested in the discussion of special cases (e.g., the complex case is quite standard, and can be treated even analytically), unless of course the inspection of a finite, small number of them leads to a general conclusion. In case of an affirmative answer, I'd appreciate much a reference to the result in its full generality. As for motivation, the question is 'naturally' related to another that I've [recently posted](https://mathoverflow.net/questions/124305/extending-riesz-lemma-to-special-classes-of-normed-groups). Thanks in advance for any help.



All and the only algebraically closed fields s.t. any regular n-by-n matrix has a k-th root for every k
wccanard's comment gives one direction: If the field has characteristic $p$, then there is no matrix $A$ with $A^p=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$.


For the other direction, let the field have characteristic $0$. We want to show that $A^k=B$ has a solution $A$ for any $B$. Without loss of generality, $B$ is a Jordan block with eigenvalue $1$, so $B-1$ is nilpotent. Then $A=\sum\_{i\ge0}\binom{1/k}{i}(B-1)^i$ is a finite sum with $A^k=1+(B-1)=B$.


