Let $n$ be a positive integer, $p$ a (positive rational) prime, and $\mathbb K$ an algebraically closed field. If ${\rm char}(\mathbb K) = 0$ then ${\rm GL}_n(\mathbb K)$ is divisible (see here). But what about ${\rm SL}_n(\mathbb K)$? Again, the answer is negative if $\mathbb K$ has finite characteristic, and again we may assume that (i) our matrices are in Jordan normal form and (ii) $\mathbb K$ is the complex field, thanks to the Lefschetz principle and the fact that, by Laplace's formula, the determinant of an $n$-by-$n$ "formal matrix" can be expressed as a wff in the first-order language $\mathcal L = (+, \cdot, -, 0, 1)$ of (the theory of) rings. Thus, let us consider a Jordan matrix of size $n \times n$, say $J = {\rm diag}(J_1, \ldots, J_m)$, where $J_i$ is a Jordan block of size $k_i \times k_i$. If $\det(J) = 1$ and $p \nmid \gcd(k_1, \ldots, k_m)$, it is not difficult to prove that there exists $A \in {\rm SL}_n(\mathbb C)$ such that $A^p = J$.

Question 1. What about the other cases?

The problem should be well-known and I feel that the answer is in the negative, but so far I couldn't either get a reference or find a counterexample by myself. EDIT 2. The $2$-by-$2$ Jordan block with eigenvalue $-1$ is a counterexample, and in some sense it is the only one possible for $n = 2$ (see the comments below). Thus, it seems natural to ask the following:

Question 2. (i) Is there any "explicit characterization" of those matrices in ${\rm SL}_n(\mathbb C)$ which fail to have a $p$-th root for each prime $p$? (ii) In particular, is the set of these matrices the union of a finite number of conjugacy classes of ${\rm SL}_n(\mathbb C)$?

Let me rely on your common sense for the actual meaning to give to the expression "explicit characterization". Last but not least:

Question 3. Does anyone know where to find a reference for this kind of questions (concerned with the divisibility of specific subgroups of ${\rm GL}_n(\mathbb K)$ when $\mathbb K$ is an algebraically closed field, either in characteristic zero or not)?

Thanks in advance for any help.

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    $\begingroup$ Did you try the obvious place to start, namely a $2\times 2$ Jordan block with $-1$s on the diagonal? $\endgroup$
    – user30035
    Mar 21, 2013 at 11:17
  • $\begingroup$ Ehr... Yes, but somehow I missed this "limit case" while working to something slightly more general: If $J = \left[\begin{array}{cc} \lambda & \mu \\ 0 & \lambda^{−1}\end{array}\right]$ for some $\lambda \in \mathbb C^\times\setminus\{−1\}$ and $\mu \in \mathbb C$, then $\left[\begin{array}{cc} a & c \\ 0 & b\end{array}\right]^2=J$ for $a=|\lambda|^{1/2}e^{i\frac{\theta}{2}}$, $b=a^{−1}$ and $c=(a+b)^{−1} \mu$, where $\theta$ is (the principal value of) the complex argument of $\lambda$. I edited the OP, fixed my mistake, and updated Q2. Thanks! $\endgroup$ Mar 21, 2013 at 13:31

1 Answer 1


$SL_n(k)$ is not divisible over any algebraically closed field $k$ and $n>1$.

This would be indeed a nice exercise in a Linear Algebra course.

Proof: Let $\zeta\neq 1$. I'll show that there is no $X\in SL_n(k)$ such that $X^n=A$ for
$$A=\scriptstyle\begin{pmatrix} \zeta & 1 & & & \newline & \zeta & 1 & & \newline & & \ddots & \ddots & \newline & & & \zeta & 1 \newline & & & & \zeta \end{pmatrix}.$$

Let $X^n=A$ and let $\mu_1,...,\mu_n$ be the eigenvalues of $X$. Clearly $\mu_i^n=\zeta$. Each eigenvector of $X$ is also an eigenvector of $A$ and since $A$ has exactly one eigenvector (up to scalar multiples), we conclude $\mu_1=\ldots=\mu_n$. Hence $\det(X)=\mu_1^n=\zeta\neq 1$. qed.

  • $\begingroup$ Clear and very nice. I really wonder if one can cluster all the matrices in ${\rm SL}_n(\mathbb C)$ which don't have a $p$-th root for some prime $p \le n$ in a finite number of conjugacy classes: Your example is still very particular, which tempts me to think that there may be only "few" exceptions. $\endgroup$ Mar 21, 2013 at 21:26
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    $\begingroup$ Of course this example requires $\zeta^n = 1$ (or else $A \not\in \operatorname{SL}_n(k)$), and so I think doesn't work when $n$ is the characteristic of $k$. $\endgroup$
    – LSpice
    Oct 23, 2015 at 2:20

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