General linear group over algebraically closed field of characteristic zero is divisible

Statement

Suppose $K$ is a field that is algebraically closed of characteristic zero and $G$ is a general linear group of finite degree $n$ over $K$, i.e., $G=GL(n,K)$. Then, $G$ is a divisible group, i.e., for any $g\in G$ and any positive integer $m$, there exists $x\in G$ (not necessarily unique) such that $x^m=g$.

Note that the characteristic zero assumption is necessary; when $K$ has finite prime characteristic $q$, then for $n\ge 2$, $GL(n,K)$ is divisible only for the $m$ that are not divisible by $q$.

Related facts

• Triangulability theorem
• Special linear group over algebraically closed field of characteristic zero is divisible precisely by those primes that do not divide its degree

Proof

The idea is to first conjugate to a Jordan canonical form matrix, then take the $m^{th}$ root of each block (fixed up to multiplication by $m^{th}$ roots of unity in each block, with the root of unity potentially varying by block).

Significance of the field being algebraically closed

The field being algebraically closed is used in two ways in the above constructions:

• Jordan normal form is available for every element because the field is algebraically closed.
• Roots of field elements can be taken because the field is algebraically closed.

Significance of characteristic zero

Characteristic zero is necessary for the original statement to hold; in finite prime characteristic $q$, the matrix below has no $q^{th}$ root even in $GL(2,K)$; similar examples can be constructed for higher orders:

$\left(\begin{array}{cc}1& 1\\ 0& 1\\ \end{array}\right)$

The way the proofs above fail when taking $q^{th}$ roots in characteristic $q$ is based on a subtle detail not explicitly spelled out in the proofs above. When calculating the value of the superdiagonal element in the $q^{th}$ root of a matrix in Jordan normal form, we get a formula that involves division by $q$, which is not a legal operation in characteristic $q$.

In finite prime characteristic $q$, the proof still works as long as $q$ does not divide $m$. Note that at $n=1$, we don't even need the assumption that $q$ does not divide $m$.