Unit group of an algebraically closed field is divisible

In the lecture notes on Valuation theory, in Ex. $1.16$ on page $11$ we are asked to show that:

If $k$ is an algebraically closed field, then $k^{\times}$ is a divisible abelian group.

Isn't $k = \overline{F_p}$, the algebraic closure of $\Bbb Z/(p)$ a counter example? We will never be able to find a $y$ such that $py = x$ for any non zero $x$ since $py = 0$.

Further, if we assume that $k$ has infinite characteristic, the property will always hold irrespective of whether or not the field is algebraically closed, correct?

• The group here is multiplicative, so instead of looking for solutions of $py=x$ you are expected to look at $y^p=x$. In other words you need roots of arbitrary order. This is clear for algebraically closed fields. – Jyrki Lahtonen Jun 30 '15 at 21:01

The claim is rather that for all $n\ge1$ and $a\in k$ with $a\ne 0$ the polynomial $X^n-a$ has a root. As $k$ is assumed algebraically closed, such root certainly exists.