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 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.