I've seen the statement "any radically closed field contains all roots of unity." Though the term "radically closed field" doesn't seem to be extremely common, I'm fairly confident that it means that the multiplicative group $K^\times$ is a divisible group.
As stated, this result is false, because if we take $K = \mathbb{F}_2$ then $\mathbb{F}_2^\times = \{1\}$ is the trivial group, which is divisible, but $\mathbb{F}_2$ does not contain all roots of unity: for instance, we have $$x^3 - 1 = (x-1)(x^2 + x + 1)$$ and $x^2 + x + 1$ is irreducible over $\mathbb{F}_2$.
However, given that this is an extremely trivial case -- the trivial group is the only finitely-generated divisible group, for instance -- I wonder if the statement can be repaired with some minor additional assumption. Unfortunately, the intended proof here isn't obvious to me.
Question: What is the correct statement here (or what are the correct definitions), and how is the statement proved?