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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?

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    $\begingroup$ I would have interpreted it to mean that the polynomials $x^n - a$ split (so, unlike your definition, one doesn't just ask it to have some root). This trivially implies the desired statement by setting $a = 1$. $\endgroup$ – Qiaochu Yuan Jun 2 '15 at 3:11
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    $\begingroup$ Notice Qiaochu's coindition is that the field does not have nontrivial radical extensions, which is a sensible meaning for "closed". $\endgroup$ – Mariano Suárez-Álvarez Jun 2 '15 at 3:20
  • $\begingroup$ Yes, this is surely correct. I'd have realized this if it weren't for the fact that I saw the erroneous "$K^\times$ is divisible" definition written down (and happened to want to consider such a situation), thereby short-circuiting my critical thinking faculties. If someone wants to write this up as an answer I'll accept it. $\endgroup$ – Daniel McLaury Jun 2 '15 at 3:23
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I think the definition is that the polynomials $x^n - a$ split (so instead of asking that they have one root we ask that they have all roots); equivalently, as Mariano points out in the comments, $K$ has no nontrivial radical extensions. The desired statement about roots of unity follows from setting $a = 1$.

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