Is there a prime $p$ and a field $k$, not real closed, with $k^\times$ not $p$-divisible, such that there exists a finite extension $l/k$ such that $l^\times$ is $p$-divisible?
This question came up since I have proved a result in which I have the hypothesis that $l^\times$ is not $p$-divisible for any finite extension of $k$. If there are no such fields as described, then I can simplify this hypothesis to simply assuming that $k^\times$ is not $p$-divisible.


Of course $\mathbb{R}^\times$ is not $2$-divisible as $-1$ is not a square, while $\mathbb{C}^\times$ is divisible. However I was not able to produce essentially different examples, so I wonder whether there are any.
If $k$ is not real closed or algebraically closed, it has infinite index in its algebraic closure. However it is not clear to me whether $k$ can have finite index in the closure of $k$ under taking $p$-th roots.


Note that the field $k$ can not be the function field of a variety (other than a point) some field $k'$, because $k'(X\_0,\dots, X\_n)$ does not satisfy the property, since taking $p^n$-th roots of $X\_0$ gives an infinite tower of field extensions. It is also clear that $k$ cannot be a number field, but it is less clear if it can be an algebraic extension of $\mathbb{Q}$ or an infinite extension of some function field.



For which fields $k$ with $k^\times$ not $p$-divisible, does there exist finite $l/k$ such that $l^\times$ is $p$-divisible?
For odd $p$ there are no such fields: If $a\in k$ is not a $p$-th power, then for every $n$ the polynomial $X^{p^n}-a\in k[X]$ is irreducible (see for example Theorem 9.1 in Chapter VI of Lang's Algebra), hence if $l/k$ is any finite extension of degree $d$, and $p^n>d$, then this $a$ has no $p^n$-th root in $l$.


For $p=2$ one has for example all the formally real Euclidean fields: Take for example the field of constructible real numbers. It is not quadratically closed, but adjoining $\sqrt{-1}$ to it will make it quadratically closed. Not sure if that is an "essentially different" example now though.


