TITLE: Every element is radical in a field extension.
QUESTION [7 upvotes]: Let $L/K$ be an algebraic field extension. Suppose for each $x\in L$, there exists an integer $n>0$ such that $x^n\in K$, where $n$ may depend on $x$. If the characteristic of $K$ is zero, does it follow that $L=K$?
Thanks.

REPLY [3 votes]: Yes, it does follow that $L=K$.
Let's start by supposing that $L\not=K$ so that we can choose some $a\in L\setminus K$. We can express $a$ in terms of roots of unity (as in Zev Chonoles' answer). The minimal polynomial over $K$ of $a$ is of degree greater than 1 so, as we are assuming nonzero characteristic, it must have at least two roots. Let $\tilde a\not=a$ be any other root, which will lie in the normal closure of $L/K$. As $a^n\in K$ for some $n > 0$, the minimal polynomial of $a$ divides $X^n-a^n$ and, hence, $\tilde a^n=a^n$. So, $\tilde a=\zeta a$ for an $n$th root of unity $\zeta$. Similarly, $(a+1)^m\in K$ for some $m$ so $\tilde a+1=\eta(a+1)$ for an $m$th root of unity $\eta\not=1$. Rearranging $\tilde a=\zeta a=\eta(a+1)-1$ gives
$$
\begin{align}
a=\frac{\eta-1}{\zeta-\eta}&&{\rm(1)}
\end{align}
$$
for roots of unity $\zeta\not=\eta\not=1$ in the normal closure of $L/K$.
As $K$ has characteristic zero, it contains the rationals and (1) shows that $a$ is algebraic over $\mathbb{Q}$. We can reduce to algebraic extensions of the rationals by setting $\tilde L=\mathbb{Q}(a)$ and $\tilde K=\mathbb{Q}(a)\cap K$. Every element of $\tilde L$ is a radical of an element of $\tilde K$, and $\tilde L,\tilde K$ (and their normal closures) are finite extensions of $\mathbb{Q}$.
Using the same argument as above, every element $b\in\tilde L\setminus \tilde K$ can be expressed as in (1) for roots of unity $\zeta,\eta$ in the normal closure of $\tilde L$. However, being a finite extension of $\mathbb{Q}$, the normal closure of $\tilde L$ only contains finitely many roots of unity, so (1) shows that $\tilde L\setminus\tilde K$ is finite. This is impossible for $\tilde L\not=\tilde K$. For example, $a+\mathbb{Z}$ is an infinite subset of $\tilde L\setminus \tilde K$, contradicting the initial choice of $a\in L\setminus K$.