Every element is radical in a field extension. 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.
 A: Here is a very incomplete idea. Perhaps someone can finish it, or point out why it won't work.
Let $a\in L\setminus K$, and let $n>1$ be the smallest number such that $a^n\in K$. Then the minimal polynomial for $a$ over $K$ divides $x^n-a^n$. Then $(x-1)^n-a^n$ is a polynomial having $a+1$ as a root, so that the minimal polynomial for $a+1$ must divide $(x-1)^n-a^n$.
But we also know that $a+1$ is the root of some polynomial of the form $x^m-(a+1)^m$, so the minimal polynomial for $a+1$ must also divide $x^m-(a+1)^m$. The roots of $(x-1)^n-a^n$ are
$$1+\zeta_n^ka,\quad 0\leq k<n$$
and the roots of $x^m-(a+1)^m$ are 
$$\zeta_m^\ell(a+1),\quad 0\leq \ell<m.$$
If 
$$(1+\zeta_n^ka)=\zeta_m^\ell(a+1)$$
then
$$a=\frac{1-\zeta_m^\ell}{\zeta_m^\ell-\zeta_n^k}.$$
If we can show that there must always be at least one $a\in L\setminus K$ that can't be expressed this way, for any algebraic extension $L/K$ of characteristic zero fields, then we're done; I suppose it seems plausible, but I don't see any way of proving such a thing.
A: 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 $a\in L\setminus K$.
A: We can suppose that $K \subset L$ is a finite extension (because any subextension still satisfies the property).
Let $L'$ be the Galois closure of $L$ and let $n=[L':K]$.
Euler's $\varphi(x)$ function diverges to $+\infty$ as $x \to \infty$, so we can pick an integer $p$ such that $\varphi(x)\le n \Rightarrow x \le p$.
Pick any $\sigma \in Aut_K(L')$.  
For all $x \in L$, by assumption, there is an integer $m$ so that $x^m \in K$. Since $x$ can have at most $n$ distinct conjugates in $L'$, we can pick $m \le p$.  Then $x^\sigma = \zeta_x x$ where $\zeta_x$ is an $m$-th root of unity,. In particular, $\zeta_x$ is also an $p!$-th root of unity.
For any $x,y \in L^*$ and $t \in K$, write $\zeta_{x+ty}(x+ty) = (x+ty)^\sigma = x^\sigma+ty^\sigma = \zeta_x x+t\zeta_y y$.
If we find a $p!$-th root of unity $\zeta$ such that $\zeta_{x+ty} = \zeta $ for more than one values of $t$, then we have just found that the polynomial $\zeta (x + Ty) - x^\sigma - Ty^\sigma$ is a degree $1$ polynomial that has more than one root in $K$, so it has to be $0$, which implies that $\zeta = \zeta_x = \zeta_y$.
But there are finitely many $p!$-th roots of unity, while $K$ is infinite. so this always happen, and forall $x \in L^*$, $\zeta_x = \zeta_1 = 1$, and so $\sigma|_L = id_L$. So $L$ is Galois and there is no nontrivial automorphism : $K = L = L'$.
