# 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.

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@Arturo Magidin, thank you. – wxu Apr 5 '12 at 17:08
To prove it false, it suffices to consider $K$ to be an algebraic extension of $\mathbb{Q}$, and $L$ finite over $K$. I think we can even reduce it to $L$ being a cyclic extension of prime order. – Hurkyl Apr 5 '12 at 17:43
(I retract my hypothesis we can reduce it to being cyclic: the argument I had in mind doesn't work) – Hurkyl Apr 5 '12 at 20:35

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 $a\in L\setminus K$.

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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.

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That does show that $L\setminus K$ must be algebraic over $\mathbb{Q}$. As we can reduce to the case where $L$ is generated by a single element, it reduces to the case where $K,L$ are finite extensions of $\mathbb{Q}$. Then, as $L$ (and its normal closure) is a finite extension of $\mathbb{Q}$, it contains only finitely many roots of unity. So your expression for $a$ can only take finitely many values. So $L\setminus K$ is finite and, as $K$ is infinite, this means that $L=K$. QED. – George Lowther Apr 5 '12 at 21:43
@George: That looks like it does it! I can incorporate that into my answer, but I really think you deserve some upvotes - maybe post that comment as a separate answer, and people can vote both of us up? – Zev Chonoles Apr 5 '12 at 21:57
Ok, I will. Maybe tomorrow morning as its getting late now. – George Lowther Apr 5 '12 at 22:30

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 : $K = L = L'$.

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I don't follow why it can be assumed to be Galois. – George Lowther Apr 5 '12 at 20:07
@ George : I think I was under the very mistaken assumption that any element of the Galois closure is a conjugate of an element of $L$. – mercio Apr 5 '12 at 20:20
Ok, next, why can you write $x^\sigma$ as a power of $\zeta$ multiplied by $x$? This would imply that $x^n\in K$. – George Lowther Apr 5 '12 at 20:28
@ George : thanks, I realize it is more complex than I thought. – mercio Apr 5 '12 at 20:57
Yes, showing that $k(x)$ is uniformly bounded seems like the tricky bit. But, I think Zev's answer + my comments achieve this. – George Lowther Apr 5 '12 at 21:00