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Let $p$ be prime and suppose $\mathrm{char}F$ is coprime to $p$ and $F$ contains the roots of unity. Why is it true that $F[a^{\frac 1p}]=F[b^{\frac 1p}]$ if and only if there's so $i$ coprime to $p$ such that $a=b^ic^p$ for $c\in F$?

extending, why's it true that $[F[a_1^{\frac1p},\dots a_k^{\frac1p}]:F]=p^k$ if and only if $a_i$ does not belong to the group generated by $(F^\times)^p$ and the other $a_j$'s $i\neq j$?

I guess the first part makes sense intuitively and the characteristic is coprime just so you can't mess up and get zeros where you don't want em.. then again.. I really got no idea how to face this..

user355165
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With basic facts about Kummer extensions in place I think the first part goes as follows.

The case were $a$ is a $p$th power is straightforward, so assume that $a\notin (F^*)^p$. Then $x^p-a$ is irreducible over $F$, and $F(a^{1/p})/F$ is a cyclic Galois extension of degree $p$ with the well known automorphisms.

Let $\zeta\in F$ be a primitive $p$th root of unity. There is a unique $F$-automorphism $\sigma$ of $F(a^{1/p})$ satisfying the relation $\sigma(a^{1/p})=\zeta a^{1/p}$. Furthermore, $\sigma$ acts semisimply on $F(a^{1/p})$, in other words the direct sum decomposition $$ F(a^{1/p})=\bigoplus_{j=0}^{p-1}F\cdot a^{j/p} $$ is also its decomposition into eigenspaces of $\sigma$. The eigenspace belonging to eigenvalue $\zeta^j$ is $F\cdot a^{j/p}$. We found $p$ distinct eigenvalues in a $p$-dimensional vector space, so the claim follows.

Let us then assume that $F(a^{1/p})=F(b^{1/p})$. This implies that $b^{1/p}\notin F$ and that the minimal polynomial of $b$ over $F$ must be of degree $p$, hence equal to $m(x)=x^p-b$. What about $\sigma(b^{1/p})$? It must also be a zero of $m(x)$. In other words there is an integer $i,0\le i<p$, such that $\sigma(b^{1/p})=\zeta^ib^{1/p}$. So $b^{1/p}$ is an element of the eigenspace $F\cdot a^{i/p}$. The claim follows from this.

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Jyrki Lahtonen
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