Isomorphism problem for two radical extensions Let $n\geq 2$ and let $a,b\in{\mathbb Q}$. Suppose that both the
polynomials $A=X^n-a$ and $B=X^n-b$ are irreducible. We want to know whether
 ( * ) there is a root $\alpha$ of $A$ and a root $\beta$ of $B$ such that 
${\mathbb Q}(\alpha)={\mathbb Q}(\beta)$, or if you prefer whether the quotient rings $\frac{{\mathbb Q}[X]}{A(X)}$ and $\frac{{\mathbb Q}[X]}{B(X)}$ are isomorphic. An obvious sufficient 
condition for ( * ) is that $\frac{b^j}{a^i}$ be a perfect $n$-th power
in $\mathbb Q$ for two positive integers $i,j$ coprime to $n$. Is it a necessary condition also ?
I also posted this question on MSE : https://math.stackexchange.com/questions/1036372/isomorphism-problem-for-two-radical-extensions-of-the-same-degree
 A: With the recent change, I believe the answer to your question is now yes, at least in the case of positive rationals.
Let $a,b\in\mathbb{Q}_{>0}$ (I'll leave the cases when one of $a$ or $b$ is negative for you to work out) and assume $X^n-a$ and $X^n-b$ are irreducible over $\mathbb{Q}[X]$, for some $n\geq 2$.  Let $\alpha,\beta$ be positive real roots of the respective polynomials, and further assume $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.  This means we can write
$$
r_0+r_1\alpha+r_2\alpha^2+\cdots + r_{m-1}\alpha^{n-1}=\beta
$$
for some elements $r_0,r_1,\ldots, r_{n-1}\in \mathbb{Q}$.  After dividing by enough copies of $\alpha$ we have
$$
(\ast) \qquad s_0 + s_1\alpha+\cdots + s_{n-1}\alpha^{n-1} = \beta/\alpha^i
$$
where $s_0\neq 0$, $0\leq i\leq n-1$, and $s_0,s_1,\ldots, s_{n-1}\in\mathbb{Q}$.
Now, I claim that ${\rm Tr}(\alpha^j)=0$ for all $1\leq j\leq n-1$, which will follow from the fact that $\alpha^j$ is a root of the irreducible polynomial $X^{n/\gcd(n,j)}-a^{j/\gcd(n.j)}$.
Important Note: Since $X^n-a$ is irreducible, $a$ is not a $p$th power for any prime $p|n$.  Thus, $a^{j/\gcd(n,j)}$ is also not a $p$th power for any prime $p|(n/\gcd(n.j))$.  By the result mentioned in the answer to this question, we know that the polynomial above is irreducible.  (This is one place where you will have to be more careful if $a<0$.)
Since $s_0\neq 0$, the trace (computed from $\mathbb{Q}(\alpha)$ down to $\mathbb{Q}$) of the left-hand side of $(\ast)$ is nonzero.  Hence ${\rm Tr}(\beta/\alpha^i)={\rm Tr}(\sqrt[n]{b/a^i})\neq 0$.  Writing $b/a^i=c^m$ where $m|n$ and $c>0$ is not a $p$th power for any prime $p|(n/m)$, we see that the minimal polynomial for $\beta/\alpha^i$ is $X^{n/m}-c$.  (This is the more crucial place where we use positivity.)  The only way the trace can be nonzero is if $n=m$, hence $b/a^i$ is an $n$th power.
