a question on 1-element field extensions of $\mathbb{Q}$ After solving some excercises in algebra, basic information on field extensions,  I have a question.
It's easy to prove that $\mathbb{Q}(\sqrt{2})\cap\mathbb{Q}(\sqrt{3})=\mathbb{Q}$, also it can be proven that $\mathbb{Q}(\sqrt[n]{2})\cap\mathbb{Q}(\sqrt[n]{3})=\mathbb{Q}$ for all $n\ge 2$.
How can it be generalised?
If $n\ge 2$, then $\mathbb{Q}(\sqrt[n]{a})\cap\mathbb{Q}(\sqrt[n]{b})=\mathbb{Q}$ for any $a,b$:
relatively prime? or maybe $a\nmid b$ and $b\nmid a$ suffices?
 A: The maximalist case is:

If $a=p_1^{a_1}\cdots p_k^{a_k}, b=p_1^{b_1}\cdots p_k^{b_k}$, then $\mathbb Q(\sqrt[n]a)\cap\mathbb Q(\sqrt[n]b)\neq \mathbb Q$ if and only if there is a pair of positive integers $q,r\in\{1,2,\dots,n-1\}$ so that for each $i$, $a_iq\equiv b_ir\pmod{n}$ some $a_iq\not\equiv 0\pmod n.$

When the numbers are relatively prime, you get one of each $a_i,b_i$ pair is zero, so when the condition is true, necessarily all of $a_iq\equiv 0\pmod n.$
$a=12,b=18,n=3$ gives a counterexample to your $a\not\mid b$ and $b\not\mid a$ question. Here, $q=1,r=2$ and you get:
$$\left(\sqrt[3]{18}\right)^2=3\sqrt[3]{12}$$
If we think of $A=(a_1,\dots,a_n)$ and $B=(b_1,\dots,b_n)$ as "vectors" over $\mathbb Z/n\mathbb Z$,[*] then for the two fields to have trivial intersection, we require that, for all $q,r\in\mathbb Z/n\mathbb Z$, $qA=rB$ if and only if $qA=rB=(0,\dots,0)$.
If we can find such a $q,r$ with $qA=rB\neq(0,\dots,0)$, we can find such a $q,r$ with $q\mid n.$ 

[*] Technically, unless $n$ is prime, this is not a vector space, but a $\mathbb Z/n\mathbb Z$-module.
