Intersections of two primitive field extensions of $\mathbb{Q}$

Question

After solving some exercises 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 generalized?

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?

Answer 1

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 $q\mid n.$

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