TITLE: Square and cubic roots in $\mathbb Q(\sqrt n)$
QUESTION [7 upvotes]: Here is my question :

Let $n$ a squarefree positive integer, $m \ge 2$ an integer and $a+b \sqrt n \in\mathbb Q (\sqrt n).$ What (sufficient or necessary) conditions should $a$ and $b$ satisfy so that $a+b \sqrt n$ has a $m$-th root in $\mathbb Q (\sqrt n)$?

Here is my attempt :
I tried the case $m=2$. If $\sqrt{a+b \sqrt n} = c+d\sqrt n$ with $c,d \in \mathbb Q$ then
$$ a=c^2+d^2n, b=2cd. $$ Assuming $b \neq 0$, I get $c^2 + n\left(\frac{b}{2c}\right)^2 = a$, and for instance $c = \pm \sqrt{\frac{a+\sqrt{a^2-nb^2}}{2}}$, so it is necessary to have $\frac{a+\sqrt{a^2-nb^2}}{2}$ is a square in $\mathbb Q$ (and then $d$ is also rational).
We may find better conditions than this one. But I don't know how to manage with the cases $m \ge 3$, because the calculations become difficult. Is there some theoretical approach (e.g. Galois theory) to treat this problem ?
Thank you !

REPLY [2 votes]: Let me come back to your question for more practical purposes.  In my former theoretical approach (I keep all my previous notations), I gave a necessary and sufficient condition for an element $\alpha\in K^*$ to be a global $m$-th power, but in practice this criterion works well only to give a negative answer, i.e. to show that $\alpha$ is not an $m$-th power, because in that case, one needs only a finite number of trials and errows to find a prime $\mathcal L_v$ outside $S$ such that $\alpha$ is not a local $m$-th power in $K_v^{*}$. But a positive answer would require an infinity of checks, which is not very satisfying in practice.
A « finite » criterion for a « positive » answer when $m$ = an odd prime $p$ (because we want to avoid some « silly special cases », @franz lemmemermer dixit) can be derived from an « interesting » (Tate’s own words) local-global principle  in chapter 7 of Cassels-Fröhlich’s book (p. 184, remark 9.3). A particular case is the following : let $E$ be a number field containing the group $\mu_p$ of $p$-th roots of unity  ; pick an $\alpha \in E$  and let $S$ be a finite set of primes of $E$  containing  (i) all archimedean primes  (ii) all primes dividing $p$ and $\alpha$   (iii) all representatives of a system of generators for the ideal class group of $E$. Then any $S$-unit of $E$ which is a local $p$-th power at all primes inside $S$ is a global $p$-th power. 
In our initial problem, $\alpha$ is in $K$ which does not necessarily contain $\mu_p$. Put $E = K(\mu_p)$ , $G = Gal(E/K)$, and try to relate $(K^*)^p$ and $(E^*)^p$. Taking $G$-cohomology of the exact sequence 1 --> $\mu_p$--> $E^*$--> $(E^*)^p$ -->1 , we get 1 -->$\mu_p^{G}$--> $K^*$ -->$ K^*\cap (E^*)^p$-->$H^1(G, \mu_p)$ ... If $K$ does not contain $\mu_p$, $G$ has order prime to $p$ and $H^1(G, \mu_p)$ is trivial. In any case we get $(K^*)^p \cong K^*\cap (E^*)^p$. Summarizing, Tate’s remark gives us a finite criterion for a positive answer (in the above sense) when $m = p$. 
We may suspect that with the general local-global principle, we are going too far beyond the simple case of a quadratic field, for which the solution should be much less elaborate. In the case of $\mathbf Q$, the solution is immediate because of the factoriality of $\mathbf Z$, so the natural idea is to replace  factoriality by the uniqueness of decomposition into prime ideals in a Dedekind domain. A necessary condition, as suggested by @Qiaochu Yuan, is obtained by taking norms down to $\mathbf Q$. But for a sufficient condition, it seems that we are definitely blocked by by the units of norm 1 in the totally real case. This is rather irritating.