number theory - Square and cubic roots in $mathbb Q(sqrt n)$

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 !

Answer

As suggested by @franz lemmermeyer, a theoretical approach would certainly consist in an adequate global-local principle (i.e. CFT in fine), but there could be technical difficulties when ramification comes into play. Take a general number field $K$, but to avoid petty trouble, assume that the given integer $m$ is odd. The global-local principle for $m$-th powers consists in studying the kernel of the natural map from the global group $K^* / (K^*)^m$ to the direct sum of all the local groups $K_v^{*} / (K_v^{*})^m$. Given a finite set $S$ of primes of $K$, an element of $K^*$ which is not divisible by any prime $\mathcal L_{v}$ outside $S$ will be called an $S$-unit. The following global-local principle is valid : "an $S$-unit $\alpha$ of $K$ is a global $m$-th power iff for any $\mathcal L_v$ outside $S$, $\alpha$ is an $m$-th power in the local field $K_v$" (Artin-Tate, chap. IX, thm. 1). The finite set $S$ is meant to give us a certain « room » adapted to the problem under study. Here we’ll choose $S$ such that it contains all the infinite primes, all the primes dividing the given $m$ and the given $\alpha$ in $K^*$,as well as all the primes dividing disc($K$). To decide if $\alpha$ is a global $m$-th power, we have only to look at its natural image in $K_v^* / (K_v^*)^m$ for any $\mathcal L_v$ outside $S$ . Using the Chinese remainder theorem, we can suppose that $m = p^r$ for some rational prime $p$. Let $l$ be the rational prime under such an $\mathcal L_v$ . By our choice of $S$, $l \neq p$, $K_v$ is an unramified extension of $\mathbf Q_l$, and $\alpha \in U_v$, the group of units of $K_v$. Let $\kappa_v$ be the residual field of $K_v$, a finite field of degree over $\mathbf F_l$ equal to the inertia index, equal here to the local degree $[K_v : \mathbf Q_l]$. It is classically known that $U_v$ is the direct product of a group $\cong(\kappa_{v})^*$ (via Hensel’s lemma) and of the group of principal units $U_1 = 1 + \mathcal L_v$ . Since $l \neq p$, raising to a $p$-primary power is an automorphism of $U_1$, hence in the end $U_v / (U_v)^{p^r} \cong \kappa_v^* / (\kappa_v^*)^{p^r}$.

Let us now switch to the case at hand, where $K$ is a quadratic field. We have only to consider two cases : either $l$ is inert in $K$, or $l$ is split. In the first case, $\kappa_l^*$ is cyclic of order $l^2 – 1$ ; in the second, $\kappa_v$ cyclic of order $l – 1$ for any of the two $v$’s above $l$. Define $W_r (l)$ to be the quotient $\kappa_l^*$ mod $p^r$ or the product of the two quotients $\kappa_v^*$ mod $p^r$ . We know explicitly $W_r (l)$ (without feeling like writing it down !), and the conclusion is : let $\alpha \in K^*$; choose $S$ as above ; then $\alpha$ is a $p^r$-th power in $K^*$ iff for any $l$ outside $S$, the natural image of $\alpha$ in $W_r(l)$ is trivial.