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 !

• You want to compute the prime factorization of $a + b \sqrt{n}$ in a suitable sense. This means understanding the prime ideals of $\mathcal{O}_{\mathbb{Q}(\zeta_n)}$ as well as the ideal class group. For rough results, a necessary but not sufficient condition is that the norm $N(a + b \sqrt{n}) = a^2 - b^2 n$ has an $m^{th}$ root in $\mathbb{Q}$. – Qiaochu Yuan Jan 8 '16 at 22:24
• Ah, sorry, when I said "ideal class group" above I meant "unit group," but you might also need to compute the ideal class group. – Qiaochu Yuan Jan 8 '16 at 22:31
• Necessary conditions are that the ideal your number generates should be an mth power, and that the number is an nth power modulo all ideals. Local-Global principles should guarantee the converse except in some silly cases of fourth or eighth powers. – franz lemmermeyer Jan 9 '16 at 13:08
• Relevant: math.stackexchange.com/q/405623 – Watson Dec 24 '16 at 16:24

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.

• Thank you very much for your answer! What do you mean by "local group" at the beginning? In particular, what does $K_{\nu}$ mean? I do not know so much about local fields… – Watson May 10 '16 at 20:18
• Given a prime ideal P_v of a global field K, one can associate to it a P_v-adic valuation v, which in turn gives an ultrametric distance, and K_v denotes the completion of K w.r.t. this distance. This K_v is usually called a P_v local field, "local" as opposed to "global". The point is that the arithmetic of a local field is much simpler, in particular its ring of integers has only one maximal ideal (more precisely, it's a discrete valuation ring). The so called local - global principle "phlosophically" means that the knowledge of (some) global properties should come from that of all the ... – nguyen quang do May 11 '16 at 6:01
• ... corresponding local ones. Some classical examples : - the theory of quadratic forms over a number field - Hasse's norm theorem for cyclic extensions - the determination of the Brauer group - the two latter examples are part of CFT from a local-global point of view . An even more far reaching illustration is the current theory of "motifs" (initiated by Grothendieck) which, in some way, "explains" how the archimedean and p-adic worlds can be related, although they are opposed by nature (connected vs. totally disconnected) . – nguyen quang do May 11 '16 at 6:10
• No, the P_v -adic additive valuation is defined as a map from K* to Z by letting v(x) = the exponent of P_v in the principal ideal (x). You can transform this into a (multiplicative) norm by defining the "absolute value" of x as being p^{-v(x)}, where p is the rational prime under P_v (there is a choice of normalization here, but it does not affect the topology). The completion of K is taken w.r.t. this absolute value.You can find all this in any (advanced) text book on ANT, such as the first two chapters of the classical Cassels-Fröhlich's book. – nguyen quang do May 11 '16 at 10:21
• 1. Yes 2. Just a prime ideal, indexed by the letter v which will denote the associated valuation 3. In the usual language of ANT, the "infinite primes" of K are the absolute values induced by the embeddings of K into an algebraic closure of Q. NB: we are suggested by the moderator to move our discussion to "chat" – nguyen quang do May 11 '16 at 19:37

The way of Quiaochu Yuan's comment is technical and hard. In elementary way you have with two arbitrary rational $r,s$ form the values given by $a+b\sqrt n=(r+s\sqrt n)^m=A+B\sqrt n$ where $$B=\binom m 1r^{m-1}s+\binom m3r^{m-3}s^3n+\binom m5r^{m-5}s^5n^2+......$$ and $$A=\text{the other terms}$$ This way you have for each couple of $r,s$ an $A+B\sqrt n$ satisfying the question.

With technical way, you'll have always this implicit condition for some couple of rationals : for each suitable $a+b\sqrt n$ there are two rational $r,s$ fulfilling the elementary condition given here.

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.

• Thank you very much! I do not know so much about cohomology yet, so do you have a reference that explains how your exact sequence is transformed by taking the $G$-cohomology? – Watson May 16 '16 at 15:04
• The main goal of group cohomology is to derive canonically from a short exact sequence of G-modules, an infinite (a priori) exact sequence of cohomology groups which extends to the right the short exact sequence (not exact at the 3rd right term) obtained by taking G-invariants. "Morally", just think of an unknown function which you wish to "approximate" by its Taylor expansion. A good self contained account is chapter 4 of Cassels-Fröhlich. – nguyen quang do May 16 '16 at 15:32
• Thank you, I will look at this chapter 4. Moreover (this is completely unrelated, a priori), do you know how Pontryagin duality can be used in number theory? I've looked quickly at Ramakrishnan, Valenza Fourier Analysis in Number Fields, but I had some trouble to understand how this duality can be related to local fields (and related algebraic objects)… [Maybe I will ask a specific question on that subject, instead of just a comment.] – Watson May 16 '16 at 19:35
• Perhaps we could switch to "chat". – nguyen quang do May 17 '16 at 6:00