Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p_1,\ldots,p_k$ be $k$ distinct primes (in $\mathbb{N}$) and $n>1$. Is it true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$? (all the roots are in $\mathbb{R}^+$)
Iurie Boreico proved here that a linear combination $\sum q_i\sqrt[n]{a_i}$ with positive rational coefficients $q_i$ (and no $\sqrt[n]{a_i}\in\mathbb{Q}$) can't be rational, but this question seems to be more difficult..

share|cite|improve this question
up vote 5 down vote accepted

Below are links to classical proofs. Nowadays such results are usually derived as special cases of results in Kummer Galois theory. See my post here for a very simple proof of the quadratic case.

Besicovitch, A. S. $\ $ On the linear independence of fractional powers of integers.
J. London Math. Soc. 15 (1940). 3-6. MR 2,33f 10.0X

Let $\ a_i = b_i\ p_i,\ i=1,\ldots s\:,\:$ where the $p_i$ are $s$ different primes and the $b_i$ positive integers not divisible by any of them. The author proves by an inductive argument that, if $x_j$ are positive real roots of $x^{n_j} - a_j = 0,\ j=1,...,s ,$ and $P(x_1,...,x_s)$ is a polynomial with rational coefficients and of degree not greater than $n_j - 1$ with respect to $x_j,$ then $P(x_1,...,x_s)$ can vanish only if all its coefficients vanish. $\quad$ Reviewed by W. Feller.

Mordell, L. J. $\ $ On the linear independence of algebraic numbers.
Pacific J. Math. 3 (1953). 625-630. MR 15,404e 10.0X

Let $K$ be an algebraic number field and $x_1,\ldots,x_s$ roots of the equations $\ x_i^{n_i} = a_i\ (i=1,2,...,s)$ and suppose that (1) $K$ and all $x_i$ are real, or (2) $K$ includes all the $n_i$ th roots of unity, i.e. $ K(x_i)$ is a Kummer field. The following theorem is proved. A polynomial $P(x_1,...,x_s)$ with coefficients in $K$ and of degrees in $x_i$ , less than $n_i$ for $i=1,2,\ldots s$ , can vanish only if all its coefficients vanish, provided that the algebraic number field $K$ is such that there exists no relation of the form $\ x_1^{m_1}\ x_2^{m_2}\:\cdots\: x_s^{m_s} = a$, where $a$ is a number in $K$ unless $\ m_i = 0\ mod\ n_i\ (i=1,2,...,s)$ . When $K$ is of the second type, the theorem was proved earlier by Hasse [Klassenkorpertheorie, Marburg, 1933, pp. 187--195] by help of Galois groups. When K is of the first type and K also the rational number field and the $a_i$ integers, the theorem was proved by Besicovitch in an elementary way. The author here uses a proof analogous to that used by Besicovitch [J. London Math. Soc. 15b, 3--6 (1940) these Rev. 2, 33]. $\quad$ Reviewed by H. Bergstrom.

Siegel, Carl Ludwig. $\ $ Algebraische Abhaengigkeit von Wurzeln.
Acta Arith. 21 (1972), 59-64. MR 46 #1760 12A99

Two nonzero real numbers are said to be equivalent with respect to a real field $R$ if their ratio belongs to $R$ . Each real number $r \ne 0$ determines a class $[r]$ under this equivalence relation, and these classes form a multiplicative abelian group $G$ with identity element $[1]$. If $r_1,\cdots,r_h$ are nonzero real numbers such that $r_i^{n_i}\in R$ for some positive integers $n_i\ (i=1,...,h)$ , denote by $G(r_1,...,r_h) = G_h$ the subgroup of $G$ generated by [r_1],...,[r_h] and by R(r_1,...,r_h) = R_h the algebraic extension field of $R = R_0$ obtained by the adjunction of $r_1,...,r_h$ . The central problem considered in this paper is to determine the degree and find a basis of $R_h$ over $R$ . Special cases of this problem have been considered earlier by A. S. Besicovitch [J. London Math. Soc. 15 (1940), 3-6; MR 2, 33] and by L. J. Mordell [Pacific J. Math. 3 (1953), 625-630; MR 15, 404]. The principal result of this paper is the following theorem: the degree of $R_h$ with respect to $R_{h-1}$ is equal to the index $j$ of $G_{h-1}$ in $G_h$ , and the powers $r_i^t\ (t=0,1,...,j-1)$ form a basis of $R_h$ over $R_{h-1}$ . Several interesting applications and examples of this result are discussed. $\quad$ Reviewed by H. S. Butts

share|cite|improve this answer
Wow, big guns to attack this question! – Lubin May 26 '12 at 19:34
Wow, they asked Feller to review Besicovitch's paper! Well, it's not as if Feller really did much in his review, but it's funny that a probabilist would be asked to write the review. – KCd May 26 '12 at 21:07
Nice bibliography, Bill:+1 – Georges Elencwajg May 27 '12 at 6:46
I accepted this answer because of the rich bibliography. Thanks to all anyway! – Mizar May 27 '12 at 10:11

Yes, it is true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$ and it was proved by Besicovitch ( a student of A.A. Markov) in 1940.

Although there are (almost) infinitely many books on field and Galois theory, the only book I know which proves Besicovich's theorem (but only for odd $n$) is Roman's Field Theory (Theorem 14.3.2, page 305 of the second edition).

share|cite|improve this answer
Yes, but the proof there is only for odd $\,n\geq 3\,$ , and the proof isn't easy at all. He mentions that the case where n is even adds no new insights and is "intricate"(!!) – DonAntonio May 26 '12 at 19:44
Dear @Donantonio, you are absolutely right: thanks a lot. I suppose brave users will have to go to Richards's article which Roman follows (and mentions in his bibliography) if they want to see the intricate case of even $n$. I have made an Edit about that caveat. – Georges Elencwajg May 26 '12 at 19:58
Another book that has a proof is Lisl Gaal's "Classical Galois Theory: With Examples", and she treats the case of all $n$, not just odd $n$. (In fact, one of her first steps is to reduce to the case when $n$ is even.) See Section 4.12, which essentially starts on p. 234. This is the place where I first saw the result when I was a student. – KCd May 26 '12 at 21:18
A link to the section of Gaal's book where the result appears is here. – KCd May 26 '12 at 21:24
Thanks for the comment and the link, @KCd. Was that the book required for your course? It doesn't seem very widespread now in American universities (but I might be mistaken: I only know about these universities through online documents [yours in particular !]) – Georges Elencwajg May 26 '12 at 21:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.