I need to find a way of proving that the square roots of a finite set of different primes are linearly independent over the field of rationals.

I've tried to solve the problem using elementary algebra and also using the theory of field extensions, without success. To prove linear independence of two primes is easy but then my problems arise. I would be very thankful for an answer to this question.

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    $\begingroup$ mathforum.org/library/drmath/view/51638.html $\endgroup$ Commented Apr 3, 2011 at 17:53
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    $\begingroup$ Also, see this: qchu.wordpress.com/2009/07/02/… $\endgroup$ Commented Apr 3, 2011 at 18:14
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    $\begingroup$ This also comes up in T..'s answer here: math.stackexchange.com/questions/6244/… $\endgroup$ Commented Apr 3, 2011 at 18:41
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    $\begingroup$ @J.M.: after Yuval's hint to the mathforum I'd like to mention, that the text of the question is 100% identical with that of the mathforum of 1996, and the fact that neither its reference was given nor anything about the existing answers there was mentioned I assume a) this is not a real question (also there was no followup interaction of "user8465") , and (see the recent meta thread on spam) b) maybe not even a real person asking but possibly an automated transfer of a somehow mathematically sounding text. Maybe that method of spam has been refined recently... $\endgroup$ Commented Oct 11, 2011 at 9:16
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    $\begingroup$ @Gottfried The question is surely not spam - see my comments here. $\endgroup$ Commented Oct 11, 2011 at 16:11

3 Answers 3


Below is a simple proof from one of my old sci.math posts, followed by reviews of related papers.

Theorem $ $ Let $\rm\,Q\,$ be a field with $2 \ne 0,\,$ and $\rm\, L = Q(S)\,$ be an extension of $\rm\,Q\,$ generated by $\rm\,n\,$ square roots $\rm\,S \!=\! \{ \sqrt{a}, \sqrt{b},\ldots \}$ of $\rm\ a,b,\,\ldots \in Q.\,$ If all nonempty subsets of $\rm S $ have product $\rm\not\in\! Q\,$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\sqrt{b}),\,\ldots$ doubles degree over $\rm Q,\,$ so, in total, $\rm\, [L:Q] = 2^n.\,$ Thus the $\rm 2^n$ subproducts of the product of $\rm\,S\, $ are a basis of $\rm\,L\,$ over $\rm\,Q.$

Proof $\ $ By induction on the tower height $\rm\,n =$ number of root adjunctions. The Lemma below implies $\rm\ [1, \sqrt{a}\,]\ [1, \sqrt{b}\,] = [1, \sqrt{a}, \sqrt{b}, \sqrt{ab}\,]\ $ is a $\rm\,Q$-vector space basis of $\rm\, Q(\sqrt{a}, \sqrt{b})\,$ iff $\:\!1\:\!$ is the only basis element in $\rm\,Q.\,$ We lift it to $\rm\, n > 2\,$ i.e. $\, [1, \sqrt{a_1}\,]\ [1, \sqrt{a_2}\,]\cdots [1, \sqrt{a_n}\,]\,$ with $2^n$ elts.

$\rm n = 1\!:\ L = Q(\sqrt{a})\ $ so $\rm\,[L:Q] = 2,\,$ since $\rm\,\sqrt{a}\not\in Q\,$ by hypothesis.

$\rm n > 1\!:\ L = K(\sqrt{a},\sqrt{b}),\,\ K\ $ of height $\rm\,n\!-\!2.\,$ By induction $\rm\,[K:Q] = 2^{n-2} $ so we need only show: $\rm\ [L:K] = 4,\,$ since then $\rm\,[L:Q] = [L:K]\ [K:Q] = 4\cdot 2^{n-2}\! = 2^n.\,$ The lemma below shows $\rm\,[L:K] = 4\,$ if $\rm\ r = \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $ all $\rm\not\in K,\,$ true by induction on $\rm\,K(r)\,$ of height $\rm\,n\!-\!1\,$ shows $\rm\,[K(r):K] = 2\,$ $\Rightarrow$ $\rm\,r\not\in K.\ \ $ $\bf\small QED$

Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $ all $\rm\not\in K\,$ and $\rm\, 2 \ne 0\,$ in $\rm\,K.$

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b}).\,$ $\rm\, [L:K] = 2\,$ by $\rm\,\sqrt{b} \not\in K,\,$ so it suffices to show $\rm\, [L(\sqrt{a}):L] = 2.\,$ This fails only if $\rm\,\sqrt{a} \in L = K(\sqrt{b})$ $\,\Rightarrow\,$ $\rm \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K,\,$ which is false, because squaring yields $\rm\,\color{#c00}{(1)}:\ \ a\ =\ r^2 + b\ s^2 + 2\,r\,s\ \sqrt{b},\, $ which is contra to hypotheses as follows:

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $\color{#c00}{(1)}$ for $\rm\sqrt{b},\,$ using $\rm\,2 \ne 0$

$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ =\ r \in K$

$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\,b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b},\, \ $times $\rm\,\sqrt{b}.\qquad$ $\bf\small QED$

In the classical case $\rm\:Q\:$ is the field of rationals and the square roots have radicands being distinct primes. Here it is quite familiar that a product of any nonempty subset of them is irrational since, over a UFD, a product of coprime elements is a square iff each factor is a square (mod unit multiples). Hence the classical case satisfies the theorem's hypotheses.

Elementary proofs like that above are often credited to Besicovitch (see below). But I have not seen his paper so I cannot say for sure whether or not Besicovic's proof is essentially the same as above. Finally, see the papers reviewed below for some stronger results.

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

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.

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

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 $\ m_i \equiv 0 \mod n_i\ (i=1,2,...,s)$. 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.

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

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 $r_1,\dots,r_h$ 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 $[r_1],\dots,[r_h]$ 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

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    $\begingroup$ This is top-notch! $\endgroup$ Commented Apr 3, 2011 at 20:59
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    $\begingroup$ @BillDubuque: I read this couple more times and given it some thouht, I agree that $\sqrt{a},\sqrt{b}$ are not in $K$ by the hypothesis that the degree of the tower with hight $n-1$ is $2^{n-1}$ but I can't figure why $\sqrt{ab}$ is not in $K$ (it is not clear by the induction hypothesis since $\sqrt{ab}$ is not adjoined in any step...). I would be greatfull if you can explain this part of the proof, it is very interesting! $\endgroup$
    – Belgi
    Commented Jul 10, 2012 at 10:31
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    $\begingroup$ As always, your mathematics is excellent and your typesetting is atrocious. :) $\endgroup$
    – Potato
    Commented Jul 10, 2015 at 8:34
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    $\begingroup$ In the above comment you mentioned that $[K(r):K]=2$. That's because by induction hypothesis $K(r)$ has height $n-1$ and $[K(r):K]$ is equal to $\dfrac{[K(r):Q]}{[K:Q]}=2^{n-1}/2^{n-2}=2$. Right? $\endgroup$
    – RFZ
    Commented Apr 5, 2019 at 23:25
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    $\begingroup$ @eraldcoil A basis of $\,\Bbb Q(\sqrt a)\,$ over $\,\Bbb Q\ \ $ $\endgroup$ Commented Sep 5, 2020 at 7:32

Assume that there was some linear dependence relation of the form

$$ \sum_{k=1}^n c_k \sqrt{p_k} + c_0 = 0 $$

where $ c_k \in \mathbb{Q} $ and the $ p_k $ are distinct prime numbers. Let $ L $ be the smallest extension of $ \mathbb{Q} $ containing all of the $ \sqrt{p_k} $. We argue using the field trace $ T = T_{L/\mathbb{Q}} $. First, note that if $ d \in \mathbb{N} $ is not a perfect square, we have that $ T(\sqrt{d}) = 0 $. This is because $ L/\mathbb{Q} $ is Galois, and $ \sqrt{d} $ cannot be a fixed point of the action of the Galois group as it is not rational. This means that half of the Galois group maps it to its other conjugate $ -\sqrt{d} $, and therefore the sum of all conjugates cancel out. Furthermore, note that we have $ T(q) = 0 $ iff $ q = 0 $ for rational $ q $.

Taking traces on both sides we immediately find that $ c_0 = 0 $. Let $ 1 \leq j \leq n $ and multiply both sides by $ \sqrt{p_j} $ to get

$$ c_j p_j + \sum_{1 \leq k \leq n, k\neq j} c_k \sqrt{p_k p_j} = 0$$

Now, taking traces annihilates the second term entirely and we are left with $ T(c_j p_j) = 0 $, which implies $ c_j = 0 $. Since $ j $ was arbitrary, we conclude that all coefficients are zero, proving linear independence.

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    $\begingroup$ That is such a cool proof. I was just wondering how you know that $L/\mathbb{Q}$ is Galios. Even if it is not Galios, does taking its normal closure do the trick since that would still be an extension of $\mathbb{Q}$ so its fixed field will be the rationals? $\endgroup$
    – user357980
    Commented Sep 11, 2016 at 18:42
  • $\begingroup$ It does do the trick, yes; however we already know that $ L/\mathbb Q $ is Galois since it is the splitting field of $ \prod_{p_k} (X^2 - p_k) $ over $ \mathbb Q $. $\endgroup$
    – Ege Erdil
    Commented Sep 11, 2016 at 23:48
  • $\begingroup$ I see, thank you. I thought that it had to be irreducible, but a simple Googling disabused me of that. Once again, that was cool. $\endgroup$
    – user357980
    Commented Sep 12, 2016 at 0:58
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    $\begingroup$ Pretty! It just feels like magic. Can you give intuition to how you came up with it (I know it was long ago). In general I love your questions and answers and would love to ask your opinion on how to approach some things in algebra if you're available. $\endgroup$
    – Andy
    Commented Mar 1, 2018 at 21:35
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    $\begingroup$ This is a nice argument (which I upvoted long time ago) : even if we don't know the degree of $L/\Bbb Q$, we know that for $d=p_ip_j$ ($i \neq j$), the subgroup $Gal(L / \Bbb Q(\sqrt d))$ has index $2$ in $Gal(L/\Bbb Q)$, so that the $L/\Bbb Q$-trace of $\sqrt d$ must vanish. $\endgroup$
    – Watson
    Commented Dec 4, 2018 at 8:53

Iurie Boreico presents several Olympiad-style proofs of this fact in the Harvard College Mathematics Review. I give a somewhat more sophisticated proof in this blog post.

The source of the sophistication is interesting. For any particular finite set of primes, there is a completely elementary proof which is found by finding a suitable prime witness $q$ relative to which all but one of the primes is a quadratic residue. But in the above I use quadratic reciprocity and Dirichlet's theorem to show that $q$ always exists in general. (I am actually not sure if Dirichlet's theorem is necessary here.)

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    $\begingroup$ The first link seems to be broken now. Can anyone provide a copy of it? $\endgroup$
    – Emolga
    Commented Apr 4, 2015 at 23:18
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    $\begingroup$ @Leullame, the paper can be found here (the full issue, not just the individual paper). $\endgroup$
    – vonbrand
    Commented Oct 25, 2015 at 11:35
  • $\begingroup$ Currently here: abel.math.harvard.edu/hcmr/issues/2.pdf, page 87. $\endgroup$ Commented Aug 3, 2020 at 11:34

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