The square roots of different primes are linearly independent over the field of rationals

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.

Solution 1:

Below is a simple proof from one of my old sci.math posts, followed by reviews of a few 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 every nonempty subset of $\rm\,S\,$ has 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 this to $\rm\, n > 2\,$ i.e. to $\, [1, \sqrt{a_1}\,]\ [1, \sqrt{a_2}\,]\cdots [1, \sqrt{a_n}\,]\,$ 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 $\rm\,[K(r):K] = 2\,$ $\Rightarrow$ $\rm\,r\not\in K.\quad$ 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\,(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 $(1)$ 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 $\rm\,\sqrt{b}\quad\quad$ 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 units). 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

Solution 2:

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.)

Solution 3:

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.