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As the title goes, I am stuck on this problem.

Prove that the set $\{\sqrt{n}:\textrm{$n$ is squarefree}\} =\{1,\sqrt{2}, \sqrt{3},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{10},\ldots\}$ is a linearly independent set.

I have some reduction. For any fixed prime $p$, any equation can be written as $$a\sqrt{p}\left(\sum \lambda_i\sqrt{n_i}\right)+b\left(\sum \mu_i\sqrt{n_i}\right)=0$$ with $n_i$ squarefree and $p\not{|}n_i$. If $a=0$, then one can use reduction. If $a\neq 0$, then one can assume $a=1$, then $$\sqrt{p}=\sum\nu_i\sqrt{n_i}$$ for some $\nu_i$, since $\mathbb{Q}[\{\sqrt{n_i}\}]$ is a field. Then we have $$p=\left(\sum\nu_i\sqrt{n_i}\right)^2$$ then the coefficients of the above all vanish. But the coefficients is a quadratic equation system on $\nu_i$, how to show it has only zero-solution?

I think that it may be solved by some valuation theory, but i am now just a beginner of the theory. Anyway, thank you for you help.

Cubic Bear
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2 Answers2

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I get a proof from my classmate, as follow.

For any nature number $m$ if $$\sqrt{m}=\sum \lambda \sqrt{n}\qquad \lambda\in \mathbb{Q}, n\in \mathbb{N},\textrm{squarefree}$$ We claim that $\lambda\sqrt{n}=0$ when $n\neq m$.

We use induction on the number of primes divides some of $n$. Pick one of the prime $p\in P$, we have $$\sqrt{m}=\sqrt{p}\sum \mu \sqrt{n}+\sum \nu \sqrt{n}$$ where $p\not{|}n$. We have $$m-p\left(\sum \mu \sqrt{n}\right)^2-\left(\sum \nu \sqrt{n}\right)^2=2\left(\sum \mu \sqrt{n}\right)\left(\sum \nu \sqrt{n}\right)\sqrt{p}$$ Then $2\left(\sum \mu \sqrt{n}\right)\left(\sum \nu \sqrt{n}\right)=0$, otherwise $$\sqrt{p}=\sum\alpha\sqrt{n}\qquad \alpha\in \mathbb{Q}, p\not{|}n$$ the number of primes divides some of $n$ reduces, which is impossible. Then $$\sqrt{m}=\sqrt{p}\sum \mu \sqrt{n}\quad \textrm{or}\quad \sqrt{m}=\sum \nu \sqrt{n}$$ That is $$\sqrt{pm}=\sum p\mu \sqrt{n}\quad \textrm{or}\quad \sqrt{m}=\sum \nu \sqrt{n}$$ the number of primes divides some of $n$ reduces, thus the hypothesis holds.

Cubic Bear
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This question has been asked many times, but with the the set {$ \sqrt p: p$ prime }. The technical difficulty when using purely linear algebra is due to the blend of addition and multiplication. A perhaps more "visible" proof consists in applying (a bit of) Kummer theory in order to use only multiplication. Recall that for a field $K$ of characteristic not dividing $m$ and containing the group $\mu_m$ of $m$-th roots of 1, Kummer's theorem states that any finite abelian extension of $K$ whose Galois group $G$ is killed by $m$ is obtained by adding $m$-th roots of a finite subgroup $A$ of $K^*/(K^*)^{m}$, more precisely $G \cong Hom (A, \mu_m)$. Usually, $K$ is called a Kummer extension and $A$ its Kummer radical .

Here $K = \mathbf Q$ and $m=2$. By Kummer and induction, the $\mathbf Q$-linear independence of $\sqrt n_1,..., \sqrt n_r$ is equivalent to saying that the Kummer extension $K_r = \mathbf Q (\sqrt n_1,..., \sqrt n_r)$ is abelian of degree $2^r$, more precisely its Kummer radical $A_r$, generated by $n_1,..., n_r$ mod $(\mathbf Q^*)^2$, is isomorphic to $(\mathbf Z /2)^r$. Note that the linear independence over $\mathbf Q$ has been transformed into a linear independence over $\mathbf F_2$, by viewing the group $\mathbf Q^*/(\mathbf Q^*)^2$ as a $\mathbf F_2$-vector space in additive notation. The rest of the proof is obvious: $\sqrt n_1,..., \sqrt n_r, \sqrt n_{r+1}$ are linearly independent over $ \mathbf Q$ iff the (squarefree) integers $n_1,...,n_r, n_{r+1}$ are multiplicatively independent mod $(\mathbf Q^*)^2$, i.e., by the UFD property of $\mathbf Z$, $n_{r+1}$ has a prime divisor $p$ not dividing $n_1,...,n_r$ .

nguyen quang do
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