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I want to prove that this system $\{1,\sqrt{m_1},\sqrt{m_2},...,\sqrt{m_n}\}$, where all $m_i$ are different square-free natural numbers, is linear independent over $\mathbb{Q}$. My teacher has asked me to prove this using induction.

  1. BASE CASE: $a_11+a_2\sqrt{m_1}$ linear independent. It's obviously.
  2. STEP CASE: Supouse $\sum\limits_i^{n-1}a_i\sqrt{m_i}\neq0$ for any $a_i\in\mathbb{Q}$.

Now I need to show that $\sum\limits_i^{n}a_i\sqrt{m_i}\neq0$. Everything I can do next is to rewrite it like this(since $a_n\neq0$): $\sqrt{m_n}=\sum\limits_i^{n-1}b_i\sqrt{m_i}$.

But I can't make any contradiction.

EDIT: $m_i\gt1, \forall i$

Kirill Losev
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  • This follows from the answers of this question. It works for different squarefree integers, not necessarily primes. For the induction, see the reference "Besicovitch". – Dietrich Burde Feb 10 '18 at 19:55
  • I would show by induction on k that the field Q(sqrt(P1), .., sqrt(pk) ) has dimension 2^k over Q. How to do it? Hint: it is enough to show that sqrt(p_k) does not belong to Q(P1,..,pk-1). Then, if you have sqrt(M1), .. sqrt(MN) distinct , this belongs to Q( sqrt(primes)). And think why this forces them to be linearly independent. – Andrea Marino Feb 10 '18 at 20:00
  • @AndreaMarino ok, ${1,\sqrt{p_1},...,\sqrt{p_k}}$ will be basis of that field. And our $\sqrt{m_1}...$ wil be multiplication of basis vectors? – Kirill Losev Feb 11 '18 at 21:52
  • No, they won't be a basis over Q. Think about Q(sqrt(2), sqrt(3) ). What is a basis? Generally if you have a basis of L over F, and a basis of F over k, how can you make a basis of L over K? Use inductively this reasoning to build up a basis of Q(sqrt(primes)) – Andrea Marino Feb 11 '18 at 23:48
  • @AndreaMarino basis in Q(sqrt2, sqrt3) will be {1, sqrt2, sqrt3, sqrt6)? – Kirill Losev Feb 12 '18 at 05:30
  • @AndreaMarino and maybe a soon as we know all primes which are in our m_i we can build Q(sqrt2...) and made change of basis? It will be subspace as our m_i are square-free? – Kirill Losev Feb 12 '18 at 06:03
  • @DietrichBurde that's the main problem. How to show that there is no linear combination for sqrt2(p_n)? It looks like my original question – Kirill Losev Feb 12 '18 at 06:21

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