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Let $m_1,m_2,\ldots, n_t$ be square-free integers $(m_i\neq 0, \pm 1)$ which are relatively prime in pairs. Show that $[\mathbb{Q}(\sqrt{m_1},\sqrt{m_2},\ldots, \sqrt{m_t}):\mathbb{Q}] = 2^t$

Hint: A careful induction on $t$ may be of use.

Hi. I have some questions regarding my possible solution.

Conjeture 1. If $m,n$ are relatively prime and $m,n$ square-free integers then $mn$ is square free integer.

Conjeture 1 seems intuitive but I can't think of how proves it.

Conjeture 2. $\mathbb{Q}(\sqrt{m_1},\sqrt{m_2},\ldots, \sqrt{m_t})=\mathbb{Q}(\sqrt{m_1},\sqrt{m_2},\ldots, \sqrt{m_{t-1}})(\sqrt{m_t})$.

Solution: For $t=1$, we know that $[\mathbb{Q}(\sqrt{m}):\mathbb{Q}]=2$. Suppose that $[\mathbb{Q}(\sqrt{m_1},\sqrt{m_2},\ldots, \sqrt{m_t}):\mathbb{Q}] = 2^t$. Now, \begin{align*} &[\mathbb{Q}(\sqrt{m_1},\ldots, \sqrt{m_{t+1}}):\mathbb{Q}]=\\ &[\mathbb{Q}(\sqrt{m_1},\ldots, \sqrt{m_t})(\sqrt{m_{t+1}}):\mathbb{Q}(\sqrt{m_1},\ldots, \sqrt{m_t})][\mathbb{Q}(\sqrt{m_1},\ldots, \sqrt{m_t}):\mathbb{Q}]\\ &=2\cdot 2^t=2^{t+1} \end{align*}

Is true the conjeture 1 and 2? is correct this solution?...

eraldcoil
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    Your solution has a big gap: you don't explain why $\sqrt{m_{t+1}}$ is not in $\mathbf Q(\sqrt{m_1}, \ldots, \sqrt{m_t})$. Without that you only know by induction that the field degree in your last calculation is at most $2^{t+1}$, not that it is equal to $2^{t+1}$. – KCd Sep 04 '20 at 04:16
  • The link in the comment two above concerns primes so may need modification when dealing with relatively primes. – Gerry Myerson Sep 04 '20 at 04:26
  • In that thread Bill Dubuque gave an answer which applies to all cases, I think. – ShBh Sep 04 '20 at 04:29
  • @KCd in general, I remember that, $\alpha$ algebraic in $F$, $F$ field then $[F(\alpha):F]=2$ if and only if $x^2-\alpha$ is the minimal polynomial.

    Therefore, I must proves that $x^2-m_{t+1}$ is the minimal polynomial. Right?...

     – eraldcoil Sep 04 '20 at 04:56
  • Now, $x^2-m_{t+1}$ is the minimal polynomial if and only if is irreducible in $\mathbb{Q}(\sqrt{m_1},\ldots, \sqrt{m_t}).$ If $x^2-m_{t+1}$ is reducible then $x^2-m_{t+1}=(x-a)(x-b)$ with $a,b\in \mathbb{Q}(\sqrt{m_1},\ldots, \sqrt{m_t})$, so $m_{t+1}=a$ or $m_{t+1}=b$. Suppose $m_{t+1}=a$. This implies that $m_{t+1}$ linear dependence with $\sqrt{m_1},\ldots, \sqrt{m_{t}}$ a contradiction. It is right? (I know that the roots are linearly independent but I don't know how proves that in a simple way (check some solutions on this website but they are somewhat difficult) – eraldcoil Sep 04 '20 at 05:07
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    Being linearly independent over $\mathbf Q$ is not enough: $\sqrt{2}, \sqrt{3}$, and $\sqrt{6}$ are linearly independent over $\mathbf Q$ but that doesn't mean $\sqrt{6} \not\in \mathbf Q(\sqrt{2},\sqrt{3})$. – KCd Sep 04 '20 at 05:24
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    You say some solutions you have found are a bit difficult, and that should be expected: this task you are working on requires some serious effort to establish. I don't think it is going to be settled just by some elementary manipulations with square roots. I have seen solutions using Kummer theory and using algebraic number theory, which both would fit under the label "somewhat difficult" if you have not seen those methods before. – KCd Sep 04 '20 at 05:27
  • I understand your argument. And if I could prove that $\sqrt{m_{n+1}}\not\in \mathbb{Q}(\sqrt{m_1}\ldots,\sqrt{m_n})$, if it would be enough to prove the problem from what I see. Right?... – eraldcoil Sep 05 '20 at 05:33

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