Let $p_{1},p_{2},\ldots,p_{n}$ be $n$ primes,$\left(p_{i},p_{j}\right)=1$ if $i\neq j$ . Prove that $\left[\mathbb{Q}\left(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}}\right):\mathbb{Q}\right]=2^{n}$
Help me some hints.
Thanks in advanced.
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See http://www.mathpuzzle.com/SumOfRoots.txt – Hagen von Eitzen Feb 20 '14 at 07:21
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2You will find Bill Dubuque's answer to this very useful. – André Nicolas Feb 20 '14 at 07:24
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I would do this inductively. The key is to show that $[Q(\sqrt{p}_1, \ldots \sqrt{p_{n+1}}) \colon \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_n})] = 2$. The extension is obtained by adjoining a root of $x^2 - p_{n+1}$ to $\mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_n})$, so the degree can only be $1$ or $2$. To show that it's $2$, you just need to prove that $x^2 - p_{n+1}$ doesn't have a root. This calls for a tiny bit of number theory.
Paul Siegel
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@chuyenvien94 I agree, but you have to prove this: note that $x^2 - p_{n+1}$ could be reducible if there were any pairwise common divisors among $p_1, \ldots, p_{n+1}$. This is why a little number theory is required. – Paul Siegel Feb 24 '14 at 00:37
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Hint-Prove that $S=\{\Pi_{i \in U}\sqrt{p_i}\ | U \subseteq \{1,2...,n\} \}$ is a linearly independent set in $\Bbb Q[\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}}]$ and that every element in $\Bbb Q[\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}}]$ is a linear combination (over $\Bbb Q$) of elements in $S$.
viplov_jain
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