0
$\begingroup$

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.

$\endgroup$

marked as duplicate by Zev Chonoles, Asaf Karagila, J. W. Perry, user127.0.0.1, Yiorgos S. Smyrlis Feb 20 '14 at 8:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $x^2 - p_{n+1}$ doesn't have a root since it's irreducible. $\endgroup$ – chuyenvien94 Feb 24 '14 at 0:05
  • $\begingroup$ @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. $\endgroup$ – Paul Siegel Feb 24 '14 at 0:37
  • $\begingroup$ Ehm... of course $\endgroup$ – chuyenvien94 Feb 24 '14 at 1:02
0
$\begingroup$

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 $S$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.