Degree of Extension: n Primes to Q

[Kummer]

Let p_1,p_2,...,p_n be distinct primes.
Show that [\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},...,\sqrt{p_n} ) : \mathbb{Q}]= 2^n

 

[StatusX]

What have you tried? It's obvious you should use induction, and then the tricky part becomes proving the irreducibility of a certain polynomial in an extension of Q. Have you gotten this far yet?

 

[Kummer]

What have you tried? It's obvious you should use induction, and then the tricky part becomes proving the irreducibility of a certain polynomial in an extension of Q. Have you gotten this far yet?

Yes. I tried induction of course.

Note that, [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}}): \mathbb{Q}] = [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})]:\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})]\cdot [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})]:\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})] \cdot 2^n

So it remains to show,
[\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})]:\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})] = 2

Which can be shown if,
\mbox{deg}(\sqrt{p_{n+1}},\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}))=2

But the difficult step is the last step. I was reading about this problem and it was solved with "Kummer Theory" but I tried to find my own "elementary" solution.

 

[StatusX]

You know \sqrt{p_{n+1}} is a root of x^2-p_{n+1}, so generates an extension of at most degree 2 over any extension of Q. It will generate an extension of degree 2 iff it is not in the original extension. In other words, you need to show there's no element in \mathbb{Q}[\sqrt{p_1},...,\sqrt{p_n}] whose square is p_{n+1}. One way to do this is to note (by induction) that every element in \mathbb{Q}[\sqrt{p_1},...,\sqrt{p_n}] can be written in the form \alpha + \beta \sqrt{p_n} for some \alpha,\beta \in \mathbb{Q}[\sqrt{p_1},...,\sqrt{p_{n-1}}], then square this and find conditions on \alpha and \beta.

 

[Kummer]

Yes you are correct, it is really not that bad to proof.