Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Degree of Extension

  1. Jun 28, 2007 #1
    Let [tex]p_1,p_2,...,p_n[/tex] be distinct primes.
    Show that [tex][\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},...,\sqrt{p_n} ) : \mathbb{Q}]= 2^n[/tex]
    Last edited: Jun 28, 2007
  2. jcsd
  3. Jun 28, 2007 #2


    User Avatar
    Homework Helper

    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?
  4. Jun 28, 2007 #3
    Yes. I tried induction of course.

    Note that, [tex][\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[/tex]

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

    Which can be shown if,

    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.
  5. Jun 29, 2007 #4


    User Avatar
    Homework Helper

    You know [itex]\sqrt{p_{n+1}}[/itex] is a root of [itex]x^2-p_{n+1}[/itex], 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 [itex]\mathbb{Q}[\sqrt{p_1},...,\sqrt{p_n}][/itex] whose square is [itex]p_{n+1}[/itex]. One way to do this is to note (by induction) that every element in [itex]\mathbb{Q}[\sqrt{p_1},...,\sqrt{p_n}][/itex] can be written in the form [itex]\alpha + \beta \sqrt{p_n}[/itex] for some [itex]\alpha,\beta \in \mathbb{Q}[\sqrt{p_1},...,\sqrt{p_{n-1}}][/itex], then square this and find conditions on [itex]\alpha[/itex] and [itex]\beta[/itex].
    Last edited: Jun 29, 2007
  6. Jun 30, 2007 #5
    Yes you are correct, it is really not that bad to proof.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?