1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove Q[sqrt 2, sqrt 3] is a field.

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove Q[sqrt 2, sqrt 3] is a field.

    2. Relevant equations
    Q[sqrt 2, sqrt 3]= {r + s[tex]\sqrt{2}[/tex] + t[tex]\sqrt{3}[/tex] + u[tex]\sqrt{6}[/tex]| r,s,t,u[tex]\in Q[/tex]}

    3. The attempt at a solution
    I know I have to show each element has an inverse, but I don't know how on these elements.
  2. jcsd
  3. Apr 9, 2011 #2
    To find the inverse of something in Q[[tex]\sqrt{2}, \sqrt{3}[/tex]] it might be helpfull to first notice the inverse of something in Q[[tex]\sqrt{2}[/tex]].

    (a + b[tex]\sqrt{2}[/tex])^-1 = (a - b[tex]\sqrt{2}[/tex])/(a^2 - 2*b^2)

    In general (a + b*sqrt n)^-1 = (a - b*sqrt n)/(a^2 - n*b^2)

    Next notice that an element of Q[[tex]\sqrt{2}[/tex],[tex]\sqrt{3}[/tex]] is also of the form a + b[tex]\sqrt{3}[/tex] where a and b are elements of Q[[tex]\sqrt{2}[/tex]] (i.e. they are each a + b*[tex]\sqrt{2}[/tex]) so an element of Q[[tex]\sqrt{2}[/tex], [tex]\sqrt{3}[/tex]] is like ((a + b[tex]\sqrt{2}[/tex]) + (c + d[tex]\sqrt{2}[/tex])[tex]\sqrt{3}[/tex]) = a + b[tex]\sqrt{2}[/tex] + c[tex]\sqrt{3}[/tex] + d[tex]\sqrt{6}[/tex].

    So the inverse of such an element is still (a - b[tex]\sqrt{3}[/tex])/(a^2 - 3*b^2) where the a and b in this case are elements of Q[[tex]\sqrt{2}[/tex]] (and so of the form a + b[tex]\sqrt{2}[/tex]).
    Last edited: Apr 9, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook