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Homework Help: 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
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