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Homework Help: Field extensions, proving an element is in a field

  1. May 18, 2010 #1
    2vj7xmx.jpg

    I can do all the question apart from the 'Conclude rt(2) in an element of Q(rt(2) + rt(3))'

    That's all I need help on.

    I tried

    rt(2) = a + b(rt(2) + rt(3))

    squaring it and equating you get

    2 = a2 + 5b2 + (some other terms of rt(2) rt(3) rt(5))

    but this cannot be true for a, b in Q, how else are you supposed to do this?

    Thanks
     
  2. jcsd
  3. May 18, 2010 #2
    Solve for the square root of 2.
     
  4. May 18, 2010 #3
    Solve what exactly?
     
  5. May 18, 2010 #4
    You know that [itex](\sqrt{2}+\sqrt{3})^2 - 2\sqrt{2}(\sqrt{2}+\sqrt{3}) - 1 = 0[/itex]. Here, we're fortunate because there is a [itex]\sqrt{2}[/itex] available that we can isolate.
     
  6. May 18, 2010 #5
    I tried that before but it left me with something pretty messy when all i wanted to show was it could be written as a + b(rt(2) + rt(3))

    i ended up with rt(2) = -(2 + rt(5)) / (rt(2) + rt(3))
     
  7. May 18, 2010 #6
    Remember that this is [itex]\mathbb{Q}(\sqrt{2}+\sqrt{3})[/itex], not [itex]\mathbb{Q}[\sqrt{2}+\sqrt{3}][/itex]. Also, I don't see from where the [itex]\sqrt{5}[/itex] comes.
     
  8. May 18, 2010 #7
    hmm whats the difference?

    guess I was slacking when jotting down notes and didn't realise when they were curvy brackets and not :(
     
  9. May 18, 2010 #8
    The former is the smallest field containing (root 2 + root 3), which means there are inverses. I've slacked before in math classes, so I know the feeling.
     
  10. May 18, 2010 #9
    Then that sort of ruins my answer for part i) of

    33cou48.jpg

    isn't k = {0, 1, T, 1+T} since k = {a + bT | a, b in {0,1}}?

    and so g is irreducible since no element of k is a root

    but you're saying k isn't {a + bT | a, b in {0,1}}
     
  11. May 18, 2010 #10
    It turns out that F2(T) contains all quotients of polynomials in F2[T]. Interpret it as the field of fractions of F2[T]. What have you learned that you can apply relating irreducible polynomials over the two?
     
  12. May 18, 2010 #11
    Probably missed your point but ill take a stab

    if it's irreducible in Fp[T], p prime, then it's irreducible in Z[X]
     
  13. May 18, 2010 #12
    It can be tough to wrap your head around, but the indeterminate variable is X. So from what you proved we know that our polynomial g(X) is irreducible in (F2[T])[X]. It is not even in Z[X]. However, we would like to say something about its irreducibility in (F2(T))[X]. This might remind you of the connection between irreducibility in Z[X] and Q[X].
     
  14. May 18, 2010 #13
    If it's irreducible in Q[X] then it's irreducible in Z[X] is gauss's lemma, and them im stumped

    Is this point crucial to the solution?
     
  15. May 18, 2010 #14
    I think it's the opposite, that irreducibility in Z[X] implies irreducibility in Q[X]. I've also just finished studying these kinds of things, so I can't be sure that this is crucial, but it seems so to me. We can use Gauss's lemma to conclude what you need passing from the ring to the field of fractions.
     
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