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Math Help - How can I prove these two field extensions are equal?

  1. #1
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    How can I prove these two field extensions are equal?

    How can I prove these two field extensions are equal?

    Q(√3, -√3, i, -i) = Q(√3+i) where Q is the field of the rational numbers.

    I got Q(√3+i) ⊆ Q(√3, -√3, i, -i), that direction is easy.

    How can I prove Q(√3, -√3, i, -i) ⊆ Q(√3+i)?

    Let m ∈ Q(√3, -√3, i, -i)

    Therefore m = q + a√3 + bi + c(-√3) + d(-i) where q,a,b,c,d ∈ Q

    Let p = a-c and r = b-d. Therefore, p,r ∈ Q

    Therefore m = q + p√3 + ri

    But I have to prove m = q + p√3 + ri = q + (SOMETHING IN Q)* (√3+i)

    Any hints?
    Last edited by Seda1; March 21st 2011 at 06:28 PM.
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  2. #2
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    Quote Originally Posted by Seda1 View Post
    How can I prove these two field extensions are equal?

    Q(√3, -√3, i, -i) = Q(√3+i) where Q is the field of the rational numbers.

    I got Q(√3+i) ⊆ Q(√3, -√3, i, -i), that direction is easy.

    How can I prove Q(√3, -√3, i, -i) ⊆ Q(√3+i)?

    Let b ∈ Q(√3, -√3, i, -i)

    Therefore b = q + p√3 + ri where q,p,r ∈ Q

    The fact that Q is a field and has additive inverses takes care of the -√3 and -i.

    But I have to prove b = q + p√3 + ri = q + (SOMETHING IN Q)* (√3+i)

    Any hints?
    Let a = "something in Q. Then b = q + p\sqrt{3} + ri = q + a(\sqrt{3}+i) = q + a \sqrt{3} + ai \in Q(\sqrt{3},~-\sqrt{3}, ~i, ~-i)

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Let a = "something in Q. Then b = q + p\sqrt{3} + ri = q + a(\sqrt{3}+i) = q + a \sqrt{3} + ai \in Q(\sqrt{3},~-\sqrt{3}, ~i, ~-i)

    -Dan
    Doesn't this assume p=r? I don't get it.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Seda1 View Post
    Doesn't this assume p=r? I don't get it.
    My bad. I thought you were trying to show that one was a subset of the other, not equal.

    -Dan
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  5. #5
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    I'm trying to prove they're equal by showing each is a subset of the other.

    I've shown Q(√3+i) ⊆ Q(√3, -√3, i, -i). Very quickly: m from Q(√3+i) can be written m = q +r(√3+i) = q + r√3 + ri, and let r = a-b,
    m= q + a(√3) +a(i) + b(-√3) + b(-i), so m is in Q(√3, -√3, i, -i). Therefore Q(√3+i) ⊆ Q(√3, -√3, i, -i)

    I'm trying to prove Q(√3, -√3, i, -i) ⊆ Q(√3+i).

    Note: Cleaned up the question a bit.
    Last edited by Seda1; March 21st 2011 at 06:28 PM.
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