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Math Help - Field Extensions

  1. #1
    DMT
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    Field Extensions

    Say:

     \alpha = \sqrt{2 + \sqrt{2}}

    And let:

     \beta = \sqrt{2 - \sqrt{2}}

    How do I show that  \beta is an element of the field extension:

     Q(\alpha):Q

    (the problem is to show that this field extension is normal, which I can do if I can show the above)

    Any help with this would be appreciated, thanks.
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  2. #2
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    We have \sqrt 2 \in Q(\alpha) and \alpha\beta = \sqrt 2.
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  3. #3
    DMT
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    Thanks! Right in front of my face the whole time.
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  4. #4
    DMT
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    Quick additional question about another method for doing the same thing.

    If I compared the fields:

    <br />
Q(\alpha):Q<br />

    and

    <br />
Q(\alpha, \beta):Q<br />

    And then showed that \alpha and \beta have the same minimum polynomial, that would be enough to show that they are the same field, right? As would showing that the order of both extensions is 4 (which amounts to the same thing, right?).
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  5. #5
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    Quote Originally Posted by DMT
    Quick additional question about another method for doing the same thing.

    If I compared the fields:

    <br />
Q(\alpha):Q<br />

    and

    <br />
Q(\alpha, \beta):Q<br />

    And then showed that \alpha and \beta have the same minimum polynomial, that would be enough to show that they are the same field, right? As would showing that the order of both extensions is 4 (which amounts to the same thing, right?).
    Your question is a very interesting one. I once asked a more general question on the forum
    http://www.mathhelpforum.com/math-he...ead.php?t=1898
    In general the answer is NO.
    But you can do something else.
    -----------------------------------
    If you can show that \mathbb{Q}\leq \mathbb{Q}(\alpha)\leq \mathbb{Q}(\beta) then by the theorem of finite extension fields we have,
    [\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]
    But since,
    [\mathbb{Q}(\alpha):\mathbb{Q}]=4=[\mathbb{Q}(\beta):\mathbb{Q}]
    Thus,
    4[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]=4
    Thus,
    [\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]=1
    Thus,
    \mathbb{Q}(\beta)=\mathbb{Q}(\alpha)

    But the trick and problem here is to show that the subfield relation \mathbb{Q}\leq \mathbb{Q}(\alpha)\leq \mathbb{Q}(\beta) is true.
    Last edited by ThePerfectHacker; March 17th 2006 at 09:14 AM.
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  6. #6
    DMT
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    Thanks, that's about what I had in mind.

    Here's a final question though ... what would be a basis for this field extension?

    I can't seem to come up with a basis with only 4 elements. My first instinct was something like:

    a + b\sqrt{2} + c\alpha + d\alpha\sqrt{2}
    a,b,c,d \in Q

    But this doesn't pick up multiples of  \beta . The only combos I can find that seem to cover all the elements of the field are too long.
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  7. #7
    DMT
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    Never mind on that one ... it just hit me that since

    \beta= \sqrt{2} / \alpha

    then it's not really a problem.

    The other thing I'm getting stuck with is the galois group for the field. It's supposed to be cyclic, but I seem to get the Klein 4-group instead.

    I'm assuming the galois group is generated by the following two maps:

    \sigma : \alpha \mapsto -\alpha

    \tau : \sqrt{2} \mapsto -\sqrt{2}

    Okay, I see now that \tau itself is order 4 and not order 2, and can generate a cyclic group of order 4, presumably the galois group. But that doesn't seem exactly right either since I don't really get \sigma in there.

    Or rather, it seems \tau^2 = \sigma but \tau and \tau^3 seem to do weird things to the coeficient assuming my above basis, and not just change the signs of the root two terms (because of how they also change \alpha into \beta).

    I'm wondering if my basis is wrong. Since

    \beta=\alpha\sqrt{2}-\alpha

    then maybe I can use \beta in my basis instead of the term with \sqrt{2}\alpha ... would that work?

    This is pushing the limit of my understanding of Galois theory which is what I am trying to learn.

    What are the correct 4 maps for the Galois group?

    Thanks.
    Last edited by DMT; March 19th 2006 at 12:06 AM.
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