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Kronecker's theorem - Find a field with roots for X^4 + 1

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider [itex]X^4 + 1 [/itex] in the field [itex] \mathbb{Q} [X] [/itex]. I used Kronecker's theorem to find a root for X^4 + 1 in the field extension [itex] \mathbb{Q} [\frac{(1+i)}{\sqrt{2}}] [/itex]. I'm asked to show that this field extension allows X^4 + 1 to be factorised completely, thus adding a single root gives all roots.

    2. Relevant equations



    3. The attempt at a solution

    I have the answer, but it uses an assumption and I'm not sure how it got. It says:

    "Since X^4 + 1 has real coefficients and has [itex] \frac{(1+i)}{\sqrt{2}} [/itex] as a root, the complex conjugate [itex] \frac{(1-i)}{\sqrt{2}} [/itex] must also be a root."

    It then carries on as if it's known [itex] \frac{(1-i)}{\sqrt{2}} [/itex] is in the field extension to show that the other two roots (the negative of both those roots) are in the field extension. But how exactly do we know that the complex conjugate of the root [itex] \frac{(1+i)}{\sqrt{2}} [/itex] is also in the field extension?

    Is it because this is a field and, letting [itex] \frac{(1+i)}{\sqrt{2}} = z [/itex];

    [itex]z \frac{\overline{z}}{|z|^2} = 1[/itex]

    so

    [itex]\frac{\overline{z}}{|z|^2} [/itex]

    Is the inverse of [itex] z [/itex], since this is a field it's inverse must be contained in the set, and so

    [itex] |z|^2 \times \frac{\overline{z}}{|z|^2} = \overline{z}[/itex]

    Is in the set? This would need to assume [itex] |z|^2 [/itex] is contained in the field too.
     
    Last edited: May 12, 2012
  2. jcsd
  3. May 12, 2012 #2

    jgens

    User Avatar
    Gold Member

    Set [itex]z = \frac{1+i}{\sqrt{2}}[/itex] and notice that [itex]|z| = 1[/itex], which means that [itex]z^{-1} = \overline{z}[/itex].
     
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