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Homework Help: Galois extensions

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex]\beta[/tex] be a real, positive fourth root of 5. Is [tex]\mathbb{Q}(\sqrt{-5})[/tex] Galois over [tex]\mathbb{Q}[/tex]? How about [tex]\mathbb{Q}(\beta + i\beta)/\mathbb{Q}[/tex] or [tex]\mathbb{Q}(\beta + i\beta)/\mathbb{Q}(\sqrt{-5})[/tex]?

    2. Relevant equations

    An extension is Galois when it is normal and separable.

    3. The attempt at a solution

    Any push whatsoever in the right direction would be much appreciated.
  2. jcsd
  3. Jan 12, 2010 #2


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    Can you tell me anything at all about those extensions?
  4. Jan 12, 2010 #3
    Do you know a condition equivalent to being Galois that involves splitting?
  5. Jan 12, 2010 #4
    well, a field [tex]A[/tex] is galois over [tex]B[/tex] if it is the splitting field of some polynomial in [tex]B[x][/tex] which has distinct roots in [tex]A[/tex].

    so then, [tex]\mathbb{Q}(\sqrt{-5})[/tex] is the splitting field of the polynomial [tex]x^2 + 5[/tex], which has coefficients in [tex]\mathbb{Q}[/tex]. this is because [tex]x^2 + 5 = (x + \sqrt{-5})(x - \sqrt{-5})[/tex], and hence the roots are also distinct, and so in the first case, we do have a galois extension.

    but i can't apply this to the second two ideas, because i can't tell anything about the field [tex]\mathbb{Q}(\beta + i\beta)[/tex]. it doesn't seem to be the same as [tex]\mathbb{Q}(i, \beta)[/tex] or anything of the sort. am i missing something obvious?

    thanks so much.
  6. Jan 12, 2010 #5
    also, thinking about the polynomial [tex](x + \beta + i\beta)(x + \beta - i\beta)(x - \beta + i\beta)(x - \beta - i\beta)[/tex] doesn't help, because it gives me the information that [tex]\mathbb{Q}(\beta + i\beta)[/tex] is galois over [tex]\mathbb{Q}(\sqrt{5})[/tex], but nothing else, and i don't see how this helps in the problem.
    Last edited: Jan 12, 2010
  7. Jan 12, 2010 #6


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    Is b-ib really in Q(b+ib)?
  8. Jan 12, 2010 #7
    no, it isn't, but it doesn't follow that it's not galois, right? i've worked for a while, and can't find explicit separable polynomials whose splitting fields are precisely those, but then what does that prove?

    i really don't see a way to disprove they're galois extensions. even if they are, i can't find the explicit polynomials, but my intuition points to a disproof. =( but how do i do that?
  9. Jan 12, 2010 #8
    oh i got the fact that it's galois over the third one -- from the polynomial x^2 - 2i*b^2.

    what about the last one?
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