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Extension fields

  1. Dec 15, 2009 #1
    Two problems from my abstract algebra class...

    Let K be the algebraic closure of a fi eld F and suppose E is a field such that  F [tex]F \subseteq E \subseteq K[/tex]. Then is K the algebraic closure of E?

    Let [tex]n[/tex] be a natural number with [tex]n\geq2[/tex], and suppose that [tex]\omega[/tex] is a complex nth root on unity. Is there a formula for [tex]\left[\mathbb{Q}(\omega) : \mathbb{Q}\right][/tex] ?


    To 1), I must be missing something really silly, because it seems to me like it is obviously the case that K is also the algebraic closure of E, and that the proof should be easy. But I simply can't think of anything.

    To 2) I would say no, but I am not exactly sure that I understand the question. For example, if n=8, then [tex]e^{i2\pi/8}[/tex] is a complex 8th root of unity such that [tex]\left[\mathbb{Q}(e^{i2\pi/8}) : \mathbb{Q}\right]=4[/tex]. However, [tex]i[/tex] is also an 8th root of unity, but [tex]\left[\mathbb{Q}(i) : \mathbb{Q}\right]=2[/tex]. Thus, for a given n, there is not necessarily a formula. Does this sound right?
  2. jcsd
  3. Dec 16, 2009 #2
    1) Well for K to be the algebraic closure of E you must show:
    a) K/E is algebraic.
    b) If [itex]g(x) \in E[x][/itex], then g(x) splits completely in K. (HINT: Remember that K is an algebraic closure of itself so [itex]h(x) \in K[x][/itex] imply that h(x) splits completely in K).

    2) Well [itex]\omega[/itex] is a specific nth root of unity so it's acceptable for your formula to behave differently when given [itex]e^{i2\pi/8}[/itex] and when given [itex]e^{i\pi/8}[/itex]. You should probably look for a formula of the form:
    [tex]\left[\mathbb{Q}\left(e^{ik\pi/n}\right) \, : \, \mathbb{Q} \right] = f(n,k)[/tex]
    so the formula can depend on both n and k, not just n.
  4. Dec 16, 2009 #3
    Ok, for 2), I've got the formula

    \left[\mathbb{Q}\left(e^{ik\pi/n}\right) \, : \, \mathbb{Q} \right] = \phi\left(\frac{n}{\gcd(n,k)}\right)

    where phi is Euler's totient funciton.

    Is that right?

    Now I just have to prove it....
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