Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Splitting Field Proof

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider [tex] f(x) = x^3-5 [/tex]
    and its splitting field [tex] K = Q(5^{1/3}, \omega) [/tex]
    where [tex] \omega = e^{2 \pi i/3} [/tex]
    Show that [tex] B = \{1, 5^{1/3}, 5^{2/3}, \omega, \omega 5^{1/3} , \omega 5^{2/3} \} [/tex]
    is a vector space basis for K over Q.

    3. The attempt at a solution

    I am just a bit confused. Since [tex] 5^{1/3} [/tex] and [tex] \omega [/tex] are in K, and K is a field, then [tex] B'= \{ \omega ^2, \omega ^2 5^{1/3}, \omega ^2 5^{2/3} \} \subseteq K [/tex]
    But how can we get any of these elements using only the shown basis B with scalars in Q? I would think that B+B' would be the vector space basis.
  2. jcsd
  3. Feb 5, 2012 #2


    User Avatar
    Science Advisor

    try building up K in 2 steps:

    first adjoin 51/3. what is [Q(51/3):Q]?

    what does a basis of E = Q(51/3) over Q look like?

    now K = E(ω). what is [K:E]? what is a basis for K over E?

    if dimQ(E) = m, and dimE(K) = n,

    what must dimQ(K) be?

    this tells you how many basis elements you have to have.

    then it's just a matter of proving linear independence over Q.


    as for how we get the elements of B', i'll show you for ω2:

    ω satisfies the polynomial x2 + x + 1 = (x3 - 1)/(x - 1).

    so ω2 = (-1)1 + (-1)ω, see?
  4. Feb 5, 2012 #3
    Yes, thanks so much I was confusing myself.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook