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Does one zero equal its algebraic conjugate?

  1. Feb 1, 2007 #1
    I'm studying abstract algebra and never got the hang of ideas like these. Currently I am trying to grasp:

    Show that the splitting field of a quadratic polynomial P(x) = Ax^x + Bx + C, with zeros [tex]\alpha[/tex] and [tex]\alpha^{'}[/tex] is [tex]V_\alpha[/tex] = Q*1 + Q * [tex]\alpha[/tex]. Q is of course the set of rationals.

    I know I must be missing something here cause I also have problems trying to figure this out:

    Suppose P(x) is a quadratic polynomial with zeroes [tex]\alpha[/tex] and [tex]\alpha^{'}[/tex], Prove that Q*1 + Q*[tex]\alpha[/tex] = Q*1 + Q*[tex]\alpha^{'}[/tex]

    To me the zeros of a polynomial are related numbers but are different and thus should yield different vector spaces. These problems say they are the same vector space. Can anyone expand upon this?

    Tex doesn't seem to work in preview but I could not see what was wrong with my notation from your tutorials. If you don't see anything an ascii version will soon follow.

    Thanks.
     
  2. jcsd
  3. Feb 1, 2007 #2
    Ok, tex worked out of preview but is kind of odd. Any alpha (except V sub alpha) should not be a superscript or a subscript. Only the prime in alpha prime is a superscript.
     
  4. Feb 1, 2007 #3

    mathwonk

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    i think its because the polynomial is nokly quadratic. i,.e,. when you adjoin all but one rot of a polynomial you get the full root field.
     
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