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Determining Galois extension based on degree of extension

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    If Char(K) = 0 and [L:K]=2, is L:K a galois extension?

    2. Relevant equations


    3. The attempt at a solution
    My gut is saying yes because if [L:K]=2 then it seems that any polynomial in K[x] with a root in L should split in L[x]. Something about how some hypothetical minimal polynomial of some element m call it m(x) where L[x] is isomorphic to the quotient field K[x]/m(x), then K(u) is isomorphic to L where given the conditions that [L:K] =2 then deg(m(x)) = 2 and so if L has one of the roots of m(x) then that means that m(x) factors linearly because it only has degree 2 and thus all the roots are in L thus the extension is normal. Am I leaving anything important out?
     
  2. jcsd
  3. Nov 24, 2016 #2

    fresh_42

    Staff: Mentor

    I've thought along the same lines. More formal we can choose an element ##u \in L-K##. Then ##\{1,u\}## is ##K-##linear independent and ##\{1,u,u^2\}## is not. This means we can write ##u^2=\alpha \cdot 1 + \beta \cdot u## which means ##u## is a root of ##x^2- \beta x - \alpha##.
    Now we can write down both solutions ##u## and ##v## and see that ##v \in K(u)##, i.e. our minimal polynomial splits and is separable. (Why?)
     
  4. Nov 27, 2016 #3
    Our minimal polynomial is separable because K is given to have Characteristic 0, and our minimal polynomial splits because both of it's roots are in K(u). Thanks!
     
  5. Nov 27, 2016 #4

    fresh_42

    Staff: Mentor

    One can also see it directly: If ##u## were a double root, then ##u=-\frac{\beta}{2} \in K## which we ruled out. And with ##u=-\frac{\beta}{2}-\sqrt{sth.}\, , \,v=-\frac{\beta}{2}+\sqrt{sth.}## we get ##v=-u-\beta \in K(u)##.
     
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