1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Galois Theory - Question about Radical Extensions

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that [itex]I\subseteq J[/itex] are subfields of [itex]\mathbb{C}(t_1,...,t_n)[/itex] (that is, subsets closed under the operations +, - , [itex]\times[/itex], [itex]\div[/itex]), and [itex]J[/itex] is generated by [itex]J_1,...,J_r[/itex] where [itex]I \subseteq J_j \subseteq J[/itex] for each [itex]j[/itex] and [itex]J_j:I[/itex] is radical. By induction on [itex]r[/itex], prove that [itex]J:I[/itex] is radical.

    2. Relevant equations

    A relevant definition: An extension [itex]L:K[/itex] in [itex]\mathbb{C}[/itex] is radical if [itex]L = K(\alpha_1, ... ,\alpha_m)[/itex] where for each [itex]j = 1, ... , m[/itex] there exists an integer [itex]n_j[/itex] such that [itex]\alpha_j^{n_j} \in K(\alpha_1,..., \alpha_{j-1})[/itex] with [itex]j \geq 1[/itex].


    3. The attempt at a solution

    I don't think I am understanding what it means when the question says "[itex]J[/itex] is generated by [itex]J_1,...,J_r[/itex]". Is the question implying that the union of all of these subsets equal [itex]J[/itex]? I think once this is clarified, the proof should be straightforward.

    My progress so far:

    [itex]J_1:I[/itex] is radical. By definition, this implies that [itex]J_1 = I(\alpha_1, ... ,\alpha_m)[/itex] where for each [itex]k = 1, ... , m[/itex] there exists an integer [itex]n_k[/itex] such that [itex]\alpha_k^{n_k} \in K(\alpha_1,..., \alpha_{k-1})[/itex] with [itex]k \geq 1[/itex].

    Because I don't understand the question, I don't know where to go from here. Help please? Thank you!
     
  2. jcsd
  3. Feb 25, 2012 #2
    Every element of J may be expressed as a finite sum/product of elements from J1,…, Jr.

    Also, fields are not closed under division. Fields excluding the 0 element do, however, form a multiplicative group, and thus are closed under division.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Galois Theory - Question about Radical Extensions
  1. Galois theory (Replies: 4)

  2. Galois Extensions (Replies: 1)

  3. Galois extensions (Replies: 7)

Loading...