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Proving an extension is simple

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Let c be a primitive 3rd root of unity in the complex numbers and b be the real root of x^4-2=0. If a = c*b, show that Q(b,c) = Q(a)


    2. Relevant equations


    3. The attempt at a solution
    So [Q(a):Q(c)]=3 and [Q(a):Q(b)]=4, and c and b contain no 'overlapping material', so [Q(a):Q)=12.

    The usual way I prove things are a simple extension is by starting off by taking (a+b)^2 all the way up to (a+b)^n-1 where n i the deg(a)*deb(b) and then playing around with these expressions trying to come up with a way to get either a or b by itself as to show that a and b are elements of Q(a+b).

    Since the degree here is 12, things could get quite messy if I use that method. Is there another way to look at this?
     
  2. jcsd
  3. Oct 15, 2016 #2

    fresh_42

    Staff: Mentor

    I would not use a statement like "no overlapping material". This actually shouts for an explanation and contains every flaw, if there is one. Furthermore, I think ##[ \mathbb{Q}(a) \, : \, \mathbb{Q}(c) ] = 4## and ##[ \mathbb{Q}(a) \, : \, \mathbb{Q}(b) ] = 3##.

    Since ##\mathbb{Q}(a) = \mathbb{Q}(b \cdot c) \subseteq \mathbb{Q}(b,c)## it only has to be shown, that ##\, b\, , \,c \in \mathbb{Q}(a)##.
    A little basic algebra on the powers of ##a=b \cdot c## should do.
     
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