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Field extensions of degree 10 over the rationals.

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex] \alpha[/itex] is real and has degree 10 over [itex]\mathbb{Q}[/itex] then
    [itex]\mathbb{Q}[\alpha]=\mathbb{Q}[\alpha^3][/itex]


    2. Relevant equations



    3. The attempt at a solution
    It is clear that [itex]\mathbb{Q}[\alpha^3]\subset \mathbb{Q}[\alpha][/itex]. This gives us the
    sequence of fields [itex]\mathbb{Q}\subset \mathbb{Q}[\alpha^3]\subset \mathbb{Q}[\alpha][/itex]. Since these are finite extensions, we then have [itex][\mathbb{Q}[\alpha]:\mathbb{Q}]=10=[\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^3]][\mathbb{Q}[\alpha^3]:\mathbb{Q}][/itex].
    Since [itex]x^3-\alpha^3\in \mathbb{Q}[\alpha^3][x][/itex] is of degree 3 and has [itex]\alpha[/itex] as a root, [itex][\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^3]]\leq 3[/itex]. Since it must also divide 10, it must be 1 or 2. The goal then is to show that it cannot be 2 (or then that [itex][\mathbb{Q}[\alpha^3]:\mathbb{Q}][/itex] cannot be 5.)
    I have tried going further on this point, but I think I'm just getting stubborn and missing something subtle. Any hints would be appreciated.
    -Theorem
     
  2. jcsd
  3. Oct 22, 2013 #2
    Okay I just got an idea. We can suppose towards a contradiction that [itex]\alpha \notin \mathbb{Q}[\alpha^3][/itex] Then the polynomial [itex]x^3-\alpha^3[/itex] is irreducible in [itex]\mathbb{Q}[\alpha^3][/itex] since
    [itex]\alpha[/itex] is real by assumption and so [itex]\mathbb{Q}[\alpha^3][/itex] is contained in [itex]\mathbb{R}[/itex] and the other two roots of the polynomial are non real (and thus it cannot be factored over [itex]\mathbb{Q}[\alpha^3][/itex]. Buth then this leads to a contradiction since then this forces [itex][\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^3]]=3[/itex]. which is impossible.
     
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