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Show that f(x) is irreducible over Q and Q(2^(1/5))

  1. Jan 31, 2012 #1
    Show that [itex] x^3 + 6x^2 - 12x + 2 [/itex] is irreducible in [itex] \mathbb{Q} [/itex] as well as in [itex] \mathbb{Q}(\sqrt[5]{2})[/itex].

    the first part i have no trouble with since it follows straight from Eisenstein's Irreducibility Criterion. For the second part i am pretty confused though. i tried looking at the solution to help me understand it but i wasn't sure what the solution did.

    They say that [itex] [\mathbb{Q}(\sqrt[5]{2}): \mathbb{Q}] = 5 [/itex] and i understand that since [itex] \sqrt[5]{2} [/itex] is the root of irreducible [itex] x^5 - 2 [/itex] so the degree of the field extension is 5. They then said that if [itex] x^3 + 6x^2 - 12x + 2 [/itex] was reducible, then it would have a linear factor and then there would be a root in [itex] \mathbb{Q}(\sqrt[5]{2}) [/itex]. Then since this root has degree 3 and since 3 does not divide 5, then [itex] x^3 + 6x^2 - 12x + 2 [/itex] is irreducible in [itex] \mathbb{Q}(\sqrt[5]{2})[/itex].

    i do not understand the part about the 3 not dividing the 5. I assume they are using the theorem that says [F:K] = [F:E][E:K] if K is a subfield of E and E is a subfield of F. i suspect that this theorem is being used but i don't know how they are using it exactly. can someone help explain? thanks
  2. jcsd
  3. Jan 31, 2012 #2
    Indeed, since we can find an [itex]\alpha\in \mathbb{Q}(\sqrt[5]{2})[/itex] that is a root of [itex]x^3+6x^2-12x+2[/itex], this means that this polynomial is a minimal polynomial of [itex]\alpha[/itex]. Thus [itex][\mathbb{Q}(\alpha),\mathbb{Q}]=3[/itex].

    Now we have that


    This means that


    Or 3 divides 5. Which is impossible.
  4. Feb 7, 2012 #3
    yes that makes sense. i was initially confused on which field to pick but i understand now why Q(a) was chosen. thank you for your reply.
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