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Prove 1+8x-12x^3+2x^4 irreducible over Q[x]

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine if 1+8x-12x^3+2x^4 irreducible over Q[x]


    2. Relevant equations
    Gauss's Lemma, Eisenstein Criterion


    3. The attempt at a solution

    If we multiply g(x)=1+8x-12x^3+2x^4 by 2^3, and then make the substitution y=(2*x), we recover a monic polynomial: h(y) = 8+32y-12y^3+y^4, but Eisenstein does not apply. h(y+/-1) doesn't help either. h(y) is of fourth degree, so it doesn't even suffice to check for absence of roots. But, I do know that if irreducibility in Z is shown, irreducibility over Q follows, for suppose such an f(x) is reducible over Q. Then, by Gauss's Lemma, there exists a nontrivial factorization of f(x) over Z into monic polynomials, but if f(x) is irreducible over Z, this is a contradiction. Thus, f(x) must be irreducible over Q.

    Still, I have tried many things, so any help is appereciated.
     
  2. jcsd
  3. Dec 9, 2011 #2

    Dick

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    Are you sure h(y+/-1) doesn't satisfy Eisenstein's criterion? It sure looks like they do to me.
     
  4. Dec 10, 2011 #3
    Thanks for the reply:

    The constant term in h(y+1) is 29, a prime, and 29|\ all other coefficients. That of h(y-1) is 11, another prime which does not divide any other coefficeints. Thus, I don't think we can use Eisenstein Criterion.

    However, if we consider b(x) = x^4*g(1/x) we obtain: b(x) = 2 -12x + 8x^3 + x^4, and we can use Eisenstein with p=2 here QED
     
  5. Dec 10, 2011 #4

    Dick

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    You're right. I misremembered the Eisenstein Criterion. And, yeah, that's a good idea!
     
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