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Irreducibility of a polynomial by eisenstein and substitution

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Show [tex]16x^4 = 8x^3 - 16x^2 - 8x + 1[/tex] is irreducible.


    2. Relevant equations

    Eisenstein's criteria, if there is n s.t. n does not divide the leading coefficient, divides all the other coefficients, and n^2 does not divide the last coefficient then the polynomial is irreducible (over the rationals)

    3. The attempt at a solution
    I want to say that consider [tex]p'(x) = p(\frac{1}{2}x) = x^4 + x^3 - 4x^2 - 4x + 1 [/tex] is irreducible by eisenstein if we use the standard trick of substituting x+1 -> x, then we get, [tex]x^4+5x^3+5x^2-5x-5[/tex] where eisenstein is immediate. What I don't know is that if I can then say that since p' is irreducible, then p is.
     
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  3. Nov 24, 2009 #2

    HallsofIvy

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    This should be [tex]16x^2+ 8x^4- 16x^2- 8x+ 1[/itex]


    p(x) is reducible if and only if p(x)= a(x)b(x) for polynomials a and b of lower degree than p. In that case p(x/2)= a(x/2)b(x/2) so, yes, if p(x) is reducible so is p(x/2). And, then, if p(x/2) is irreducible, so is p(x).
     
  4. Nov 28, 2009 #3
    First of all, thank you for your help!

    That makes sense. I was stuck because I had a counter example that in general one cannot make substitutions; for example, p(u) = u^2 + 8u + 36 is irreducible over Q, but under the substitution, [t^2 -> u] p(t) splits into quadratic factors. So perhaps, is it possible to claim that any linear substitution over the single variable will be acceptable (i.e. [a*x + b -> x])?

    Is there a theory that deals with substitutions in the abstract sense?
     
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