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Monic Polynomials

  1. Jun 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Show x^2 + 2 in Z_5[x] is irreducible. This is before the section on the factor theorem (j is a root -> (x-j) is a factor). So I'm not so sure I want to start checking for zero's since "its not available" per se.


    2. Relevant equations


    3. The attempt at a solution

    Suppose it was reducible. Then x^2 + 2 = (ax + b) (a^-1x + c). First, I have a feeling that the products are monic, so i dont have to worry about the a, and a^-1.

    But why would the products would be monic (if it were reducible, even though its not)?

    I know if a poly can be factored into monics, the poly itself must be monic, (which is clear since the leading coefficient will turn out to be 1 upon collection of like powers.) When I say monic factors, i mean "just" monic factors, with no non-zero coefficients out in front.


    But I can't say that x^2 + 2 *can be factored into monics* to begin with, since I can't assume this. Please help me with showing why its okay to assume if x^2 + 2 over Z_5 were reducible, it would have monic factors.
     
  2. jcsd
  3. Jun 17, 2007 #2
    Nevermind, they are all units in Z_5. I can't believe i overlooked this.

    x^2 + 2 = (ax + b) (a^-1x + c) = a(x + b/a) a^-1(x + ca) = (x + b/a)(x + ca).
     
  4. Jun 17, 2007 #3

    morphism

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    It doesn't have to have monic factors.

    Take for example x^2 + 1. Over Z_5, it factors as (2x+1)(3x+1).
     
  5. Jun 17, 2007 #4
    (2x + 1) = 2(x + 3)
    (3x + 1) = 3(x + 2)
    = 6(x + 3)(x + 2) = (x+3)(x+2) = x^2 + 5x + 6 = x^2 + 1

    so x^2 + 1 = (x + 2)(x + 3).
     
  6. Jun 17, 2007 #5

    morphism

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    Right. You can assume without loss of generality that the factors are monic. I was just saying that you can write down a non-monic factorization.
     
  7. Jun 17, 2007 #6
    Oh okay. Heh.
     
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