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Math Help - Irreducible polynomials

  1. #1
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    Irreducible polynomials

    Hi, does anyone know how to do the following questions?

    In each of the following cases, find all roots of f(x) in F and factor f(x) as the product of irreducible polynomials in F[x].

    (a) F=R, f(x) = (x^6)-1 in R[x]
    (b) F=Z3, f(x) = (x^5)+x+1 in Z3[x]
    (c) F=Z5, f(x) = (x^3)+2x+3 in Z5[x]

    Thanks very much guys....
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  2. #2
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    Quote Originally Posted by suedenation
    Hi, does anyone know how to do the following questions?

    In each of the following cases, find all roots of f(x) in F and factor f(x) as the product of irreducible polynomials in F[x].

    (a) F=R, f(x) = (x^6)-1 in R[x]
    You have, x^6-1=0 thus find all real values such as x^6=1. This only happens for x=-1,1 (note the other 4 roots are complex we are only concerned with real). Thus, x+1 and x-1 are factors. Dividing x^6-1 by (x-1)(x+1)=x^2-1 we get,
    x^4+x^2+1. Thus, the polynomial x^6-1 fractors as
    (x-1)(x+1)(x^4+x^2+1) in \mathbb{R}[x]

    Quote Originally Posted by suedenation
    (b) F=Z3, f(x) = (x^5)+x+1 in Z3[x]
    To factor x^5+x+1 in \mathbb{Z}_3[x] we simply see what are its zero's. This is easy to check because they are only 3 elements in this field. Checking we see that 1 is a zero because, 1^3+1+1=0 under addition and multiplication modulo 3. Thus, x-1 is a factor. Dividing these polynomials we get, (working modulo 3).
    x^4+x^3+x^2+x+2
    Checking the zero's of this polynomial we find that 1 is it zero. Thus, x-1 is a factor. Dividing these polynomials we get, (working modulo 3).
    x^3+2x+1, notice this polynomial has no zero and because it has a degree of 3 it is irreducible (this is a theorem about the irreducibility of 2 and 3 degree polynomials). Thus, it factors as,
    (x-1)(x-1)(x^3+2x+1) or another way of writing this, (x+1)(x+1)(x^3+2x+1

    I do part (c) latter unless you understand and I could stop.
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