Factor a given polynomial into irreducibles over different rings

Completely factor p(x) = x^8 - 1 and q(x) = x^6 - 1 over \mathbb{Z}, \mathbb{Z}/(2), and \mathbb{Z}/(3).


Note that over any ring with 1 we have p(x) = (x^4+1)(x^2+1)(x+1)(x-1) and q(x) = (x^2+x+1)(x^2-x+1)(x+1)(x-1).

Over \mathbb{Z}, x^4+1 is irreducible since (x+1)^4 + 1 = x^4 + 4x^3 + 6x^2 + 4x + 2 is Eisenstein at 2. Also, x^2+1 is irreducible since (x+1)^2 + 1 = x^2 + 2x + x is Eisenstein at 2. Thus p(x) = (x^4+1)(x^2+1)(x+1)(x-1) is completely factored into irreducibles.

Similarly, x^2 + x + 1 is irreducible since (x+1)^2 + (x+1) + 1 = x^2 + 3x + 3 is Eisenstein at 3, and x^2 - x + 1 is irreducible since (x-1)^2 - (x-1) + 1 = x^2 - 3x + 3 is Eisenstein at 3. Thus q(x) = (x^2 + x + 1)(x^2 - x + 1)(x+1)(x-1) is completely factored into irreducibles.

In this previous exercise, we computed the irreducible monic polynomials of degree at most 3 over \mathbb{Z}/(2) and \mathbb{Z}/(3).

Over \mathbb{Z}/(3), r(x) = x^4 + 1 has no linear factors since r(0) = 1 and r(1) = f(2) = 2. Suppose now that r(x) has a quadratic factor; say x^2 + ax + b. Using the polynomial long division algorithm, we see that r(x) = (x^2 + ax + b)(x^2 - ax + (a^2 - b)) + (2ab - a^3)x + (1 - a^2b + b^2). Thus 2ab - a = a(2b-1) = 0 and 1 - a^2b + b^2 = 0. If a = 0, then the second equation gives b^2 = -1, a contradiction. Thus 2b - 1 = 0, so that b = 2. Now we have a^2 = 1, so that a \in \{1,2\}. Indeed, x^4 + 1 = (x^2 + x + 2)(x^2 + 2x + 2), and these factors are irreducible.

Evidently, p(x) = (x + 1)^8 and q(x) = (x^2 + x + 1)^2(x+1)^2 are completely factored over \mathbb{Z}/(2) and p(x) = (x^2 + x + 2)(x^2 + 2x + x)(x^2 + 1)(x+1)(x+2) and q(x) = (x+1)^3(x+2)^3 are completely factored over \mathbb{Z}/(3).

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Comments

  • Blaze  On November 18, 2011 at 8:27 am

    The 7th row of the proof has a typo. Should be x^2 + 3x + 3.

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