Show that a given polynomial is irreducible over the rationals with √(-2) adjoined

Prove that p(x) = x^4 - 4x^2 + 8x + 2 is irreducible over the quadratic field \mathbb{Q}(\sqrt{-2}).


We saw previously that \mathbb{Z}[\sqrt{-2}] is a Euclidean domain with field of fractions \mathbb{Q}(\sqrt{-2}). Thus it is a unique factorization domain, and thus by Gauss’ lemma, p(x) is irreducible over \mathbb{Q}(\sqrt{-2}) if and only if it is irreducible over \mathbb{Z}[\sqrt{-2}].

Suppose p(x) is reducible in \mathbb{Z}[\sqrt{-2}][x].

Suppose p(x) has a linear factor; say x - a. Then we must have that a divides 2 in \mathbb{Z}[\sqrt{-2}]. Recall that the Euclidean norm in this ring is given by N(a+b\sqrt{-2}) = a^2 + 2b^2; thus N(a) divides 4, so that N(a) \in \{ \pm 1, \pm 2, \pm \sqrt{-2}\}. Evidently, however, we have p(1) = 7, p(-1) = -9, p(2) = 18, p(-2) = -14, p(\sqrt{-2}) = 14 + 8\sqrt{-2}, and p(-\sqrt{-2}) = 14 - 8\sqrt{-2}. So p(x) does not have a linear factor.

Suppose instead that p(x) has a quadratic factor, say x^2 + ax + b. Using the long division algorithm for polynomials, we have p(x) = (x^2 + ax + b)(x^2 - ax + (a^2 - b - 4)) + (8 - a^3 + 2ab +4a)x + (2 - a^2b + b^2 + 4b). Since quotients and remainders in the Euclidean algorithm in F[x] are unique, we have 8 - a^3 + 2ab + 4a = 0 and 2 - a^2b + b^2 + 4b = 0. Moreover, we have b \in \{ \pm 1, \pm 2, \pm \sqrt{-2} \} as in the linear factor case, and similarly a^2 - b - 4 \in \{ \pm 1, \pm 2, \pm \sqrt{-2}. Note that 8 = a(a^2 - 2b - 4); computing norms, we see that N(a) divides 64. If b = 1, then the second equation gives 2a^2 = -5. Computing norms, we have that 4 divides 25, a contradiction. If b = 1, then we have -2a^2 = 3, which is similarly a contradiction. If b = 2, then we have a^2 = -5. Then N(a) is either 1, 5, or 25; since N(a) divides 64, N(a) = 1 and so a = \pm 1. Then a^2 = 1, a contradiction. If b = -2, then we have a^2 = -1, a contradiction since no such element exists in \mathbb{Z}[\sqrt{-2}]. If b = \sqrt{-2}, then the second equation yields \sqrt{-2}(a^2 + 2) = 1, a contradiction since the only units in \mathbb{Z}[\sqrt{-2}] are 1 and -1. If b = -\sqrt{-2}, then we likewise have \sqrt{-2}(a^2 + 2) = -1, again a contradiction. Thus p(x) does not have a quadratic factor.

Thus p(x) is irreducible over \mathbb{Z}[\sqrt{-2}], and thus over \mathbb{Q}(\sqrt{-2}).

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Comments

  • Kreizhn  On July 15, 2011 at 11:34 am

    Problem statement should be Q adjoined \sqrt{-2} but it is currently listed with a positive.

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