Compute the splitting field of x⁴+2 over QQ

Compute the splitting field of p(x) = x^4+2 over \mathbb{Q} and its degree.


The roots of p(x) exist in \mathbb{C} (if we don’t know this yet, just assume some roots are in \mathbb{C}). Let \zeta = a+bi be such a root; then \zeta^4 = -2.

Evidently, \zeta^4 = [a^4-6a^2b^2 + b^4] + [4ab(a^2-b^2]i. Comparing coefficients, we see that either a = 0, b = 0, or a^2 = b^2. If a = 0, then b^4 = -2, a contradiction in \mathbb{R}. Likewise, if b = 0 we get a contradiction. Thus a^2 = b^2. Substituting, we have b^4 - 6b^4 + b^4 = -2, so -4b^4 = -2, and so b^4 = 1/2. There is a unique positive 4th root of 1/2, which we denote by 2^{-1/4}; so b = \pm 2^{-1/4} and a = \pm 2^{-1/4}, and hence \zeta = \pm 2^{-1/4} \pm 2^{-1/4}i. There are 4 such roots, and so we have completely factored p(x). (WolframAlpha agrees.)

So the splitting field of p(x) is \mathbb{Q}(\pm 2^{-1/4} \pm 2^{-1/4}i). Note that if \zeta_1 = 2^{-1/4} + 2^{-1/4}i and \zeta_2 = \pm 2^{-1/4} - 2^{-1/4}i, then (\zeta_1 + \zeta_2)/2 = 2^{-1/4}, and (\zeta_1 - \zeta_2)/22^{-1/4} = i. Thus \mathbb{Q}(\pm 2^{-1/4} \pm 2^{-1/4}i) = \mathbb{Q}(i,2^{-1/4}).

Note that 2^{-1/4} is a root of q(x) = 2x^4 - 1, and that the reverse of q(x) is t(x) = -x^4+2. Now t(x) is irreducible over the UFD \mathbb{Z}[i], since it is Eisenstein at the irreducible element 1+i. So t(x) is irreducible over \mathbb{Q}(i), the field of fractions of \mathbb{Z}[i] as a consequence of Gauss’ Lemma. As we showed previously, the reverse of t, namely q, is also irreducible over \mathbb{Q}(i). So \mathbb{Q}(i,2^{-1/4}) has degree 4 over \mathbb{Q}(i), and thus has degree 8 over \mathbb{Q}.

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