## 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 $\mathbb{Q}$.