## Compute the splitting field of x⁶-4 over QQ

Compute the splitting field of $p(x) = x^6-4$ over $\mathbb{Q}$ and its degree.

Note that $x^6-1$ factors as $x^6-1 = (x+1)(x-1)(x^2+x+1)(x^2-x+1)$. (Using only the difference of squares and sum of cubes formulas familiar to middle schoolers.) Using the quadratic formula (again familiar to middle schoolers) we see that the roots of $p(x)$ are $\pm 1$ and $\pm \dfrac{1}{2} \pm \dfrac{i\sqrt{3}}{2}$.

Now if $\zeta$ is a 6th root of 1, then $\sqrt[6]{4}\zeta$ is a root of $p(x)$, where $\sqrt[6]{4} = \sqrt[3]{2}$ denotes the positive real 6th root of 4 (aka the positive cube root of 2). (Middle schoolers could verify that.) Evidently, then, the splitting field of $p(x)$ is $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2}(1+i\sqrt{3}, \sqrt[3]{2}(1-i\sqrt{3}) = \mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$.

Now $\mathbb{Q}(\sqrt[3]{2})$ has degree 3 over $\mathbb{Q}$, and $\mathbb{Q}(i\sqrt{3})$ has degree 2 over $\mathbb{Q}$. (Use Eisenstein for both.) By Corollary 22 on page 529 of D&F, then, $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$ has degree (as a middle schooler could compute) $3 \cdot 2 = 6$ over $\mathbb{Q}$.

So we have proved not only that the splitting field of $p(x)$ has degree 6 over $\mathbb{Q}$, but, in a metamathematical twist, also that a middle schooler could prove this as well.