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.

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