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Is the following reasoning sound?

Note that $$X^4 - 2 = (X^2-\sqrt{2})(X^2+\sqrt{2}) = (X+i \sqrt[4]{2})(X - i\sqrt[4]{2})(X+\sqrt[4]{2})(X - \sqrt[4]{2}) ∈ ℂ[X],$$ so the zeroes of this polynomial generate $$ℚ(-i\sqrt[4]{2}, i\sqrt[4]{2}, -\sqrt[4]{2}, \sqrt[4]{2}) = ℚ(\sqrt[4]{2}, i) = ℚ(\sqrt[4]{2})(i).$$ Now $X^2 +1 ∈ ℚ(\sqrt[4]{2})$ has $i$ as a root, so $[ℚ(\sqrt[4]{2})(i):ℚ(\sqrt[4]{2})] = \deg(X^2+1) = 2$. Also $[ℚ(\sqrt[4]{2}): ℚ] = \deg(X^4-2) = 4$, so we conclude

$$[Ω^{X^4-2}_{ℚ} : ℚ] = [ℚ(\sqrt[4]{2})(i):ℚ(\sqrt[4]{2})][ℚ(\sqrt[4]{2}): ℚ]= 4 · 2 = 8. $$

(I believe there is a less 'ad hoc' way to do this (not using our knowledge of ℂ, bascially), which I would be interested in seeing.)

user26857
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Jos van Nieuwman
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    The answer by Adam Hughes (first five lines) is exactly what you had in mind, isn't it? Since the splitting field is $\Bbb Q(\sqrt[4]{2},i)$ we should "use our knowledge of $\Bbb C$". – Dietrich Burde Mar 30 '20 at 19:21

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