Galois group of $X^4 + 20 ∈ ℚ[X]$

Question

I believe this one hasn't been answered yet. To me, it's a tough one.

Let $f := X^4 + 20 ∈ {\Bbb Q}[X]$, $L := Ω^f_{\Bbb Q}$.

Now $f$ factors over $\Bbb C$ as

$$ f = (X + \sqrt[4]{-20})(X - \sqrt[4]{-20})(X + i\sqrt[4]{-20})(X - i\sqrt[4]{-20}).$$

At first I thought that gave splitting field of degree 16 over ${\Bbb Q}$, but it must also be the order of a subgroup of $S_4$, so its degree must divide 24, which 16 does not do.

Once I know the correct splitting field, I would just like a hint for constructing the Galois group at first, instead of a full answer.

Answer 1

There is complex conjugation and an obvious 4-cycle (which is reversed by conjugation) so the Galois group will be $D_8$.