# Determine Gal$(\mathbb{Q}(\sqrt{7},i)/\mathbb{Q}(\sqrt{7}))$ and Gal$(\mathbb{Q}(\sqrt{7},i)/\mathbb{Q}(\sqrt{-7}))$.

Let $$K=\mathbb{Q}(\sqrt{7},i)$$, let $$F_1 = \mathbb{Q}(\sqrt{7})$$ and let $$F_2=\mathbb{Q}(\sqrt{-7})$$.

(a) Prove $$K$$ is Galois over $$F_1$$ and over $$F_2$$, and determine $$[K:F_1]$$ and $$[K:F_2]$$.

(b) Determine Gal$$(K/F_1)$$ and Gal$$(K/F_2)$$.

I first thought that $$K/\mathbb{Q}$$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $$K=\mathbb{Q}(\sqrt{2},i)$$. Now I'm having trouble to show $$K$$ is Galois over both $$F_1$$ and $$F_2$$, I got stuck on finding the separable minimal polynomial over $$\mathbb{Q}(\sqrt{7})$$ or over $$\mathbb{Q}(\sqrt{-7})$$. I feel like that $$[K:F_1]=8=[K:F_2]$$, since $$\mathbb{Q}(\sqrt{7})\subseteq\mathbb{Q}(\sqrt{7})\subseteq\mathbb{Q}(\sqrt{7},i),$$ but I need the minimal polynomials to justify I think. I would appreciate for any help!

If $$F$$ has characteristic zero, and $$K=F(\sqrta,i)$$ where $$a\in F$$, then $$K$$ is Galois over $$F$$, being the splitting field of $$x^4-a$$. Your first example has $$a=\sqrt7$$.
The Galois group of $$K/F_1$$ has order $$8$$, and computing it is very similar to standard examples such as $$\Bbb Q(\sqrt2,i)/\Bbb Q$$. The group is dihedral of order $$8$$.
I don't think $$K/F_2$$ is Galois.
• You only need $\operatorname{char}(F) \ne 2$ – Kenny Lau Jan 4 at 22:08
• I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois? – Alex Jan 5 at 7:31