Let $K=\mathbb{Q}(\sqrt{-13+2\sqrt{13}})$. $K$ is normal over $\mathbb{Q}$
Prove there is an $A \in \mathbb{Q}$ such that $K(i)=\mathbb{Q}(i, \sqrt[4]{A})$
So we need to show that $K(i)=\mathbb{Q}(i,\sqrt{-13+2\sqrt{13}})=\mathbb{Q}(i, \sqrt[4]{A})$
So does this mean we must show that $[\mathbb{Q}: K]=[\mathbb{Q} : \sqrt[4]{A})]$?
To me it seems intuitive that such an $A$ exists since there are two square roots in the extension (one embedded in the other). How could we find an $A$ that satisfies this?
I was wondering if we could use the fact that if $C \subset B \subset A$ then $[A:C]=[A:B] \times [B:C]$
Note that $ K $ contains a root of $ x^4 + 26x^2 + 117 $ over $ \mathbb{Q} $, which factors as $$ (x^2 + 13 - 2\sqrt{13})(x^2 + 13 + 2\sqrt{13}) $$ It is irreducible by Eisenstein, therefore the extension is of degree $ 4 $. To see normality of $ K/\mathbb{Q} $ it is sufficient to note that
$$ \sqrt{-13 + 2\sqrt{13}} \cdot \sqrt{-13 - 2\sqrt{13}} = 3\sqrt{13} $$
so that $ \sqrt{-13 - 2\sqrt{13}} \in K $, and $ K $ is therefore the splitting field of $ x^4 + 26x^2 + 117 $ over $ \mathbb{Q} $.
It is easy to check directly that the Galois group of $ K/\mathbb{Q} $ is cyclic of order $ 4 $, therefore $ i \notin K $ and $ K(i) $ has degree $ 8 $ over $ \mathbb{Q} $. But then $ K(i) $ is an abelian extension of $ \mathbb{Q} $, and any extension of the form $ \mathbb{Q}(i, \sqrt[4]{A}) $ which is of degree $ 8 $ over $ \mathbb{Q} $ is not. Therefore, we cannot have $ K(i) = \mathbb{Q}(i, \sqrt[4]{A}) $.