Determine $[\mathbb{Q}(\sqrt[4]{11}+i\sqrt[4]{11}):\mathbb{Q}]$, and determine if $\mathbb{Q}(\sqrt[4]{11}+i\sqrt[4]{11})/\mathbb{Q}$ is normal

Question

Determine $[\mathbb{Q}(\sqrt[4]{11}+i\sqrt[4]{11}):\mathbb{Q}]$, and determine if $\mathbb{Q}(\sqrt[4]{11}+i\sqrt[4]{11})/\mathbb{Q}$ is normal.

If we let $x = \sqrt[4]{11}+i\sqrt[4]{11} = \sqrt[4]{11}(1+i)$ then take the 4th power on both sides we have $x^4=-44\implies x^4 +44=0$. But I got stuck to showing that this polynomial is irreducible over $\mathbb{Q}$. I also tried to show that $\mathbb{Q}(\sqrt[4]{11}+i\sqrt[4]{11}) = \mathbb{Q}(i\sqrt[4]{11})$ but I don't know how. If this were true, then $\mathbb{Q}(\sqrt[4]{11}+i\sqrt[4]{11})/\mathbb{Q}$ would be normal. Someone please help me out here! Much appreciated.

Answer 1

Eisenstein with the prime $11$ shows that $x^4+44$ is irreducible.

Is $\Bbb Q(\sqrt[4]{11}(1+i))$ a splitting field for this polynomial? No, it is not.

If the field contained $\sqrt[4]{11}(-1+i)$, then it would contain $$\sqrt{11}=\frac{(\sqrt[4]{11}(1+i))^3}{\sqrt[4]{11}(-1+i)}$$ And therefore it would contain $$\sqrt[4]{11}=\frac{(\sqrt[4]{11}(1+i))^2+\sqrt{11}}{\sqrt[4]{11}(1+i)}$$ But $\Bbb Q(\sqrt[4]{11})/\Bbb Q$ is a degree-4 extension which doesn't contain $\sqrt[4]{11}(1+i)$, so $\Bbb Q(\sqrt[4]{11}, \sqrt[4]{11}(1+i))/\Bbb Q$ cannot have degree $4$ and therefore cannot be our field.