What's the degree of $[\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2}):\mathbb{Q}]$?
I know that $\sqrt[4]{2},i\sqrt[4]{2}$ are roots of $P(X)=X^4-2$ and $P$ is irreducible. So I figured that the degree should be $4$.
On the other hand I have $[\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2}):\mathbb{Q}(\sqrt[4]{2})]\cdot [\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]$
But $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]$ should also be $4$ which leaves $[\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2}):\mathbb{Q}(\sqrt[4]{2})]=1$ which is impossible because then they would be the same field.
Where am I thinking wrong?
To clarify, $x^4-2$ is irreducible, over $\mathbb{Q}$. If we add$\sqrt{2}$, then the polynomial splits into two irreducible polynomials, $$x^4-2=(x^2-\sqrt{2})(x^2+\sqrt{2})$$ Now we can add a root of one of these, say $\sqrt[4]{2}$ then this will split the first factor, but not the second. So this an example of where adding a root, $\sqrt[4]{2}$ of $x^4-2$ does not split the polynomial. Your field is the splitting field of $P$. To split the polynomial we have to add two roots, one of $x^2-\sqrt{2}$ and another of $x^2+\sqrt{2}$