I have been dumb stuck on the following problem for quite a while: $$\text{Prove by finding minimal polynomial that: }$$ $$\sqrt{\sqrt[3]{2}-i} \text{ is algebraic over }\mathbb{Q}.$$
My attempt:
Through squaring and cubing, I have been able to find that $$p(x)=x^{12}+3x^8-4x^6+3x^4+12x^2+5$$ has the above described element as its root. Now the only caveat is to exhibit that $p(x)$ is minimal. As one could imagine this entails proving that $p(x)$ is irreducible over $\mathbb{Q}$. Now, I tried applying Eisenstein criterion and Gauss’s Lemma, but they don’t seem to work straight away, and I don’t know how to get around that, unfortunately. So my approach was to prove that $\mathbb{Q}(\sqrt{\sqrt[3]{2}-i})$ has dimension 12 over $\mathbb{Q}$. Hence, $\sqrt{\sqrt[3]{2}-i})$ has degree 12 over $\mathbb{Q}$. Thus, I conclude that the polynomial, which I found, that is also of degree 12 and is monic, must be the minimal polynomial we were looking for.
I know that this is very sloppy, but I have no other approach. Some people proposed different solutions, which use Galois theory or some very convoluted ways of proving the irreducibility of $p(x)$, but I could not make sense out of those arguments.
Best Answer
If you are happy with extensions over finite fields, here's one method. Over $\mathbb F_3$ we have $p(x)=q(x^3)=q(x)^3$ where $q=x^4-x^2-1$. Then $q$ is irreducible: it has no linear factor, and any factorisation into squares must be $(x^2+ax+b)(x^2-ax+b)$ to cancel odd degrees, so $b^2=-1$, a contradiction. We deduce that every irreducible factor of $p$ has degree a multiple of $4$, so the degree of the extension is one of $1,4,8,12$.
On the other hand, $\alpha^2=\sqrt[3]2-i$ lies in the splitting field, and hence so does its complex conjugate, so $\sqrt[3]2,i$ both lie in the splitting field. Thus the degree of the extension is divisible by both $2$ and $3$.
The only possibility is therefore that the splitting field extension has degree $12$.