Finding Minimal Polynomial over Rationals

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.

• This method of proof is fine, subject to the correctness of your field-degree calculation. The big difficulty must be to show that $\sqrt[3]2-i$ is not a square in $\Bbb Q(\sqrt[3]2,i)$, I suppose? Commented Apr 11, 2019 at 6:50
• $p(x)$ is irreducible mod $7$, but that's not something you want to do by hand.
– lhf
Commented Apr 11, 2019 at 11:24

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 $$12$$.
• Just showing that $x^4-x^2-1$ is irreducible over $\Bbb F_3$ is not enough to claim that $p(X)$ is irreducible over $\Bbb Q$? Commented Jul 2, 2019 at 14:19