Field $\mathbb{Q}(\alpha)$ with $\alpha=\sqrt[3]7+2i$

Question

(1) Prove that $\alpha=\sqrt[3]7+2i$ is algebraic over $\mathbb{Q}.$
(2) Prove that both $\sqrt[3]7, 2i$ are elements of $\mathbb{Q}(\alpha)$.
(3) Compute $[\mathbb{Q}(\alpha):\mathbb{Q}]$.
(4) Find the minimal polynomial in $\mathbb{Q}[x]$.

Here is my attempt on each question

1. $x=\sqrt[3]7+2i$
   $x-2i=\sqrt[3]7$
   $(x-2i)^3=7$
   $x^3-12x-7=6ix^2-8i$
   $(x^3-12x-7)^2=(6ix^2-8i)^2$
   $x^6+12x^4-14x^3+48x^2+168x+113=0$

2. Let $\beta=\sqrt[3]7-2i$
   $\alpha+\beta=\sqrt[3]7$
   $\alpha-\beta=4i$
   I tried to use the fact $(\alpha+\beta)^3=7, (\alpha+\beta)^2=-16\in \mathbb{Q}$
   To show $\beta \in \mathbb{Q}(\alpha)$, but it wasn't successful so far

3. From (2), $\sqrt[3]7, i \in \mathbb{Q}(\alpha)$ and it's trivial that any new elment of field $\mathbb{Q}(\alpha)$ generated by $\alpha$ can be written with {$1, \sqrt[3]7, \sqrt[3]{7^2}, i, \sqrt[3]7i , \sqrt[3]{7^2}i$} and vice versa, i.e. $S=\mathbb{Q}(1, \sqrt[3]7, \sqrt[3]{7^2}, i, \sqrt[3]7i , \sqrt[3]{7^2}i) = \mathbb{Q}(\alpha)$
   Also,{$1, \sqrt[3]7, \sqrt[3]{7^2}, i, \sqrt[3]7i , \sqrt[3]{7^2}i$} are linearly independent over $\mathbb{Q}$.
   Therefore, $[\mathbb{Q}(\alpha):\mathbb{Q}]=6$

4. $p(x)=x^6+12x^4-14x^3+48x^2+168x+113$ from (1) is the minimal polynomial.
   $p(x)$ is a monic polynomial over $\mathbb{Q}$. To show it is irreducible over $\mathbb{Q}$,
   reducing mod 3 yields $x^6-2x^3+2$. This factors over $\mathbb{Z}_3$ if and only if
   it has a root in $\mathbb{Z_3}$, but it doesn't have a root. Hence, $p(x)$ is irreducible over $\mathbb{Z_3}$, so it is irreducible over $\mathbb{Q}$.

Any idea on question 2 or is there anything to fix in other questions?

Answer 1

For $2:$

Given that: $$x^6+12x^4-14x^3+48x^2+168x+113=(x^3-12x-7)^2+(6x^2-8)^2,$$ you have that:

$$\left(\frac{\alpha^3-12\alpha-7}{6\alpha^2-8}\right)^2=-1,$$

at least as long as you can show that $6\alpha^2\neq 8.$

This means that $$\frac{\alpha^3-12\alpha-7}{6\alpha^2-8}=\pm i$$ and hence $i\in\mathbb Q(\alpha)$.

For (3-4): Since $\mathbb Q(i)\subseteq \mathbb Q(\alpha)$ and $\mathbb Q(\sqrt[3]7)\subseteq \mathbb Q(\alpha)$, you have that $[Q(\alpha):\mathbb Q]$ must be divisible to $3$ and $2$, and thus be divisible by $6$.

But your polynomial also gives us that $[\mathbb Q(\alpha):\mathbb Q]\leq 6$.

So $\mathbb Q(\alpha):\mathbb Q]=6$, and thus the minimal polynomial of $\alpha$ must be degree $6$.

Answer 2

Your proof of the irreducibility of $f(x)=x^6 +12x^4-14x^3 +48x^2+168x+113$ is not correct. Here’s a way of doing it, though. Over $\Bbb Q(i)$, we have the factorization $$ f(x)=(x^3-6ix^2-12x+8i-7)(x^3+6ix^2-12x-8i-7)\,, $$ which can be found by various methods. These cubic polynomials are $\Bbb Q(i)$-irreducible, since their roots are not in $\Bbb Q(i)$, so this is the $\Bbb Q(i)$-factorization into irreducibles, unique.

Now assume $f$ had a nontrivial $\Bbb Q$ factorization. This would be a $\Bbb Q(i)$-factorization, and hence the product of two quadratics in the displayed formula. But these are not $\Bbb Q$-polynomials, contradicting our assumption. So $f$ is irreducible.