<s>[INST] Find minimal polynomial of this element? Let $f(x)=x^3+x+1$. $\alpha_1, \alpha_2, \alpha_3 - $ roots of $f$. The task is to determine the minimal polynomial of $\frac{\alpha_1}{\alpha_2}$ over $\Bbb Q $ and $\Bbb Q(\alpha_1)$.
My thoughts (not sure if everything is correct)
There is only one real root of $f$, since $f'=3x^2+1>0$. So $x^3+x+1=(x-b)(x-w)(x-\overline{w})$ for some $w \in \Bbb{C} \setminus \Bbb{R}$ and $b \in \Bbb{R}$. If we open brackets then we get $$ \begin{cases}b=-w-\overline{w}\\ b(w+\overline{w})+w\overline{w}=1\\aw\overline{w}=-1\end{cases}$$
This set of equations might somehow lead to the fact that $\Bbb Q (\frac{\alpha_1}{\alpha_2}) = \Bbb Q(\alpha_1, \alpha_2, \alpha_3)$ and $[\Bbb Q (\frac{\alpha_1}{\alpha_2}):\Bbb Q] = 6$ (which should be if we take a look at first equation), regardless what roots we call $\alpha_1, \alpha_2, \alpha_3$. So we get that the minimal polynomial of $\frac{\alpha_1}{\alpha_2}$ has degree 6. And since $[\Bbb{Q}(\alpha_1):\Bbb{Q}]=3$, then the minimal polynomial over $\Bbb{Q}(\alpha_1)$ must have degree 2. But how can I figure out what are they?
Thanks in advance!
 [/INST] $$a^3+a+1=0 \tag1$$
$$b^3+b+1=0 \tag2$$
Moving the $1$ to the RHS and multiplying
$$\implies a^3b^3+ab^3+ba^3+ab=1\tag3$$
Substracting $(2)$ from $(1)$
$$\implies a^3-b^3=b-a\implies a^2+ab+b^2=-1\implies a^3b+a^2b^2+ab^3+ab=0\tag4$$
Substracting $(4)$ from $(3)$
$$a^3b^3-a^2b^2-1=0\tag5$$
Let $t=\frac{a}{b}$.Then, from $a^2+ab+b^2=-1$, we have that
$$t^2+t+1=-b^{-2}\implies (t^2+t+1)^3=-b^{-6}\tag6$$
But also
$$t^2+2t+1=-b^{-2}+\frac{a}{b}\implies t^4+2t^3+t^2=-\frac{a^2}{b^{4}}+\frac{a^3}{b^{3}}\tag7$$
Summing up $(6)$ and $(7)$
$$(t^2+t+1)^3+(t^2+t)^2=\frac{-1-a^2b^2+a^3b^3}{b^6}$$
And here comes $(5)$ to the rescue :)
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