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!
 A: The Galois group of your polynomial is $S_3,$ and so all of the $\beta_{ij} = \alpha_i/\alpha_j$ are conjugate. So, the coefficients of the minimal polynomial are the various symmetric functions of the $\beta_{ij}.$ Notice that since the roots of this polynomial come in inverse pairs $\beta_{ij} \beta_{ji} = 1,$ the polynomial is reciprocal (that is the $k$-th coefficient equals the $6-k$th coefficent. The constant term is, therefore, $1.$ The rest is just some tedious computation, as long as you remember that the symmetric functions of the $\alpha_i$ are the coefficients of your original polynomial.
A: From the coefficients you get $\alpha_1+\alpha_2+\alpha_3 = 0$ and $\alpha_1\alpha_2\alpha_3=-1$. So $1=(\alpha_1+\alpha_2)\alpha_1\alpha_2$ giving you $\alpha_2^{-1}= (\alpha_1+\alpha_2)\alpha_1$ and $$\frac{\alpha_1}{\alpha_2} = \alpha_1^2\alpha_2-\alpha_1-1.$$
On the other hand, $\alpha_2^3 + \alpha_2 = -1$ and therefore $\alpha_2^{-1} = -(\alpha_2^2+1)$, leading to $$\frac{\alpha_1}{\alpha_2} = -\alpha_1(\alpha_2^2+1).$$
From there you can easily calculate the minimal polynomial of $\frac{\alpha_1}{\alpha_2}$ over $\Bbb Q(\alpha_1)$.
A: $$a^3+a+1=0 \tag1$$
$$b^3+b+1=0 \tag2$$
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 :)
