Finding irreducible polynomial over the rationals Question : Let $f(x)=x^3-3x+3$ on $\mathbb{Q}$, and $\alpha$ is a complex root of $f(x)$.
For $\beta=1-\alpha+{\alpha}^2 $, find the minimal polynomial of $ \beta$ over $\mathbb{Q}$, $\ irr(\beta,\mathbb{Q})$.
My attempt was this ;
Let $\mathbb{Q}(\alpha)$, simple extension of $\alpha$ over $\mathbb{Q}$.
Since $\beta =1-\alpha+{\alpha}^2 \in \mathbb{Q}(\alpha)$, $\mathbb{Q}(\beta) \subset \mathbb{Q}(\alpha)$, so $\mathbb{Q}(\beta) \leqslant \mathbb{Q}(\alpha)$.
A$\ f(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein, so$\ [ \mathbb{Q}(\alpha): \mathbb{Q}]=deg(f(x)=3 $.
$\beta \notin \mathbb{Q}$, and 3 is a prime,
So [$\mathbb{Q}(\alpha): \mathbb{Q}(\beta)$]=1 which includes $\mathbb{Q}(\alpha)= \mathbb{Q}(\beta)$ so $\ irr(\beta,\mathbb{Q})=f(x)$.
But it was wrong... What am I missing?
I'm guessing the part I got $\mathbb{Q}(\beta) \leqslant \mathbb{Q}(\alpha)$ is wrong, but can't seem to find the reason why. Thanks in advance!
 A: $\newcommand{\QQ}{\mathbb Q}$The property $\QQ(\beta)=\QQ(\alpha)$ doesn't imply that $\beta$ and $\alpha$ have the same minimal polynomial. For a concrete example, $\QQ(\sqrt 2)=\QQ(\sqrt 2-1)$, but the first has minimal polynomial $x^2-2$ while the second has minimal polynomial $(x+1)^2-2=x^2+2x-1$.
A hint as for how to continue: You've already shown that the minimal polynomial of $\beta$ is degree $3$, so its roots must be the three Galois conjugates $\{\beta_1,\beta_2,\beta_3\}$ of $\beta$ which correspond to the three roots $\alpha_1,\alpha_2,\alpha_3$ of $f(x)$. See if you can use Vieta's formulas to compute the coefficients of the minimal polynomial of $\beta$ in terms of symmetric functions of the $\beta_i$, and then compute these using symmetric functions of the $\alpha_i$.
A: You can use the isomorphism of $\mathbb{Q}(\alpha)$ to the field generated by the identity matrix and $A$ where $A$ is the companion matrix of $f$, so
$$A=\left( \begin{array}{rrr}
0 & 0 & -3 \\ 
1 & 0 & 3 \\
0 & 1 & 0 \\ 
\end{array}\right).$$
To get the minimal polynomial of $\beta$ calculate $B=1-A+A^2$ and note that it is not diagonal, so since the order of the field $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is prime, $B$ and hence $\beta$ have a minimal polynomial of degree $3$. Now by the Cayley-Hamilton theorem we yield that the minimal polynomial of $\beta$ is the characteristic polynomial of $B$.
We get
$$B=\left( \begin{array}{rrr}
1 & -3 & 3 \\ 
-1 & 4 & -6 \\
1 & -1 & 4 \\ 
\end{array}\right)$$
which has as characteristic polynomial $x^3-9x^2+12x-7$.
