Let $\alpha \in \mathbb{C}$ be a root of $f(x)=X^{3}+3X-3 \in \mathbb{Q}$. Find the minimal polynomial of $\beta=\alpha ^2-\alpha +1$ Let $\alpha \in \mathbb{C}$ be a root of $f(x)=X^{3}+3X-3 \in \mathbb{Q}$. Find the minimal polynomial of $\beta=\alpha ^2-\alpha +1$
I've tried working with $f(\alpha)=0$ and tried "forcing" $\beta$ or one of its powers into it, but I've been trying for the last hour and I'm up already trying to compute a 5th degree polynomial by multiplying $\alpha^{3}+3\alpha-3$ by $\alpha^{2}$ and trying to have an expression of only $\beta$ which equals $0$ but have been unsuccessfull (I've for every other degree under 5). Should I continue (meaning the degree of the minimal polynomial is greater than 5 (which I doubt)) or did I make an error and the degree it's less than 5?
 A: Approach 1: With $ \alpha^3 + 3 \alpha  -3 = 0$,  show that

*

*$\beta = \alpha^2 - \alpha + 1 $

*$ \beta^2 =  7\alpha  - 5 $

*$ \beta^3 = -12\alpha^2 - 9 \alpha + 16$

*Hence, $\beta^3 + 3 \beta^2 + 12 \beta -13 = 0  $.

*This is indeed minimal.

Notes

*

*A priori, it should be obvious that the degree of the minimal polynomial of $\beta$ is at most 3. (EG If you're familiar with field theory, then $F(\beta) \subset F(\alpha)$.) Hence we didn't need to search up to degree 5.

*Otherwise, it should follow from this (EG Finding the kernel of the matrix), that the degree of the minimal polynomial of any "polynomial of $\alpha$" is at most 3.


Approach 2: Using $ \beta = \frac{\alpha^3 + 1 } { \alpha + 1 } = \frac{ - 3\alpha + 4 } { \alpha + 1 } = - 3 + \frac{7}{\alpha + 1 }$, so $ \alpha = -1 + \frac{7}{\beta+3} = \frac{ - \beta + 4 } { \beta + 3 }$.
Substituting this into $ \alpha^3 + 3\alpha  - 3 = 0$ and multiplying out by the denominator $(\beta+3)^3$, we get $\beta^3 + 3\beta^2 + 12\beta - 13 = 0 $.
Again, verify that this is indeed minimal.
Notes:

*

*The subsitution need not be easy to find. In this case, were lucky to be working with a depressed cubic, that allows for the "linear rational" substitution. If quadratics were involved, there could be more work.


Approach 3. Using vietas formula, determine

*

*$ \sum (\alpha_i ^2 - \alpha_i + 1)$

*$ \sum (\alpha_i ^2 - \alpha_i + 1) (\alpha_j ^2 - \alpha_j + 1)$

*$\prod (\alpha_i ^2 - \alpha_i + 1)$

*Hence determine the cubic for $\beta$.
I didn't do the calculations, but you should end up with the same equation above.

A: The computational minded approach is to view this as a linear algebra problem.
Consider the ring/field $\mathbf{Q}[\alpha] = \operatorname{span}_{\mathbf{Q}} \{1, \alpha, \alpha^2\} = \{a + b\alpha + c\alpha^2 : a, b, c \in \mathbf{Q}\}$. This is a $3$ dimensional $\mathbf{Q}$-vector space. For any $\beta \in \mathbf{Q}[\alpha]$ we can consider the $\mathbf{Q}$-linear map $L_\beta(x) = \beta x$. And we have some properties: for all $k \in \mathbb{Q}$ and $\beta, \gamma \in \mathbf{Q}[\alpha]$,
$$L_\beta + L_\gamma = L_{\beta + \gamma}, \quad L_\beta \circ L_\gamma = L_{\beta\gamma}, \quad kL_{\beta} = L_{k\beta}$$
This implies that if $P(x)$ is any polynomial, then $P(L_\beta) = L_{P(\beta)}$. It follows that $P(\beta) = 0$ if and only if $P(L_\beta) = 0$. Therefore $\beta$ and $L_\beta$ have the same minimal polynomial.
So now we can just compute $L_\beta$ in the basis $1, \alpha, \alpha^2$:
\begin{align}
\beta \cdot 1        &= 1 - \alpha + \alpha^2 \\
\beta \cdot \alpha   &= 3 - 2\alpha - \alpha^2 \\
\beta \cdot \alpha^2 &= -3 + 6\alpha - 2\alpha^2
\end{align}
So the minimal polynomial of $\beta$ is the minimal polynomial of
\begin{pmatrix} 1 & 3 & -3 \\ -1 & -2 & 6 \\ 1 & -1 & -2 \end{pmatrix}
That happens to be equal to the characteristic polynomial in this case.
You can also compute this from $L_\alpha$ since $L_\beta = I - L_\alpha + L_\alpha^2$ and
$$L_\alpha = \begin{pmatrix} 0 & 0 & 3 \\ 1 & 0 & -3 \\ 0 & 1 & 0 \end{pmatrix}.$$

Computation approach #2, you can rephrase this problem as an elimination problem as dxiv pointed out. Let $X = V(x^3 + 3x - 3, t - (x^2 - x + 1))$ and compute the projection of $X$ onto the line $x = 0$. Which can be computed using a resultant/Gröbner bases.
So in Macaulay2 for instance:
i1 : R = QQ[x, t, MonomialOrder => Eliminate 1]
i2 : I = ideal"x3 + 3x - 3, t - x2 + x - 1"
i3 : eliminate(x, I)

            3     2
o3 = ideal(t  + 3t  + 12t - 13)

i4 : groebnerBasis I

o4 = | t3+3t2+12t-13 7x-t2-5 |

i5 : resultant(I_0, I_1, x)

      3     2
o5 = t  + 3t  + 12t - 13

depending on whether you want to compute the elimination via a resultant or via a Gröbner basis.
I recommend the book "Ideals, Varieties, and Algorithms" by Cox, Little and O'Shea if you want to learn more about this method.
