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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?

  • Hint for one approach: Express $1, \beta, \beta^2, \beta^3$ in terms of $ 1, \alpha, \alpha^2$. Then solve the system of equations. $ \quad$ Hint for different approach: Convert the equation to one whose roots are $ \alpha^2 - \alpha + 1$. – Calvin Lin Nov 15 '21 at 20:23
  • Note: I'm not sure what you mean by "root of $f(x) = \ldots \in \mathbb{Q}$". Did you mean "root of $f(x) = \ldots = 0 $? – Calvin Lin Nov 15 '21 at 20:26
  • I suspect without knowing for sure, that the problem composer intends that you use that $\displaystyle \frac{\alpha^3 + 1}{\alpha + 1} = \alpha^2 - \alpha + 1.$ – user2661923 Nov 15 '21 at 20:33
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    An alternative (that I know Calvin Lin knows about :-) ) is the following. The number $\alpha-1/2$ is a zero of $$g(x)=f(x+1/2)=x^3+\frac32x^2+\frac{15}4x-\frac{11}8.$$ As $$-g(x)g(-x)=x^6+\frac{21}4x^4+\frac{291}{16}x^2-\frac{121}{64}=h(x^2),$$ where $$h(x)=x^3+\frac{21}4x^2+\frac{291}{16}x-\frac{121}{64},$$ we see that $(\alpha-1/2)^2$ is a zero of $h(x)$. But $\beta=(\alpha-1/2)^2+3/4$, so $\beta$ must be a zero of $$j(x)=h(x-\frac34)=x^3+3x^2+12x-13.$$ – Jyrki Lahtonen Nov 15 '21 at 22:16
  • (cont'd) The tricks here are simple, but the numbers are not very convenient. I think I prefer the methods in Calvin's answer for that reason. Observe that $g(x)g(-x)$ is even, and hence a polynomial in the variable $x^2$. That is a recurring trick for finding a polynomial whose zeros are the squares of the zeros of a given polynomial. – Jyrki Lahtonen Nov 15 '21 at 22:19
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    Yet another possible approach: represent $\alpha$ by the companion matrix $\begin{bmatrix} 0 & 0 & 3 \ 1 & 0 & -3 \ 0 & 1 & 0 \end{bmatrix}$; calculate the $3\times 3$ matrix corresponding to $\beta = \alpha^2 - \alpha + 1$; and then calculate the minimal polynomial of that $3\times 3$ matrix. – Daniel Schepler Nov 15 '21 at 22:26
  • Related: 1, 2 – Viktor Vaughn Nov 15 '21 at 23:27

2 Answers2

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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.
Calvin Lin
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  • +1: I don't even know anything about Field Theory, and still, I feel compelled to upvote this answer. – user2661923 Nov 15 '21 at 21:15
  • @user2661923 It's not needed. Because OP mentioned that they checked up to degree 5, I wanted to include an immediate explanation for why the degree is at most 3 (and since they tagged it with abstract algebra, I assumed field theory was alright). $\quad$ Looking at the coefficients as per above also yields this result (EG via linear algebra - because we have linear dependence). – Calvin Lin Nov 15 '21 at 21:18
  • +1 for going through the alternatives. Maybe also add approach 4 using resultants (better suited for a CAS) resultant(x^3+3x-3, z-(x^2-x+1), x). – dxiv Nov 15 '21 at 21:27
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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.

Trevor Gunn
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