Lagrange Method of Solving Cubic Equations

Question

Let $K$ be a field (for simplicity, one may assume $K\subseteq\mathbb{C}$), and let $d\in K$. Denote $w=\sqrt[3]{d}$, and $\zeta=(-1+\sqrt{-3})/2$. If $f(x)=x^3+ax^2+bx+c$ is the minimal polynomial of $x_{1}=\alpha+\beta w+\gamma w^2$, show that:

1. The other two roots of $f$ are $x_{2}=\alpha+\zeta\beta w+\zeta^2\gamma w^2$ and $x_{3}=\alpha+\zeta^2\beta w+\zeta\gamma w^2$.
2. $\alpha = (x_1+x_2+x_3)/3$, $\beta w=(x_1+\zeta x_2 +\zeta^2 x_3)/3$ and $\gamma w^2=(x_1+\zeta^2x_2+\zeta x_3).$

I tried expanding the term $f(x_1)$ (see below), using the fact that $1+\zeta+\zeta^2=0$ and $\zeta^3=1$, however I found that under the transformation $\beta \mapsto\zeta\beta, \gamma \mapsto\zeta^2\gamma$ the coefficients $a_j$ of $w^j$ ($0\le j\le 6$) are multiplied by $\zeta^j$ too. How can I show that this transformation does not change the result and deduce that $f(x_2)=f(x_3)=0$, without explicitly solving the original equation?

The expansion of $f(x_1)$ in terms of $w$ is $$f(x_1 )=(γ^3 ) w^6+(3βγ^2 ) w^5+(aγ^2+3αγ^2+3β^2 γ) w^4 +(6αβγ+2aβγ+β^3 ) w^3+(2aαγ+aβ^2+3αβ^2+bγ+3α^2 γ) w^2 +(2aαβ+3α^2 β+bβ)w+(aα^2+αb+c+α^3 ),$$ however I do not see how it helps. Any elegant suggestions?

Answer 1

We may assume $\omega$ is not in $K$. Let $L=K(\omega,\zeta)$; then $L$ is the splitting field for $f$ over $K$. The Galois group of $L$ over $K$ is either cyclic of order three (if $\zeta$ is in $K$) or it's the symmetric group on three letters; in either case, the conjugates of $\omega$ over $K$ are $\sigma(\omega)=\zeta\omega$ and $\sigma^2(\omega)=\zeta^2\omega$ for an element $\sigma$ of order three in the Galois group. Then the conjugates of $x_1$ over $K$ are $\sigma(x_1)=\sigma(\alpha+\beta\omega+\gamma\omega^2)=\alpha+\beta\zeta\omega+\gamma\zeta^2\omega^2=x_2$ and $\sigma^2(x_1)=\sigma^2(\alpha+\beta\omega+\gamma\omega^2)=\alpha+\beta\zeta^2\omega+\gamma\zeta\omega^2=x_3$. So, those are the other two roots of $f$.

Answer 2

For the second part, it can be shown merely by calculation: $$\begin{align}x_1+x_2+x_3&=(α+βw+γw^2 )+(α+ζβw+ζ^2γw^2 )+(α+ζ^2βw+ζγw^2)\\& =3α+βw(1+ζ+ζ^2 )+γw^2 (1+ζ+ζ^2 ),\end{align}$$ and by using $1+ζ+ζ^2=0$, as it is a root of $\Phi_3(x)=1+x+x^2=\frac{x^3-1}{x-1}$: $$α=\frac{1}{3} (x_1 + x_2 + x_3).$$ In a similar fashion: $$\begin{align}x_1+ζx_2+ζ^2x_3&=(α+βw+γw^2)+ζ(α+ζβw+ζ^2 γw^2)+ζ^2 (α+ζ^2 βw+ζγw^2 )\\&=α(1+ζ+ζ^2 )+βw(1+ζ^2+ζ^4 )+γw^2 (1+ζ^3+ζ^3 )\\&=3γw^2,\end{align}$$ and $$\begin{align}x_1+x_2ζ^2+x_3ζ&=(α+βw+γw^2)+ζ^2 (α+ζβw+ζ^2 γw^2)+ζ(α+ζ^2 βw+ζγw^2)\\&=α(1+ζ+ζ^2 )+3βw+γw^2 (1+ζ^2+ζ^4)\\&=3\beta w.\end{align} $$

Answer 3

Adding to Gerry's suggestion, another way to explicitly calculate the minimal polynomial is by using the Cayley–Hamilton theorem.

Assume $w∉K$. In this case, $B=(1,w,w^2)$ is a base of $K(w)$ over $K$. Consider the linear map $T_1: K(w)\to K(w), x\mapsto x_1x$ of multiplication in $x_1$. The representation of $T_1$ according to $B$ is $$\left[T_1\right]_B=\left(\begin{matrix}α&γd&βd \\ β&α&γd \\ γ&β&α\end{matrix}\right).$$

From this we obtain the characteristic polynomial $$χ(x)=x^3-3αx^2+(3α^2-3βγd)x+(3αβγd-γ^3 d^2-β^3 d-α^3)$$ of $x_1$. From the Cayley–Hamilton theorem it follows that for every $b\in K(w)$ we have $(χ(T_1))(b)=χ(x_1 b)=0$. In particular, for $b=1$ we see that $\chi(x_1)=0$. The extension $K(w)/K$ is of degree $[K(w) : K] = 3=\deg(\chi)$, hence $\chi$ is the minimal polynomial of $x_1$ over $K$. It is easy to see that $\chi$ is invariant under the transformation $\beta\mapsto\zeta\beta, \gamma\mapsto\zeta^2\gamma$, and since $x_1$ maps to $x_2$ under this transformation (and $x_1\mapsto x_3$ by applying it twice), the minimal polynomials of $x_2$ and $x_3$ are $\chi$ as well.