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# Galois-theoretic derivation of the cubic formula

We are trying to find the roots $r_{1},r_{2},r_{3}$ of the polynomial $x^{3}+ax^{2}+bx+c=0$. From the equation

$(x-r_{1})(x-r_{2})(x-r_{3})=x^{3}+ax^{2}+bx+c$ |

we see that

$\displaystyle a$ | $\displaystyle=$ | $\displaystyle-(r_{1}+r_{2}+r_{3})$ | ||

$\displaystyle b$ | $\displaystyle=$ | $\displaystyle r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}$ | ||

$\displaystyle c$ | $\displaystyle=$ | $\displaystyle-r_{1}r_{2}r_{3}$ |

The goal is to explicitly construct a radical tower over the field $k=\mathbb{C}(a,b,c)$ that contains the three roots $r_{1},r_{2},r_{3}$.

Let $L=\mathbb{C}(r_{1},r_{2},r_{3})$. By Galois theory we know that $\operatorname{Gal}(L/\mathbb{C}(a,b,c))=S_{3}$. Let $K\subset L$ be the fixed field of $A_{3}\subset S_{3}$. We have a tower of field extensions

$\xymatrix{\makebox[10.0pt][l]{$L=\mathbb{C}(r_{1},r_{2},r_{3})$}\ar@{-}[d]_{{A% _{3}}}\\ \makebox[10.0pt][l]{$K=\ ?$}\ar@{-}[d]_{{S_{3}/A_{3}}}\\ \makebox[10.0pt][l]{$k=\mathbb{C}(a,b,c)$}}$ |

which we know from Galois theory is radical. We use Galois theory to find $K$ and exhibit radical generators for these extensions.

Let $\sigma:=(123)$ be a generator of $\operatorname{Gal}(L/K)=A_{3}$. Let $\omega=e^{{2\pi i/3}}\in\mathbb{C}\subset L$ be a primitive cube root of unity. Since $\omega$ has norm 1, Hilbert’s Theorem 90 tells us that $\omega=y/\sigma(y)$ for some $y\in L$. Galois theory (or Kummer theory) then tells us that $L=K(y)$ and $y^{3}\in K$, thus exhibiting $L$ as a radical extension of $K$.

The proof of Hilbert’s Theorem 90 provides a procedure for finding $y$, which is as follows: choose any $x\in L$, form the quantity

$\omega x+\omega^{2}\sigma(x)+\omega^{3}\sigma^{2}(x);$ |

then this quantity automatically yields a suitable value for $y$ provided that it is nonzero. In particular, choosing $x=r_{2}$ yields

$y=r_{1}+\omega r_{2}+\omega^{2}r_{3}.$ |

and we have $L=K(y)$ with $y^{3}\in K$. Moreover, since $\tau:=(23)$ does not fix $y^{3}$, it follows that $y^{3}\notin k$, and this, combined with $[K:k]=2$, shows that $K=k(y^{3})$.

Set $z:=\tau(y)=r_{1}+\omega^{2}r_{2}+\omega r_{3}$. Applying the same technique to the extension $K/k$, we find that $K=k(y^{3}-z^{3})$ with $(y^{3}-z^{3})^{2}\in k$, and this exhibits $K$ as a radical extension of $k$.

To get explicit formulas, start with $y^{3}+z^{3}$ and $y^{3}z^{3}$, which are fixed by $S_{3}$ and thus guaranteed to be in $k$. Using the reduction algorithm for symmetric polynomials, we find

$\displaystyle y^{3}+z^{3}$ | $\displaystyle=$ | $\displaystyle-2a^{3}+9ab-27c$ | ||

$\displaystyle y^{3}z^{3}$ | $\displaystyle=$ | $\displaystyle(a^{2}-3b)^{3}$ |

Solving this system for $y$ and $z$ yields

$\displaystyle y$ | $\displaystyle=$ | $\displaystyle\left(\frac{-2a^{3}+9ab-27c+\sqrt{(2a^{3}-9ab+27c)^{2}+4(-a^{2}+3% b)^{3}}}{2}\right)^{{1/3}}$ | ||

$\displaystyle z$ | $\displaystyle=$ | $\displaystyle\left(\frac{-2a^{3}+9ab-27c-\sqrt{(2a^{3}-9ab+27c)^{2}+4(-a^{2}+3% b)^{3}}}{2}\right)^{{1/3}}$ |

Now we solve the linear system

$\displaystyle a$ | $\displaystyle=$ | $\displaystyle-(r_{1}+r_{2}+r_{3})$ | ||

$\displaystyle y$ | $\displaystyle=$ | $\displaystyle r_{1}+\omega r_{2}+\omega^{2}r_{3}$ | ||

$\displaystyle z$ | $\displaystyle=$ | $\displaystyle r_{1}+\omega^{2}r_{2}+\omega r_{3}$ |

and we get

$\displaystyle r_{1}$ | $\displaystyle=$ | $\displaystyle\frac{1}{3}(-a+y+z)$ | ||

$\displaystyle r_{2}$ | $\displaystyle=$ | $\displaystyle\frac{1}{3}(-a+\omega^{2}y+\omega z)$ | ||

$\displaystyle r_{3}$ | $\displaystyle=$ | $\displaystyle\frac{1}{3}(-a+\omega y+\omega^{2}z)$ |

which expresses $r_{1},r_{2},r_{3}$ as radical expressions of $a,b,c$ by way of the previously obtained expressions for $y$ and $z$, and completes the derivation of the cubic formula.

## Mathematics Subject Classification

12D10*no label found*

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