PlanetMath: Galois-theoretic derivation of the cubic formula

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (210)
Orphanage
Unclass'd (1)
Unproven (473)
Corrections (37)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

Galois-theoretic derivation of the cubic formula

(Proof)

We are trying to find the roots of the polynomial . From the equation

$\displaystyle (x-r_1)(x-r_2)(x-r_3) = x^3 + ax^2 + bx + c$

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

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

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ \makebox[1em][l]{$L = \mathbb{C}... ...\ar@{-}[d]_{S_3/A_3} \ \makebox[1em][l]{$k = \mathbb{C}(a,b,c)$} } } \end{xy}$

which we know from Galois theory is radical. We use Galois theory to find

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 and $y^3 \in K$ , thus exhibiting as a radical extension of .

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

$\displaystyle \omega x + \omega^2 \sigma(x) + \omega^3 \sigma^2(x);$

then this quantity automatically yields a suitable value for

provided that it is nonzero. In particular, choosing

yields

$\displaystyle y = r_1 + \omega r_2 + \omega^2 r_3.$

and we have

with $y^3 \in K$ . Moreover, since $\tau := (23)$ does not fix

, it follows that $y^3 \notin k$ , and this, combined with

, shows that

Set $z := \tau(y) = r_1 + \omega^2 r_2 + \omega r_3$ . Applying the same technique to the extension , we find that with $(y^3-z^3)^2 \in k$ , and this exhibits as a radical extension of .

To get explicit formulas, start with and , which are fixed by and thus guaranteed to be in . 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

and

yields

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

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

$\displaystyle r_1$	$\displaystyle =$	$\displaystyle \frac{1}{3} (-a + y + z)$
$\displaystyle r_2$	$\displaystyle =$	$\displaystyle \frac{1}{3} (-a + \omega^2 y + \omega z)$
$\displaystyle r_3$	$\displaystyle =$	$\displaystyle \frac{1}{3} (-a + \omega y + \omega^2 z)$