PlanetMath: Galois-theoretic derivation of the quartic formula

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Galois-theoretic derivation of the quartic formula

(Proof)

Let be a general polynomial with four roots , so . The goal is to exhibit the field extension $\mathbb{C}(r_1,r_2,r_3,r_4)/\mathbb{C}(a,b,c,d)$ as a radical extension, thereby expressing in terms of by radicals.

Write for $\mathbb{C}(r_1,r_2,r_3,r_4)$ and for $\mathbb{C}(a,b,c,d)$ . The Galois group $\operatorname{Gal}(N/F)$ is the symmetric group , the permutation group on the four elements $\{r_1,r_2,r_3,r_4\}$ , which has a composition series

$\displaystyle 1 \lhd \mathbb{Z}/2 \lhd V_4 \lhd A_4 \lhd S_4,$

where:

is the alternating group in , consisting of the even permutations.
$V_4 = \{1, (12)(34), (13)(24), (14)(23)\}$ is the Klein four-group.
$\mathbb{Z}/2$ is the two-element subgroup $\{1, (12)(34)\}$ of .

Under the Galois correspondence, each of these subgroups corresponds to an intermediate field of the extension . We denote these fixed fields by (in increasing order) , , and .

We thus have a tower of field extensions, and corresponding automorphism groups:

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ \text{Subgroup} & \text{Fixed fi... ...\ar@{-}[d] \ V & L \ar@{-}[d] \ A_4 & K \ar@{-}[d] \ S_4 & F } } \end{xy}$

By Galois theory, or Kummer theory, each field in this diagram is a radical extension of the one below it, and our job is done if we explicitly find what the radical extension is in each case.

We start with . The index of in is two, so is a degree two extension. We have to find an element of that is not in . The easiest such element to take is the element $\Delta$ obtained by taking the products of the differences of the roots, namely,

$\displaystyle \Delta := \prod_{1 \leq i < j \leq 4} (r_i - r_j) = (r_1-r_2) (r_1-r_3) (r_1-r_4) (r_2-r_3) (r_2-r_4) (r_3-r_4).$

Observe that $\Delta$ is fixed by any even permutation of the roots

, but that $\sigma(\Delta) = -\Delta$ for any odd permutation $\sigma$ . Accordingly, $\Delta^2$ is actually fixed by all of

, so:

$\Delta \in K$ , but $\Delta \notin F$ .
$\Delta^2 \in F$ .
$K = F[\Delta]$ = $F[\sqrt{\Delta^2}]$ , thus exhibiting as a radical extension.

The element $\Delta^2 \in F$ is called the discriminant of the polynomial. An explicit formula for $\Delta^2$ can be found using the reduction algorithm for symmetric polynomials, and, although it is not needed for our purposes, we list it here for reference:

$\displaystyle \Delta^2$	$\displaystyle =$	$\displaystyle 256 d^3 - d^2 (27 a^4 - 144 a^2 b + 128 b^2 + 192 a c) -$
		$\displaystyle c^2 (27c^2 - 18 abc + 4a^3 c + 4 b^3 - a^2 b^2) -$
		$\displaystyle 2d (abc (9a^2 - 40 b) - 2b^3 (a^2-4b) - 3c^2 (a^2-24b)).$

Next up is the extension , which has degree 3 since . We have to find an element of which is fixed by but not by . Luckily, the form of almost cries out that the following elements be used:

$\displaystyle t_1$	$\displaystyle :=$	$\displaystyle (r_1+r_2)(r_3+r_4)$
$\displaystyle t_2$	$\displaystyle :=$	$\displaystyle (r_1+r_3)(r_2+r_4)$
$\displaystyle t_3$	$\displaystyle :=$	$\displaystyle (r_1+r_4)(r_2+r_3)$

These three elements of are fixed by everything in , but not by everything in . They are therefore elements of that are not in . Moreover, every permutation in permutes the set $\{t_1, t_2, t_3\}$ , so the cubic polynomial

$\displaystyle \Phi(x) := (x-t_1) (x-t_2) (x-t_3)$

actually has coefficients in

! In fancier language, the cubic polynomial $\Phi(x)$ defines a cubic extension

which is linearly disjoint from

, with the composite extension

equal to

. The polynomial $\Phi(x)$ is called the resolvent cubic of the quartic polynomial

. The coefficients of $\Phi(x)$ can be found fairly easily using (again) the reduction algorithm for symmetric polynomials, which yields

$\displaystyle \Phi(x) = x^3 - 2b x^2 + (b^2+ac-4d) x + (c^2+a^2d - abc).$

(1)

Using the cubic formula, one can find radical expressions for the three roots of this polynomial, which are

, and

, and henceforth we assume radical expressions for these three quantities are known. We also have

, which in light of what we just said, exhibits

as an explicit radical extension.

The remaining extensions are easier and the reader who has followed to this point should have no trouble with the rest. For the degree two extension , we require an element of that is not in ; one convenient such element is , which is a root of the quadratic polynomial

$\displaystyle (x-(r_1+r_2)) (x-(r_3+r_4)) = x^2 + a x + t_1 \in L[x]$

(2)

and therefore equals $(-a + \sqrt{a^2 - 4 t_1})/2$ . Hence $M = L[r_1+r_2] = L[(-a + \sqrt{a^2 - 4 t_1})/2]$ is a radical extension of

Finally, for the extension , an element of that is not in is of course , which is a root of the quadratic polynomial

$\displaystyle (x-r_1)(x-r_2) = x^2 - (r_1+r_2) x + r_1 r_2.$

(3)

Now,

is known from the previous paragraph, so it remains to find an expression for

. Note that

is fixed by

, so it is in

but not in

. To find it, use the equation

, which gives

$\displaystyle (x - r_1 r_2) (x - r_3 r_4) = x^2 - \frac{(t_2 + t_3 - t_1)}{2} x + d$

and, upon solving for

with the quadratic formula, yields

$\displaystyle r_1 r_2$	$\displaystyle =$	$\displaystyle \frac{(t_2 + t_3 - t_1) + \sqrt{(t_2 + t_3 - t_1)^2 - 16d}}{4}$	(4)
$\displaystyle r_3 r_4$	$\displaystyle =$	$\displaystyle \frac{(t_2 + t_3 - t_1) - \sqrt{(t_2 + t_3 - t_1)^2 - 16d}}{4}$	(5)

We can then use this expression, combined with Equation (3), to solve for

using the quadratic formula. Perhaps, at this point, our poor reader needs a summary of the procedure, so we give one here:

Find , , and by solving the resolvent cubic (Equation (1)) using the cubic formula,
From Equation (2), obtain

$\displaystyle r_1+r_2$ $\displaystyle =$ $\displaystyle \frac{(-a + \sqrt{a^2 - 4 t_1})}{2}$

$\displaystyle r_3+r_4$ $\displaystyle =$ $\displaystyle \frac{(-a - \sqrt{a^2 - 4 t_1})}{2}$
Using Equation (3), write

$\displaystyle r_1$ $\displaystyle =$ $\displaystyle \frac{(r_1+r_2) + \sqrt{(r_1+r_2)^2 - 4 (r_1 r_2)}}{2}$

$\displaystyle r_2$ $\displaystyle =$ $\displaystyle \frac{(r_1+r_2) - \sqrt{(r_1+r_2)^2 - 4 (r_1 r_2)}}{2}$

$\displaystyle r_3$ $\displaystyle =$ $\displaystyle \frac{(r_3+r_4) + \sqrt{(r_3+r_4)^2 - 4 (r_3 r_4)}}{2}$

$\displaystyle r_4$ $\displaystyle =$ $\displaystyle \frac{(r_3+r_4) - \sqrt{(r_3+r_4)^2 - 4 (r_3 r_4)}}{2}$

where the expressions and are derived in the previous step, and the expressions and come from Equation (4) and (5).
Now the roots of the quartic polynomial have been found, and we are done!