Lagrange quartic resolvent $x_1+ix_2-x_3-ix_4$

Question

Suppose we want to solve the "reduced" quartic equation $x^4+px^2+qx+r=0$ by means of Lagrange resolvent. I denote the roots by $x_1, x_2, x_3, x_4$; we have $x_1+x_2+x_3+x_4=0$.

In many texts one typically reads: the "generic" Lagrange resolvent would be $R=x_1+ix_2-x_3-ix_4$; however, Lagrange found easier resolvents, for example $(x_1+x_2)(x_3+x_4)$.

That's fine; but I still want to see how $R$ works.

When one permutes the roots in all possible ways, $R^4$ has six (rather than three) distinct values. So we get an equation $\prod (x-\ldots)$ of degree 6 (its coefficients are some polynomials in $p,q,r$). At this point some texts say "with some tricks, it can be reduced to degree 3". I computed this equation: its coefficients are not pleasant, and have no clear pattern. So I guess that the tricks should not be applied on the explicit form.

Hence the question: What are the tricks? How one reduces this equation to degree 3?

Answer 1

When I studied Lagrange's method, I had the same question as Caligari posted here. Although Michael's answer is correct, I would like to clarify it further with some more details.

First of all, the resolvent should be

$$ R =: \left((x_1-x_3) +i(x_2-x_4) \right)^4 $$

As we permute the roots, this resolvent will produce 6 different values but each value will appear with its conjugate therefore we can reduce the number of variables to 3, and then from $(X-t_1)(X-t_2)(X-t_3)$ we can compute $t_1,t_2,t_3$ in terms of coefficients as Michael explained. However, this is not super helpful. Because if you simplify the expressions you get something like this

$$ \begin{align*} t_1&= \left(a^2 - 2b - 2(x_1x_4+x_2x_3)\right)^4\\ t_2&= \left(a^2 - 2b - 2(x_1x_4+x_2x_3)\right)^4\\ t_3&= \left(a^2 - 2b - 2(x_1x_2+x_3x_4)\right)^4 \end{align*} $$ where $a$ and $b$ are coefficients of $x^4$ and $x^3$ in the original equation. Therefore, even if you find $t_1,t_2,t_3$, this process will give you $x_1x_4+x_2x_3$, $x_1x_4+x_2x_3$, and $x_1x_2+x_3x_4$. Observing that it makes sense that one would define the resolvent simply as $$ \tilde{R} := x_1x_2+x_3x_4 $$ Indeed, this resolvent takes only 3 different values as roots are permuted, and we can compute the same quantities ($x_1x_4+x_2x_3$, $x_1x_4+x_2x_3$, and $x_1x_2+x_3x_4$) more directly with this resolvent. Most likely that is why Lagrange didn't use a fourth root of unity in his resolvent for quartic.

Finally, one (as Peter Pesic did in his book on Abel's proof) could also work with
$$ \hat{R} \equiv \left(x_1+x_2 -x_3 -x_4\right)^2 $$ But in this case too, it can be shown that it is equal to $$ \hat{R} = a^2 - 2b + 4(x_1x_2 + x_3x_4) $$ and therefore again the same expression, $x_1x_2 + x_3x_4$, appears and the same story...

Answer 2

I tried to query the original work of Lagrange for answers. Even though I do not speak french, I believe I found the solution.

The lagrange resolvent for the quartic is $R = x_1 + i x_2 - x_3 -i x_4$. You can write this as a complex number

$$R = (x_1 - x_3) + i(x_2 - x_4)$$

This lagrange resolvent has $4!=24$ permutations. One of these permutations is

$$ \bar{R} = (x_1 - x_3)-i(x_2 - x_4) $$

the complex conjugate of $R$.

Note that

$$R\bar{R} = (x_1-x_3)^2+(x_2-x_4)^2$$

only has three images under all $24$ permutations of $x_1,x_2,x_3,x_4$:

$$\begin{alignat*}{5} R\bar{R} &={}&(x_1-x_3)^2+(x_2-x_4)^2 &=:{}&t_1 \\ &&(x_1-x_2)^2+(x_3-x_4)^2 &=:{}&t_2 \\ &&(x_1-x_4)^2+(x_2-x_3)^2 &=:{}&t_3 \\ \end{alignat*}$$

These are fixed by the klein four group $V_4$ and can be found as the solution of a cubic polynomial

$$(X-t_1)(X-t_2)(X-t_3).$$

Answer 3

I would like to add even more detail to the answer by @UWS. I, too, have realized that one should examine the orbit of $R\bar{R}$ and found the $x_1x_2+x_3x_4$ expressions. But then it's still not immediately clear how to actually recover the roots from these expressions. So here's my attempt:

• Consider $(x_1+x_2)^2 + (x_3+x_4)^2 = x_1^2 + x_2^2 +x_3^2 + x_4^2 + 2(x_1x_2+x_3x_4)$
• Because we have $x_1+x_2+x_3+x_4 = 0$, we get $(x_1+x_2)^2 + (x_3+x_4)^2 = (x_1+x_2)^2 + (-x_1-x_2)^2 = 2(x_1+x_2)^2$
• $x_1^2 + x_2^2 +x_3^2 + x_4^2$ is symmetric in the roots, therefore it equals some S which is a polynomial expression in the coefficients.
• From the resolvent cubic we found that $x_1x_2+x_3x_4$ equals some Y_1 (again a function of the coefficients). Similarly for the other expressions in the orbit.
• Rearranging we get $$ x_1+x_2 = \sqrt{\frac{S}2 + Y_1} \\ x_3+x_4 = -\sqrt{\frac{S}2 + Y_1} \\ x_1+x_3 = \sqrt{\frac{S}2 + Y_2} \\ x_2+x_4 = -\sqrt{\frac{S}2 + Y_2} \\ x_1+x_4 = \sqrt{\frac{S}2 + Y_3} \\ x_2+x_3 = -\sqrt{\frac{S}2 + Y_3} \\ $$
• From here we can easily solve for each $x_i$