I'm trying to understand the technique used by Lagrange to solve cubic and quartic equations. I have read that the Lagrange resolvent for the cubic is

$$ x_1+\omega x_2+ \omega^2 x_3 $$

where $\omega$ is the principal cubic root of 1.

My question is: Why isn't the resolvent for the quartic

$$ x_1+\omega x_2 +\omega^2 x_3 +\omega^3 x_4 $$

where $\omega$ is the principal quartic root of 1?

Why did Lagrange use $x_1-x_2+x_3-x_4$?

Is there an intuitive explanation?

More generally, is there an intuitive way to understand Lagrange resolvent?


Recall that if $P\in K_n[x]$ admits the roots $(\alpha_i)_{i\leq n}$, then its discriminant is $discrim(P)=\Pi_{i<j}(\alpha_i-\alpha_j)^2\in K$.

A Lagrange (1736-1813) resolvent is an expression in the form $\sum_{i\leq n}\omega^j\alpha_j$ where $\omega^n=1$.

Part $1$: Degree $3$. Let $P(x)=x^3+px+q, j=\exp(2i\pi/3),K=\mathbb{Q}(p,q,j)$. Let $L=K(\alpha_1,\alpha_2,\alpha_3)$ be the decomposition field of $P$. Generically, $L$ is an algebraic extension of $K$ of degree $6$: $[K,L]=6$.

More precisely, $K\subset M=K(\sqrt{D})\subset L$ with $[M:L]=3$. We consider the Lagrange resolvents: $u=\alpha_1+j\alpha_2+j^2\alpha_3,v=\alpha_1+j^2\alpha_2+j\alpha_3$. We can show that $L=M(u)=M(v)$. The key is that $u^3,v^3$ are roots of this polynomial $\in K_2[x]$: $x^2+27qx-27p^3$. The previous polynomial was found by Del Ferro (1515), Tartaglia (1535) and Cardan (1545).

Thus we obtain, $u^3,v^3,u,v$ and the required roots.

Part $2$: Degree $4$. Let $P(x)=x^4+px^2+qx+r,K=\mathbb{Q}(p,q,r,i)$. Let $L=K(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ be the decomposition field of $P$. Generically, $L$ is an algebraic extension of $K$ of degree $24$: $[K,L]=24$. Note that, since Galois (1811-1832), we know that $\Sigma_4$ acts on the set of roots; there are several resolvent polynomials. The idea is to find an expression (that depends on the roots) which has only $3$ values under the action of $\Sigma_4$ (recall that, since few minutes, we know how to solve an equation of degree $3$). Behind these constructions lie the properties of the symmetrical functions of the roots.

$1$) $u=(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)$, under the action of $\Sigma_4$, takes only two other values $v=(\alpha_1+\alpha_3)(\alpha_2+\alpha_4),w=(\alpha_1+\alpha_4)(\alpha_2+\alpha_3)$. Then they are roots of a polynomial $\in K_3[x]$: $x^3-2px^2+(p^2-4r)x+q^2$. Then we calculate $u,v,w$ and deduce the roots $(\alpha_i)$.

$2)$ $u=\alpha_1\alpha_2+\alpha_3\alpha_4$ has the same property. The elements of the orbit are roots of $x^3-px^2-4rx+4pr-q^2$. This polynomial was found by Ferrari (vroom) (1540) and Cardan (1545); they did not use the symmetrical functions of the roots because these functions were discovered in 1629 by Girard.

$3)$ The Lagrange resolvent $t_1=\alpha_1+i\alpha_2-\alpha_3-i\alpha_4\;(\times i)$ with $t_2=\alpha_1-\alpha_2+\alpha_3-\alpha_4\;(\times i^2)$ and $t_3=\cdots\;(\times i^3)$; we can work with $t_1^4,t_2^4,t_3^4$; yet, Lagrange found a simpler method.

Indeed, $t_2^2$ takes $2$ other values under the action of $\Sigma_4$: ${t'}_2^2=(\alpha_1+\alpha_2-\alpha_3-\alpha_4)^2,{t''}_2^2=(\alpha_1-\alpha_2-\alpha_3+\alpha_4)^2$. Thus we can calculate $t_2^2,{t'}_2^2,{t''}_2^2$ as roots of an equation of degree $3$. In a second time, we deduce the $(\alpha_i)$.

Remark. Galois read (with interest) the Lagrange's papers; of course, Lagrange used implicitly or explicitly the symmetric functions of the roots. May be Galois had, during his readings, the idea of introducing his famous group.

  • $\begingroup$ Thank you for your answer. How is $\Sigma_4$ exactly defined? $\endgroup$ – zar Apr 6 '18 at 9:06
  • 1
    $\begingroup$ @zar , $\Sigma_4$ denotes the group of permutations of the $(\alpha_i)$; a better notation is $\mathfrak{S}_4$. $\endgroup$ – user91684 Apr 6 '18 at 9:26
  • $\begingroup$ i'm trying to do an example using technique 3). Suppose i have a polynomial $(x-1)(x-2)(x-3)(x-4)$, then resolvent $t_1$ gives me $-64,-4, 28+96i, 28-96i$, which are four different values, whereas $t_2$ gives me $256,16,0$. Does this mean $t_1$ does not work? $\endgroup$ – enochk. May 7 '19 at 21:36

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