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 $\mathfrak{S}_4$. $\endgroup$ – loup blanc 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 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.