Clarifications regarding Lagrange resolvent 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?
 A: 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.
A: 
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 cubic root of $1$?

It is. The lagrange resolvent of a quartic is by definition $x_1 + \omega x_2 + \omega^2 x_3 + \omega^3 x_4$. But this is just a naming convention to honor Lagranges achievements.
One can solve the quartic using $x_1 + \omega x_2 + \omega^2 x_3 + \omega^3 x_4$. And Lagrange did this. See my answer here to understand how. Doing this, one necessary crosses a resolvent which is only fixed by the klein four group $V_4$ and has three images under the $4!=24$ permutations of the roots.
Lagrange noticed that there are more, easier such resolvents, which have this probability and which can be used to solve the quartic, like
$$x_1 x_2 + x_3 x_4,$$
$$(x_1 + x_2)(x_3 + x_4),$$
or
$$(x_1 + x_2-x_3-x_4)^2.$$
