Resolvent of the Quintic...Functions of the roots

Last year Mathlover posted a very good question about Galois theory: specifically, the existence of funtions of roots which map to each other under permutations of those roots. You can see his question here: Number of distinct $f(x_1,x_2,x_3,\ldots,x_n)$ under permutation of the input

The question was answered quite well by Generic Human but I am still unable to close the loop on the question of the resolvent of the fifth degree equation. I know that if A, B, C and D are the roots of the fourth degree, then there are three functions which are the roots of a cubic equation:

$AB+CD = p$

$AC+BD = q$

$AD+BC = r$

I know how to construct the cubic equation whose roots are $p$, $q$, and $r$ and I assume this is what people mean when they talk about the resolvent of the quartic being a third-degree equation.

I also know that people say the resolvent of the quintic is a sixth-degree equation. What I don't know how to do is construct six functions in $A$, $B$, $C$, $D$ and $E$ which, by analogy with my $p$, $q$ and $r$ for the 4th-degree, are mapped to each other by permutations of the five roots.

I've almost convinced myself that no such functions exist, but then people say there is something called the "resolvent", which I'm assuming must be analogous to what I know for the case of the fourth-degree. So is there or isn't there...can anyone write out those six functions? (Or at least just one of them from which I can infer the other five by analogy?)

Solution 1:

I initially intended this to post it as a comment, but it seems to get a bit bigger to do that. The answer is positive. In fact, there are lots of such resolvents. For example,

Runge's resolvent:

$$F(x_1,x_2,x_3,x_4,x_5,x_6)=(x_1-x_2)^2(x_2-x_3)^2(x_3-x_4)(x_4-x_5)^2(x_5-x_1)^2+(x_1-x_3)^2(x_3-x_5)^2(x_5-x_2)^2(x_2-x_4)^2(x_4-x_1)^2$$

Dummit's resolvent:

$$F(x_1,x_2,x_3,x_4,x_5,x_6)=x_1^2x_2x_5+x_1^2x_3x_4+x_2^2x_1x_3+x_2^2x_4x_5+x_3^2x_2x_4+x_4^2x_1x_2+x_4^2x_3x_5+x_5^2x_1x_4+x_5^2x_2x_3$$

Malfatti's resolvent:

$$F(x_1,x_2,x_3,x_4,x_5,x_6) = (x_1x_2 + x_2x_3 + x_3x_4 + x_4x_5 + x_5x_1 - x_1x_3 - x_3x_5 - x_5x_2 - x_2x_4 - x_4x_1)^2$$

Conjugates for all of them can be obtained by the permutations $(1, 2, 3), (1, 3, 2), (1, 2), (2, 3)$ and $(1, 3)$