*those*and adding all the resulting 480 terms togethers. You collect all the terms symmetric and throw them away, and what's left is Dummit's Resolvent.

And it almost works...but not quite. Lagrange's Resolvent is linear in the roots, so all the terms you get by taking fifth powers are fifth-order in the roots. But Dummit's Resolvents are made up of terms that are

*fourth-*order in the roots:

AA(BE + CD) + BB(CA + DE).... etc.

And so my method can't work exactly. But it's close. The leftover terms when you follow my system look a lot like Dummit's:

AAA(BE + CD) + BBB(CA + DE)...etc; and also,

A(BBEE + CCDD) + B(CCAA + DDEE)....etc.

Since for the sake of Galois Theory we are only interested in the permutation properties of the resolvents, we don't need to keep the whole term...obviously the terms in A^cubed etc. are going to have permute the same way as Dummit's terms in A-squared. (But differently if you reduced it to linear in A, because you need to distinguish the terms outside the brackets from the ones inside!)

It's not even that hard to see why Dummit's resolvents "work", if you think about it. It's an awful lot of work to take the fifth power of all those Lagrange Resolvents and add them together: but if you take a step backwards and look at the big picture, you can see there are only a handful of different

*forms*that the resulting terms can take:

AAAAA + BBBBB + CCCCC....etc; (five terms of this type)

AAAAB + BBBBC....etc; (twenty terms of this type)

AAABB etc. (twenty terms)

AAABC etc. (thirty terms)*

AABBC etc. (thirty terms)*

AABCD etc. (twenty terms)

There are only six different

*types*of terms that you get; and only the fourth and fifth (I've marked them with an asterik) survive the symmetrization. And the reason is kind of funny. The terms in AAAAA etc. are obviously symmetric, so they add up to an integer value which you can throw away. The terms in AAAAB etc....well, when you take the fifth powers, these all have a fifth root of unity stuck to them. (Try it if you don't believe me.) There are twenty terms...five of them have a multiplier of w, five of them w^2, etc. You know the sum of the fifth roots of unity is -1; so when you add all twenty terms together, you get:

-(AAAAB + BBBBC +.....+ CCCCA +...etc)

which is also symmetric, so you throw them away.

And the same thing happens whith the third and sixth terms. But the terms I've marked off are a little different. For them, you get all the terms with with a multiplier of a fifth roots of unity: but you

*also*get real terms. You collect the terms with the fifth roots, and they add up to a negative term...the other terms add up positive. You get:

-(AAABC + AAACD +.... + CCCDA +...) + (AAABE + AAACD....)*

You get all thirty terms, but some of them are negative and some of them are positive....the ones I've marked with an asterik. You know how to symmetrize this! It's just

-(all thirty terms) + 2*(AAABE + AAACD)

You throw away the symmetric terms and what's left are the (modified) Dummits Resolvents.

And it's easy to make a sixth degree equation out of them. Remember, there were 120 possible Lagrange resolvents based on the order you choose the roots of the fifth: but when you take fifth powers, this reduces to 24 different ones because some of them are just rotations of others. Then I grouped them according to ones that are related via legal permutations of the roots of unity....this gave me six groups of four. Each group of four generates, by the reduction I've shown above, a bunch of symmetric term (which you throw away) plus a leftover group of ten terms which have the form of Dummit's Resolvent.

It's not hard to convince yourself that while Dummits Resolvents are not symmetric in A,B,C,D and E, their generalized sums and products certainly are. It's a bit of work but before long you can derive therfore integer values for all the elementary symmetric polynomials in the conjugate Dummit Functions.

And those are the coefficients of the sextic equation which is called the "Resolvent Equation" of your original quintic equation.

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