Saturday, January 18, 2014

The Galois Group of the Resolvent

Okay, I thought I was done but I'm not. We generated a sixth-degree equation from the general quintic. People say "that's why you can't solve the quintic...because the resolvent you generate is of higher order than the equation you started with."

That's not the reason you can't solve the quintic. There are solvable sextic equations...maybe our resolvent is one of them? The question is: what is the Galois Group of the resolvent sextic?

I'm thinking it has to be S5. Because there are absolutely no permutations of the roots of the sextic that aren't generated by permutations of the roots of the quintic. And furthermore...every single permutation of the roots of the quintic affects some of the roots of the sextic. Does that make it a one-to-one correspondence?

Because if it doesn't....if there are two different elements of S5 that give the same permutation of the resolvents...then the order of the Galois group is less than 120...which makes it 60 or less. And unless it's A5, then all bets are off...you may indeed have a solvable sextic. So the problem is to show explicitly just what is the Galois Group of the sextic. I'm thinking it has to be S5. But I'm having trouble constructing it.

The Galois Group in question comes from the reshuffling of the six resolvents functions, so it if obviously at most a subgroup of S6. And it's easy to find S5 as a subgroup of S6...just take one letter and hold it fixed while the others get shuffled around. But that's not what the resolvent functions do. They all move under permutation, and they do so in a way that is somehow...symmetric.

Furthermore, there are 5-cycles in the Galois group...permutations that, when repeated 5 times, take us back to where we started. There is only one way to make a five-cycle is S6...hold one element fixed and let the others move. So without loss of generality, we must have

(1 2 3 4 5)

as a permutation, and also:

(2 3 4 5 6)
(1 3 4 5 6)
etc....

....except we can't pick and choose all of them willy-nilly, because there are 120 5-cycles of this kind in S6 and that's just too many! We need exactly 24, because that's how many there are in S5...so we have to pick them carefully. It's not that easy.

So that's what I'm working on: to find (and explicitly describe!) the group of permutations of six letters which is isomorphic to S5, and show that this is the Galois Group of the resolvent sextic.