I learned something new about the fifth-degree equation over the last few days. The other day I told you that the reason you can't solve the quintic has something to do with the fact that there exists no set of functions on five letters which "map to themselves under all permutations". Or something like that. Well, it turns out there is...only there isn't. Here's what I learned.
I had constructed these three functions on four letters (corresponding to the four roots of a quartic equation:)
AB + CD = p
AC + BD = q
AD + BC = r
We can see that no matter how you re-shuffle A, B, C and D, you get back the same three functions p, q and r. I said that the reason you can't solve the quintic was that there is no comparable set of four functions on five letters (A, B, C, D and E) which is similarly preserved on taking permutations of letters.
And that's true. But what I didn't know is that you can construct a set of six functions on those five letters which have the desired property...sort of.
I had almost concluded that there were no such functions, and I posted a question on stackexchange.com here to see if I was right. I wasn't. One Balarka Sen pointed out that there were a number of such functions, including one called Dummit's Resolvent, which seems to have been discovered only twenty years ago by a math professor from the U of Vermont. Dummit's functions look something like this:
AA(BE + CD) + BB(CA + DE) + CC(DB + EA) + DD(EC + AB) + EE(AD + BC)
There are actually six of these functions...you can generate the other five by suitable permutations of the A's, B's and C's, as Sen explains in his answer to my question on stackexchange.
Now, what people say next is a little misleading: they say that these six functions generate a sixth-degree equation, and therefore they're not going to help you solved a fifth-degree. That's not quite true. It might be a sixth-degree equation of a simpler form than the fifth-degree which you started from. It's certainly not a sixth-degree equation of the most general type, because the permutations of the roots cannot be more complicated than the permutations of the five letters which generated them. In other words, the Galois Group of the sixth degree equation cannot be bigger than S5, the permutation group on five elements. But could it be a simpler Galois Group? Maybe it could...
But here's where group theory actually gives us some guidance. I said that Dummet's functions are similar by analogy to my p, q and r which gave me a third-degree resolvent to the general fourth-degree equation. But in one crucial respect they fail to do what p, q and r do. It's true that unlike my p, q and r, Dummits functions are more numerous than the five roots that generated them. That's a problem, but it's not decisive. The real problem is that there are no (non-trivial) permutations on the five letters which leave all six of Dummit's functions in place.
I'm going to leave it for another day to tell you why I consider this circumstance to be critical. Suffice it to say (for now) that there are indeed permutations of A, B, C and D which leave my p, q, and r in place, and the set of all such permutations forms what they call a normal subgroup of S4. But before we get there, I have to tell you one more thing. In one way or another, I've been working on this problem off and on for most of forty years. It's a great problem and I don't think it gets its due in the undergraduate curriculum....I've talked about that previously. So when I posted the question the other day on stackexchange, asking about the existence of these functions, I wasn't sure what I'd get, and I was happy to get a pretty clear answer from that fellow Sen. Now here's the kicker...it turns out that out Balarka Sen is a thirteen-year-old kid from India.
Thirteen years old????