Tuesday, January 7, 2014

Galois Group C5: Constructing a Polynomial

I don't have any great illusions that I'm doing significant math here, but having started down the road, it's funny where I've ended up. Using recursive relations of the form

A2 = A1^2 -n

and requiring A4 to equal A1, we were able to generate any number of cubic equations with cyclic Galois groups. Of course we always got a cyclic relationship between A1, A2 and A3, but we didn't always get cubic equations with integer coefficients. The surprising thing was that sometimes we did...for n=2, 4, 8 and (apparently) 14.

It wasn't surprising that we got the cyclic relationship for n=2, because that comes straight from the properties nth-cyclotomic polynomial. The sum of any nth root of unity with its conjugate, squared, differs by two from some other nth roots of unity. And specifically for the 7th roots of unity, you get the roots of a cubic by summing the 1st and 6th, the 2nd and 5th, and the 4th and 7th.

So I thought I'd try and generate the 5th-degree equation having a cyclic galois group by doing the same trick on the 11th roots of unity. I had already figured out that I didn't need to be a genius at factoring...I could plot the cyclic relationship in Excel and monitor the 5 elementary symmetric polynomials in the roots to get the coefficients of the equation. Sure enough, I verified that the following equation had its roots in a cyclic relationship:

x^5 +x^4 -4x^3 - 3x^2 +3x +1 = 0

Now, when we did the cubic, we got an extra polynomial "for free"...remember how? Applying the cyclic relationship three times gave us an eigth degree equation...we factored out the trivial quadratic corresponding to the case of the three roots being equal, leaving us a sextic. The cubic generated by the seventh roots of unity factored cleanly into the sextic, giving us a brand new cubic, also cyclic in the roots.

So what about the fifth degree?

Well, when we apply the recursive relationship five times, we get a thirty-second degree equation, which, upon factoring out the trivial quadratic, gives us thirty roots. Thirty?

I started hunting for roots in Excel, and it seemed I was finding all kinds of 5-cycles wherever I looked, and for the most part they didn't give me equations in integer coefficients. Remember, I had Excel displaying the values of the elementary symmetric functions...that's where I got the nice fifth-degree equation I showed you earlier. But most of the 5-cycles I generated corresponded to garbage polynomials with irrational coefficients. Where were they coming from?

Eventually, I figured it out. They came from the 31st roots of unity, and the 33rd roots of unity. When you apply the recursive relationship to the sum of any root and its conjugate, the new power of the root you generate is just double the one you started with. So operating on the eleventh roots of unity, you get the powers:

1, 2, 4, 8, 16 (which is the same as 5) and 10 (which is the same as -1).

So five iterations brings you around to the conjugate of the root you started with. It works the same if its the root itself or the conjugate, because either way your going to be taking sums.

So what about the 31st roots of unity? Well here, the sequence is:

1, 2, 4, 8, 16, 32(=1)

so adding these to their conjugates give you a 5-cycle. But you can also take these alternate choices for the roots:

3, 6, 12, 24, 17, 34(=3)      or     5, 10, 20, 9, 18, 36(=5) 

and in either case, adding them to their conjugates gives you a five-cycle in the roots. None of which, by the way, give you fifth-degree equations in integer coefficients.

So that's counting the ring generated by the eleventh roots of unity, that's four. But we had a thirtieth-degree equation to account for...there should be six rings of roots. That's where the 33rd-degree equation comes in.

The cycle {1, 2, 4, 8, 16, 32(=-1)} gives us five. The cycle {5, 10, 20, 8, 16, 32(=-1)} gives us five more. And the cycle {3, 6, 12, 24, 14, 28(=-3)} is the same cycle we already generated from the 11th degree!

Those are the six five-cycles accounting for all the roots of the 30th-degree equation. Only one of the 5-cycles gives us a 5th-degree in integer coefficients; but in any case, there they are.

I don't know what you want to take away from this discussion, but if you ever had any doubts that group theory applies to the theory of equations, well, those doubts should pretty much be laid to rest.

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