I mentioned the other day that I had “known” for a long time that the three cube roots of two were algebraically indistinguishable. When I say I “knew” it, I mean that I had read somewhere that the Galois Group of x^3-2 was S3, the symmetric group. In other words, all permutations of the roots were allowed. This bothered me: I knew you could swap the complex roots, but how could you swap one of the complex roots with the real root? I wrote about this the other day.
This question led to another question: is it always the case that the roots of an irreducible cubic equation are algebraically indistinguishable, or is it possible for the splitting field to have a more specific structure? Specifically, can you write a cubic equation whose Galois group is C3, the cyclic group?
This turns out to be a pretty interesting question, which I was able to answer in the affirmative. I recently stumbled across some old scribbled notes of mine relating to this problem, and I think I’d like to sketch out what I did for the sake of posterity. So here goes…
You start by considering the seventh roots of unity...yes, those seven points in the complex plain located 51.4 degrees apart. We can throw away the obvious root and realize that the remaining six are the roots of the sixth-degree equation:
x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0.
We can take the first complex root and call it omega, and then the other roots are simply omega raised to the power of 2, 3, 4, 5 and 6. And of course the seventh power brings you back to one.
I said you start with the first complex root, but of course the beauty of algebra is that it doesn't matter which one you pick. None of them are distinguishable from any of the others. Yet despite this indistinguishability, there is an order to the roots which flows from their cyclic relationship. It's all very suttle.
What we're going to do is take combinations of these roots to create the roots of a cubic equation. And we'll see that those roots will inherit some of the cyclic nature of the omegas which begat them. You should be able to guess where we're going from here...we're just going to add together complex conjuates, so the (hoped-for) roots of the cubic are:
w^1 + w^6 = A
w^2 + w^5 = B
w^3 + w^4 = C
And I'm claiming that A, B and C are the roots of a cubic equation (with integer coefficients). It's not hard to see why this has to be so. You know that the coefficients of any equation are generated by the elementary symmetric polynomials in the roots:
A + B + C
AB + BC + CA
How can we caluculate these? It looks like it's going to be a fair bit of work, but it turns out to be a piece of cake. The sum of the roots is the most obvious...the sum of A, B and C is just the same as the sum of the omegas, which is just -1. The other sums and products are only slightly harder...because the products of expressions in powers of omega are themselves just powers of omega, and because of the near-symmetries in the A's B's and C's, they collapse with little difficulty. It's not hard to show that A, B and C satisfy the equation:
x^3 +x^2 -2x -1 = 0
Now all we have to do is show that there is a cyclic relationship between the roots...in other words, if you replace A with B, then you have to replace B with C and C with A. Let's leave that for tomorrow...