## Monday, December 30, 2013

### The Seventh Roots of Unity

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
ABC

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... Balarka said...

Hello again,

I just wanted to encourage everyone to think about the relation between finding a cubic with Galois group C3 and the plausibility of swapping a complex root with a real root, which lies significantly deep in the theory.

Take any cubic with rational coefficients with only one real root in x. Let x = p + q be the real root [note how p and q are of degree 3 over Q]. Then the fact that (x + a + b)(x + w * a + w^2 * b)(x + w^2 * a + w * b) = x³ - 3abx + a^3 + b^3, imply every such polynomial has splitting field Q[2^(1/3), w)] where w is the cube root of unity.

Let's compute the degree of extension by multiplicativity of [..]: [Q[2^(1/3), w] : Q] = [Q[2^(1/3)] : Q] * [Q[w] : Q] = 3 * 2 = 6.

As Q[2^(1/3), w] is Galois over Q, which follows from the fact that it is the splitting field of a polynomial over Q, we have that |Aut(Q[2^(1/3), w]/Q)| = [Q[2^(1/3), w] : Q] = 6 by Artin's theorem but then Aut(...) is in turn Gal(...), so we have that Galois group of any polynomial over Q with only one real root has Galois group of degree 6, implying that the Galois group is not C3 [C3 is of degree 3].

So the only other possibility is the discriminant is positive, i.e., the polynomial has three distinct real roots [as it is easy enough to exclude the possibility of having non-distinct roots].

Thus, not only having to permute a real and a complex root of a cubic in the complex plane imply that the Galois group is not cyclic but replacing that with "having a complex root" does the same.

As a remark, I'd like to note that it is a necessary condition [i.e., having no complex roots] for a cubic to have cyclic Galois group, but not sufficient enough. Looking for such polynomials are especially tough. Balarka said...

Oh, and as a note, the polynomial is supposed to be irreducible over Q, which I forgot to add.

Marty Green said...

Balarka, I have to observe that the essence of your explanation is that the complex roots give you a two cycle, the overall order of the splitting field is six, and therefor there has to be a three-cycle in the Galois group. I'm not entirely sure why C6 is ruled out as a possiblity, but maybe that falls our of an examination of where the three roots a,b and c have to go.

My real issue with this type of analysis is that while it may logically prove that S3 is the Galois group, it fails to show in a constructive way just why it is that you're entitled to swap a real root with a complex one. That's why I was so pleased to notice that for every such asymmetrical swap, there is a corresponding swap with the "opposite" asymmetry. If you know what I mean.

Anyhow, I'm going to continue in my next post with a discussion of the C3 Galois group... Balarka said...

C6 is not actually a Galois group of a cubic, as it is a permutation on 6 objects and S3 is permutation on 3, implying C6 is not a subgroup of S3.

You're right, it does not really shows what happens for S3, but it does ensures that no such swapping is allowed where the Galois group is C3, as there is no complex root for the later case. What is really superior understanding here is the rotational symmetry, as you have mentioned in the previous blogpost.

I am eager to see your next blogpost on C3, hope you'll post it soon.