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

I promised to show you that the roots of this equation had a cyclic relationship. Let's remember how we got them. We started with the seventh roots of unity, and generated this cubic equation by letting its roots be

A = w^1 + w^6

B = w^2 + w^5

C = w^3 + w^4

It's very easy to see from these relationships that A^2 = B + 2, and similar equations relate B to C and C back to A. Those are the cyclic relationships.

You can also notice that the splitting field of the original equation is just the Q(A), or the rationals extended by appending A. Or B, or C...you just have to append one of the roots, and the resulting field contains all three. That's different from a typical cubic where the three extensions of Q are quite disjoint. It's obvious if you have two imaginary roots and one real root, the field you get by adjoing the real root is not going to include the comples roots. So no simple equation can link one of the roots to the next, as we obtain in the present case. In short, for a cubic to have a Galois group which is cyclic, it pretty much means that all three roots live in the same simple extension, and hence functions of each other.

Does this give us a recipe for constructing cyclic cubics? In the present case, we had the relationship A^2-2=B. What if we say we're going to look for a field where, for example, A^2-4=B? Will we generate another cubic equation with a cyclic relation among the roots?

I'm not

*totally*sure this will work. A funny thing happens if we try...but let's

*first*try with the case where we know it has to work...with A^2 - 2 = B. If the roots have a cyclic relationship, then we ought to also have:

B^2 - 2 = C; and,

C^2 - 2 = A

But this brings us back full circle...we can substitute one equation into another, and eleminate B and C, giving us:

A^8 - 8A^6 + 20A^4 - 16A^2 -A + 2 = 0

It's not a cubic at all...it's an eight-degree equation. But wait...maybe that makes sense. After all, one "trivial" solution to our cyclic relationship between A, B and C is simply that

A^2-2=A (because A = B = C)

So this should theoretically factor out of our eight-degree equation...and indeed it does, leaving us the sixth-degree equation:

A^6 + A^5 - 5A^4 - 3A^3 + 7A^2 + A - 1 = 0

Still not the cubic we're looking for...or is it? Maybe the cubic we already found is actually buried inside this sixth-degree equation. And in fact, we find that it is...on synthetic division, we find that it does indeed go factor through, leaving the residue:

A^3 - 3A + 1 = 0

And there it is...a whole new cubic, generating a different set of A's, B's and C's than the ones we already had, but with the same cyclic relationship amongst themselves. It's not obvious to me that these new ABC's are especially connected to the seventh roots of unity, but I stand to be corrected. For now, all I can say is...there they are. (I actually found them by approximation....it's not to hard to verify that the numbers (-1.9, 0.35, 1.8) satisfy not only the cubic equation written above, but also the cyclic relationship wherewith they were generated.

So can you start with

*any*similar cyclic relationship and generate a pair of cubic equations with cyclic roots? Sadly, it does not appear so easy. I tried starting with A^2 - 4 = B, generated an eight-degree equation, factored out the quadratic giving me the sixth-degree, but then....I couldn't see any way to factor the sixth degree into two cubics.

And that's about as far as I'm likely to get down this road in the immediate future. Unless anyone out there has any better ideas...

## 3 comments:

Ingenious! You excluded the factor of the sextic that contains the well-known cyclic cubic and get another; very nice, quite nice indeed.

The fact that a sextic gives rise to two cubics satisfying the same cyclic relationship, I think this has something to do with the fact that the cyclic relations are all quadratic. Saying this, I don't think it is hard afterwards to prove why such phenomena occurs.

Now, about A^2 - 4 = B, etc., I get the octic x^8 - 16x^6 + 88x^4 - 192x^2 - x + 140 and after the elimination of trivial quadratic

this reduces to a sextic that gives rise to two cubics x^3 - x^2 - 6x + 7 and x^3 + 2x^2 - 3x - 5, both having the alternating group of order 3, as desired. I can't take all credits for this apparently tedious factorization, btw : GP did it all. [:-p]

So the sextic factors for x^2-4?? Well, that changes everything! Maybe we do have a machine for generating cyclic cubics after all. By the way, you have a sign error on the second factor...it shoould be plus 3 for the x term, not minus 3. No, I didn't multiply it out...I figured it out a different way, quite a funny way actually, which I'm going to explain in my next post...

Yes, that does. I half assumed it does not, but still went ahead to take a look at because even if the sextic wasn't reducible, it would have been nice to look at it's Galois group.

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