Sunday, June 24, 2012

The Fifth Degree: Part V

I should come clean. I started this topic last week because I thought maybe if I forced myself to write it up, I would work out some of the troubling issues that have bothered me for years. The truth is, I really don't know why the fifth degree equation is unsolvable. I have bits and pieces of the picture, odd bits of insight, and not much more. I know about the Galois group of a polynomial...that's the set of allowable permutations of the algebraic field formed by adjoining the roots of that polynomial to the field of rationals...the so-called "splitting field" of the polynomial. And I know that we're supposed to look for a correspondence between the structure of the Galois Group and the field structure. If the equation is solvable, we're supposed to be able to write the field as a series of extensions of the rationals, where each extension is formed by adjoining a simple nth root of an element of the previous field.

This part makes a little sense. When you write the square root of this and the cube root of that, what are you doing but extending your field by adjoining the root of something you already have? The odd thing, the thing I remarked on at the end of my last post, is that when you've done with all these nice field extensions, you don't reach the splitting field...not even for the cubic equation. You reach a field  bigger than the splitting field. The splitting field of the cubic is a subfield of the field you get by adjoining a square root, then six conjugate cube roots (three for each of two conjugate square roots) to the rationals.

The tragedy is that with all this agonising over field theory, I still haven't shown you how nicely the cubic equation solves. It's a very cool thing and I don't know what else to do than just write it out. Here is how it works. We assume we have the three solutions, alpha beta and gamma. Of course we don't have them...they are the things we are looking for. But assuming we knew them, then we could combine them together in certain picturesque combinations to give us these three expressions:

Now, in general, the cube root of unity is not an element of the splitting field of our cubic. So the three expressions I've listed are outside the splitting field. Except for the third one, that obviously reduces to the sum of alpha, beta, and gamma, which is just a rational number. In fact it's the coefficient of the x-squared term in our cubic, which we can easily force to zero by a fairly simple linear shift.

Notice that if we add these three expressions together, the betas and gammas all go away, and we're left with alpha. That's the root we're looking for. Now we're about to show how we get it. The three expressions I've written are all the cube roots of things we can evaluate! The third one (r) trivially so, since it's the cube root of zero, or at worst the cube root of an ordinary number. But try cubing either p or q. You get almost the same result, whichever one you cube. You get terms in alpha-cubed, beta-cubed, and gamma-cubed. Because those are symmetric in alpha, beta, and gamma, they evaluate out to ordinary rational numbers. You get terms in alpha-beta-gamma....again, these are symmetric terms which are easily evaluated as simple multiples of the zero-order coefficient of the cubic. And then...then, you get terms in alpha-squared beta and beta-squared gamma. These terms are mixed in with the omegas....but when you work them out, you can group them as a term expressible as the symmetric sum of alpha-squared betas etc....hence a rational number....and a term which is the difference  of alpha-squared betas take-away beta-squared alphas etcetera. This is the w expression I talked about the other day. It's the square root of an ordinary number, and that number is easily evaluated by writing it out in terms of the elementary symmetric polynomials.

In other words, p and q are just the cube roots of an ordinary number plus the square root of an ordinary number. And I wrote this out explicitly a couple of posts ago:

At that time I simply declared that the solution of the cubic had this form. Now I'm showing you why it works. By working out their algebraic structure, I've identified our p's and q's with those funny combinations of alpha, beta and gamma mixed together with our cube roots of unity. We can see that they have this structure by cubing them out and verifying directly that the pieces making them up are nothing more than ordinary numbers and the square roots of ordinary numbers.

And that's how you solve the cubic equation. Anyone want to see how the fourth degree works out?

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