Wednesday, September 12, 2012

The Galois Group of x^5-2

I made a very bad mistake yesterday when I was talking about the fifth roots of two. I said if you moved the first one to the second one, you had to move the second to the third etc. all around the circle in order that the quotients of successive roots remained the fifth root of unity. My mistake comes down to the use of the definitie article: there is no such thing as the fifth root of unity. What would have been true is if I had said that the quotients of successive terms must all equal a fifth root of unity.

Let me explain. For the sake of convenience, I have labelled the fifth roots of two according to their position in the complex plane:
Drawing them out like this makes it appear as though the real root is somehow special. It's not. We can re-shuffle the roots in any number of ways whereby the real root gets mixed in with the complex roots, and there is no algebraic way to tell which is which. It's just convenient for us to have one special configuration to start off with, and this is the obvious choice. We can number the roots one through five, starting with alpha.

I started off yesterday by saying that you couldn't just exchange the first two roots, because the quotient of succesive roots must equal "the" fifth root of unity. Based on this, I said if you move the first root to the second, then the second must go to the third, etc;
You can see that on the left, succesive quotients are all equal the fifth root of unity; and the same holds true on the right hand side.

My mistake was to ignore that successive quotients on the left hand side must all equal the same fifth root of unity as those on the right. But that would be identifying one of those omegas as special...which it isn't. It could be omega squared or omega cubed....any of the fifth roots of unity would do just as well, as long as they are consistent. So moving the first root to the second doesn't restrict our second move at all. We can, for instance move the first root to the second, and the second to the fifth:

The quotient of the first two terms on the left is omega, and on the right it's omega-cubed. But that's long as we make sure all the subsequent quotients on the right are the same:

It turns out wherever you send the first two elements...even if you just swap them with each can always preserve the constancy of the successive quotients by making the correct placements for the final three. Since there are five choices for where the first one goes, and four choices for the second move, there are exactly twenty permissible re-shufflings of the fifth roots of two.

I should point out that I haven't really shown that all these permutations are preserve algebraic consistency. What I've shown is that they don't blatantly violate consistency in any obvious way. It turns out, although I won't show it here, that these permutations are in fact algebraically sound, and the permutiation group as a whole is therefore the Galois Group of the splitting field of x^5-2=0.

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