Thursday, December 10, 2009

Algebraic integers vs Complex numbers

One of the weirdest things is where the ring of algebraic integers can be shown to contradict with the field of complex numbers.


7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

divide off the 7.

The problem is finding how the 7 multiplied, and it's easier to figure out in the complex plane.

So from the complex plane, first step is to normalize the functions, that is, have functions that equal 0, when x=0.

Looking at x=0, gives a2 + a = 0, so a1(0) = 0, or -1, and a2(0) = -1, or 0.

So I can let a1(0) = 0, and then introduce a new function b2(x), where:

b2(x) = a2(x) + 1, so a2(x) = b2(x) - 1, and making that substitution:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

where now a1(0) = b2(0) = 0, and with normalized functions, it's now clear that the first factor, which is (5a1(x) + 7), is the product of a factor multiplied by 7.

It might seem trivial, but notice that if the OTHER factor were such a product then you'd have:

7(175x2 - 15x + 2) = (5a1(x) + 1)(5b2(x)+ 14)

and if the 7 split up, so that each had sqrt(7), then:

7(175x2 - 15x + 2) = (5a1(x) + sqrt(7))(5b2(x)+ 2sqrt(7))

as I keep the appearance of the functions the same with those hypothetical's as the functional notation can hide things, but the 2 cannot.

So from the complex plane it is clear we need one more function--the other b--as we actually have something like:

7(175x2 - 15x + 2) = (5(7)b1(x) + 7)(5b2(x)+ 2)

And dividing off the 7 is easy then:

175x2 - 15x + 2 = (5b1(x) + 1)(5b2(x)+ 2)

Which it should be! Constant factors are supposed to be trivially removable.

But notice, you needed a1(x) = 7b1(x) for it to BE trivial.

Going back to the challenge then, and considering the ring of algebraic integers, it seems logical that in agreement with the field of complex numbers the ring of algebraic integers would give the same result, but as the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

that would indicate that just one root would have 7 as a factor. But there is the contradiction as provably if the roots are non-rational with an integer x, then NEITHER of the roots can have 7 as a factor in the ring of algebraic integers!

So maybe there is a mistake in the proof over the complex plane?

Not if 7*1 = 7, which it does even on the complex plane. The important thing here is not the functions--it's the two constants.

With normalized functions:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

can only be obtained one way. If you don't believe me, try some other way.

So how could a factor result emerge from the field? Because it's the distributive property, and is equivalent to something like:

7(x2 + 3x + 2) = (7x + 7)(x+2)

But now we can get to the truly weird.

The key here at this point is the source of the a's, as they are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

But with x=1, that has roots 3 + sqrt(-26) and 3 - sqrt(-26), and you now know that one of them is a product of 7 and some other number, but which one?

There's no way to determine.

And even weirder, it's not possible to say that one in particular does as the square root is ambiguous. To understand THAT result consider:

5 + sqrt(4)

and that is 7 and 3, but 3 and 7 work too as sqrt(4) = -2 or 2.

Maybe, you say, no? Well (-2)(-2) = 4, so it is proven that -2 IS a solution to sqrt(4).

Historically some time ago some math people decided they didn't like that the square root gives two results, so they defined one away and people get taught that you take the positive, but that is a convention that usually doesn't seem to matter, but here it does.

It is one of those annoying math errors where a person may believe that sqrt(4) is just 2.

But sqrt(4) is 2 or -2 no matter what people believe.

With 3 + sqrt(-26) and 3 - sqrt(-26) in both cases you have two numbers, but the order simply shifts, where you just can't get rid of the square root, but you know that for one case 7 should be a factor.

How? Because of the proof above in the field of complex numbers that's how!

But the ring of algebraic integers will not allow 7 to be a factor of either root.

So you have just learned of a fascinating contradiction at the heart of modern number theory. A place where the ring of algebraic integers dares to contradict the field of complex numbers.

James Harris

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