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Wednesday, September 28, 2011

Under the hood

This post will be primarily mathematical and is to show the building of the factorization I often like to use to show a problem with the ring of algebraic integers, where that ring contradicts what you can prove from the field of complex numbers.

The ring is the ring of objects.

See: http://somemath.blogspot.com/2005/03/object-ring.html

I will be using tautological spaces.

The tautological space is an identity. With it I use what I call a conditional, as unlike the identity, it is not always true.

The conditional is x2 + xy + y2 = z2. It was chosen by me for several reasons but for this discussion the primary importance is that it be in the ring of objects, which it is.

The tautological space is x+y+vz = 0(mod x+y+vz).

x+y = -vz (mod x+y+vz), squaring both sides gives: x2 + 2xy + y2 = v2 z2 (mod x+y+vz), subtracting the conditional gives

xy = (v2 - 1)z2 mod (x+y+vz), so (v2 - 1)z2 - xy = 0(mod x+y+vz).

From the tautological space itself: x = -y-vz (mod x+y+vz), so I can substitute to get:

(v2 - 1)z2 + (y+vz)y = 0(mod x+y+vz), so (v2 - 1)z2 + vzy + y2 = 0(mod x+y+vz).

Now let v = -1 + mf, then I have:

(1 - 2mf + m2 f2 - 1)z2 + (-1+mf)zy + y2 = 0(mod x+y+vz)

which is:

(m2 f2 - 2mf )z2 + (mf - 1)zy + y2 = 0(mod x+y+vz)

Now let P(m) = (m2 f2 - 2mf )z2 + (mf - 1)zy + y2.

It's easier to look at it as:

P(m) = y2 + (mf - 1)zy + (m2 f2 - 2mf )z2

My usual example has y = a, m = x, f = 7 and z=1, which is:

P(x) = a2 + (7x - 1)a + (72 x2 - 2(7)x )

And I've shifted the '7' in front of the variable.

Next I derive the actual factorization. Some of what I do is a bit abstruse, but the point is to make sure everything is being done in the object ring without assumptions.

Now multiply both sides by 4, and complete the square:

4P(m) = 4y2 + 4(mf - 1)zy + (mf - 1)2 z2 - (mf - 1)2 z2+ 4(m2 f2 - 2mf )z2

so:

4P(m) = (2y + (mf - 1)z)2 - [ (mf - 1)2 - 4(m2 f2 - 2mf )]z2

Now I need the result that in general x2 - y2 = (x+y)(x-y), which can be constructed from within the tautological space as an intrinsic property:

x+y+vz = x+y+vz, x= -y - vz + (x+y+vz), squaring both sides, gives:

x2 = y2 + 2yvz + v2 z2 - 2(y+vz)(x+y+vz) + (x+y+vz)2

x2 - y2 = 2yvz + v2 z2 - 2(y+vz)(x+y+vz) + (x+y+vz)2, and setting v = 0, I have:

x2 - y2 = - 2y(x+y) + (x+y)2, which is: x2 - y2 = (x+y)(-2y + x +y),

so: x2 - y2 = (x+y)(x-y).

4P(m) = (2y + (mf - 1)z)2 + [(mf - 1)2 - 4(m2 f2 - 2mf )]z2

4P(m) = (2y + (mf - 1)z + sqrt[(mf - 1)2 - 4(m2 f2 - 2mf )]z)(2y + (mf - 1)z - sqrt[(mf - 1)2 - 4(m2 f2 - 2mf )]z)

and now I can flip things around so that z is first in each factor and divide the 4 back off:

P(m) = (((mf - 1) + sqrt[(mf - 1)2 - 4(m2 f2 - 2mf )]z/2 + y)(((mf - 1) - sqrt[(mf - 1)2 - 4(m2 f2 - 2mf )]z/2 + y)

and I have a1(m) = ((mf - 1) + sqrt[(mf - 1)2 - 4(m2 f2 - 2mf )]/2, and

a2(m) = ((mf - 1) - sqrt[(mf - 1)2 - 4(m2 f2 - 2mf )]/2

where P(m) = (a1(m)z + y)(a2(m)z + y).

Note that with y = uf, P(m) has a factor of f.

And that's a complete derivation from conditional all the way through, doing everything in the object ring.

The reason that derivation comes up as an issue is that one objection some people raise is that you can show another contradiction by breaking the construction, for instance with something like:

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

where they just break the original factorization by deleting off a key 7, so that x=1, will zero out the 'a' coefficient.

With what's shown above that's equivalent to just changing:

P(m) = y2 + (mf - 1)zy + (m2 f2 - 2mf )z2

into:

P(m) = y2 + (m - 1)zy + (m2 f2 - 2mf )z2

But that is NOT constructible in the object ring. So it's like just changing something with a derived equation.

This post goes through some of the underlying machinery in order to show a full construction within the object ring, primarily to handle that type of objection.

Also I think it's just way cool.

Figuring out details like the above helped me gain confidence in something I introduced, as I discovered tautological spaces. So I don't have the guide of a lot of history, but had to feel my way mostly on my own.


James Harris
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