Saturday, August 09, 2003

 

Problem with Algebraic Integers: Detailed Exposition

A mathematical proof begins with a truth, and proceeds by logical steps to a conclusion which then must be true.

I've pulled a detailed exposition of a short argument that quickly shows a problem with algebraic integers. It starts after the reference.

Message-ID: <3c65f87.0308081016.460d766c@posting.google.com>

Now here's a math proof. Those who doubt that fact can believe it's a claim of proof, but it's verified to be a proof by tracing the argument out.

In this case, I begin with an expression. The expression exists, so that is the truth from which you start.

Consider, in the ring of algebraic integers,

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f).

That is, I have the identity which defines P(m) in terms of various symbols, and it's all in the ring of algebraic integers, which means that the symbols can only represent numbers that are algebraic integers.

Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization

P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)

where w_1 w_2 w_3 = f, and

b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),

and at m=0

P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),

so two of the b's must equal 0, which means

P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)

which is

P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)

proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3.

Now that was a lot of steps, but each was a logical one.

First I introduced b_1, b_2, b_3, w_1, w_2, and w_3, which are defined by the factorization

P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)

then I set m=0, and used the definition of P(m) to get P(0).

That told me that at m=0 two of the b's are 0, because then

P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),

where the "u^2" couldn't get there unless two of the b's are 0.

Then using that result I get from

P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)

that

P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)

and multiplying through by w_1 w_2 I have

P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f)

which with

P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),

tells me that w_1 w_2 = 1, when m=0.

Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3.

Now I'm focusing on what has been revealed to be an area of confusion. Apparently some people believe that when I divide off f^2 that it can divide off as a function of m, so that m=0 might be a special case. I'm now starting the argument to address that belief by noting again that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is, when f doesn't have 3 as a factor.

But that was an arbitrary choice, so let f=3.

That is, I said f is coprime to 3 but in considering this possibility it's worth it to relax that restriction and now consider what would happen if it equals 3.

Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.

Seeing that is as simple as looking at

P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

with f=3 as then you have

P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3

so every coefficient has a factor that is 3, as you can tell by looking.

So with

P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)

each of the b's and each of the w's has a factor that is 3^{1/3}, while the b's can have additional factors in common with 3, the w's cannot, as when 3 is separated out, notice you have

P(m)/3^2 = 3((m^3 3^3 - 3m^2 3 + m) x^3 - (-1+m3^2 )x u^2 + u^3).

But before at m=0, they were coprime to f, now they are not when f=3, as they are constant. Clearly, they are constant in both cases with respect to m, without regard to the value of f. Which makes sense as f^2 is not a function of m, and it is what is being divided off.

That is, if they were functions of m, so that w_1 w_2 = 1 at m=0 as a function of m, then it wouldn't matter if f had a factor of 3 or not, you'd STILL get w_1 w_2 = 1 at m=0, without regard to the value of f.

But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at m=0, which only works if the w's are independent of m, which they are.

It makes sense that they are anyway, as f^2 isn't a function of m, but I've seen that for some people the idea can take hold after seeing m=0 highlighted.

But if the w's were functions of m, then w_1 w_2 would equal 1, without regard to the value of f, but it does not.

Therefore, the factorization is

P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers.

And here I've packed in a lot of information as well.

First, with f coprime to 3, I now know that the factorization is

P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

as the w's are constant with respect to m, so I can just check at m=0, which revealed that w_1 w_2 = 1. Now that doesn't necessarily force w_1 and w_2 to each equal 1, but even if they were factors of 1, i.e. unit factors, that would only change b_1 and b_2.

So I have my factorization without regard to m in terms of where the f goes, and then I point out that you can actually check my work using m=1, f=sqrt(2), as then you get a polynomial which you can factor rather simply. So you can actually get the values for the b's and check them, and see that they are all algebraic integers, and all are
coprime to 2.

However, usually, for f values that are coprime to 3, you don't get b's that are algebraic integers, which shows a problem with the ring of algebraic integers.

Now the nice thing about a mathematical proof is that if someone disagrees they have to find some misstep.

Unfortunately, people can say that proof is not a proof, even when it is, just like if you tried to say you were human, and not a dog, someone might dispute any proof you might give, claiming it false.





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