Friday, December 20, 2002

 

Disproof of assertion about non monic roots

For quite some time I've faced assertions that a non monic primitive polynomial with algebraic integers that is irreducible over Q cannot have algebraic integer roots. Some have even claimed to have proven or seen proofs in textbooks of what I know is a false assertion.

It occurred to me that I had the means for disproving it.

The disproof depends on an expression which come from my work on Fermat's Last Theorem for p=3, though here I will not be using the FLT equation.

The expression is

(v^3 + 1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3

where v,x,y and z are nonzero integers and the ring is algebraic integers.

The expression is not a polynomial because I'm going to be varying v, so I call it polynomial-like. I'll be considering factorizations of it.

For instance, from Gauss' FTA in algebraic integers I have that

(v^3 + 1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3 =

(a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

where the a's and b's are algebraic integers.

That factorization is one that's often been seen, but the next one comes after changing my v slightly.

First I'll further restrict x by letting x = uf, where f is a prime integer, u is a nonzero integer, and f is coprime to 3, u, y and z.

Next, let v = -1 + mf^2, where m is some nonzero integer.

Then I have

((-1+mf^2)^3 + 1)z^6 - 3(-1+mf^2)(uf)^2 y^2 z^2 - 2(uf)^3 y^3

and it can be seen that it has a factor of f^2.

Therefore

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy)

has a factor of f^2, which shows that the a's have non unit algebraic integer factors of f.

Now going back to

((-1+mf^2)^3 + 1)z^6 - 3(-1+mf)(uf)^2 y^2 z^2 - 2(uf)^3 y^3

and expanding out

(-3mf^2(-1+mf^2)+m^3 f^6)z^6 +3(uf)^2 y^2 z^2 - 3mu^2 f^3 y^2 z^2-2(uf)^3 y^3

and collecting with an eye on m gives

f^6 z^6 m^3 - 3f^4 z^6 m^2 + 3(f^2 - u^2 f^3 y^2 z^2) m + 3 (uf)^2 y^2 z^2 - 2 u^3 f^3 y^3.

Now separating off the factor of f^2, gives

f^2(f^4 z^6 m^3 - 3f^2 z^6 m^2 + 3(1 - u^2 f y^2 z^2) m + 3 u^2 y^2 z^2 - 2 u^3 f y^3)

and if we consider the expression as a polynomial P(m) then the constant term is

3 u^2 y^2 z^2 - 2 u^3 f y^3

which clearly then does not have a factor of f.

So now from Gauss' FTA in algebraic integers I have

f^4 z^6 m^3 - 3f^2 z^6 m^2 + 3(1 - u^2 f y^2 z^2) m + 3 u^2 y^2 z^2 - 2 u^3 f y^3 = (c1 m + d1)(c2 m + d2)(c3 m + d3)

where the c's and d's are algebraic integers.

Therefore, I have

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy) = f^2 (c1 m + d1)(c2 m + d2)(c3 m + d3).

But I have that a1 a2 a3 = -3mf^2(-1+mf^2) + m^3 f^6

so the a's have factors of m, and they also clearly have a factors of f.

Now suppose that separating off the f^2 on the left side could happen without any of the a's having f itself as a factor, that would leave additional factors of f multiplied times uy because each factor like

(a1 z^2 + b1 ufy)

has a factor of f, to the right of that plus sign.

But the constant term from the perspective of m, with P(m) after the f's have been separated off is

3 u^2 y^2 z^2 - 2 u^3 f y^3 = u^2 y^2(3z^2 - 2ufy)

where we see uy from two of the expressions, but there are no non unit factors of f.

You may suppose that the m and f factors might somehow separate and distribute themselves in odd ways, but clearly no more than two of the a's can have some factor of m, as we only see u^2 y^2 as that constant term, and not u^3 y^3.

Given that only two of the a's can have non unit factors of m, there still is the problem of the f factors that are seen to the right of each plus sign in

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy)

as if only two of the a's have non unit factors of m, then for the constant term of P(m) to not have non unit factors of f, a factor of f has to be separated off in each case.

Therefore, exactly two of the a's have a factor that is f.

Notice I didn't have to refer to FLT for the proof, as it follows from your ability to see two factorizations in an expression which differs from the typical in that it is polynomial-like, as its coefficients can vary.

So how does that disprove the assertion that I started with?

Well in arguments with me claiming that the short FLT Proof is false, certain posters have used the p=3 case to construct a supposed counterexample, where they stop when they get a non monic primitive polynomial irreducible over Q, which must have an algebraic integer root.

They then claim to have proven the assertion and say that they have a counterexample to the short FLT Proof.

However, as I've pointed out that would require an inconsistency within mathematics itself, and in looking over supposed proofs of their assertion, I've noticed that none are indeed proofs.

I've emphasised the need to be able to look at the expression from two perspectives but one poster dedicated in arguing with me started claiming I was talking about a "new polynomial" as if the main expression changed, when I was looking at the same expression factorized in two different ways.

When I tried to explain to this person they started spewing math statements like an octopus spews ink, and the talks bogged down, though that person kept claiming that the short FLT Proof was wrong. You see, they either couldn't or wouldn't see the truth, but they kept making their false claims.

The short FLT Proof is still available as it has been at

http://www.msnusers.com/AmateurMath

and if you still believe that two of the a's cannot have a factor that is f, then I have to think that your mind is limited in its ability to shift perspectives, and it may be the case that not everyone can achieve that, as the mental wiring may not be there for the task.





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