Re: JSH: The "Published" paper he dosen't what you to know about.


jst...@xxxxxxx wrote:
marcus_b wrote:
jstevh@xxxxxxx wrote:
marcus_b wrote:

<deleted>

One of the simpler cases to which your result should
apply is when f = sqrt(5). In that case, the polynomial
equation above becomes

a^3 + 12 a^2 - 65 = 0.

This is a monic polynomial with integer coefficients.
It is irreducible over the rationals. The roots
are algebraic integers. Dedekind's theorem
and Galois theory both say that none of the roots are
coprime to f^2 = 5 in the ring of algebraic integers.
You however infer that one of the roots IS coprime
to 5. And since you are saying that Galois theory is
wrong, you must be saying that it is wrong in the ring
to which it applies: the ring of algebraic integers.
From which we must conclude: you think one of roots is
coprime to 5 *** in the ring of algebraic integers.***
This is inescapable.


Nope.

Why go in circles? I already note that the result isn't true in the
ring of algebraic integers, so why come back and claim that I'm saying
it's true in the ring of algebraic integers?

It's trivial math too.

You're doing what other posters have tried to do which is claim I'm
saying what I'm not, and denying the reality that I acknowledge that
none of the roots can be coprime to 5 in the ring of algebraic
integers.


Your paper says Galois theory disagrees with your algebra.
The Galois theory in question applies to the ring of algebraic
integers, not to arbitrary rings. Therefore you must be saying
that your results - particularly your claim that one of a1(m),
a2(m) or a3(m) is coprime to f - must apply in the ring of
algebraic integers. Otherwise you have no reason to say
Galois theory is incorrect.

Marcus.


It's false implication.

One can correctly say that 2 is coprime to 6 in evens, right?

But what if some mathematician proclaims that means that 2 is coprime
to 6 in a larger sense?

No one would, of course, because you know that 2*3 = 6.

But what I've done is prove a similiar situation where it's not so
trivial to point to the error, so mathematicians who if they admit the
error have no accomplishments in their entire careers can dance around
the proper conclusion.

Have you still not understood that for the algebraic integers

-if a and b are coprime they remain coprime in any
ring that contains the algebraic integers.

No, the algebraic integers are not like the evens. If something
is coprime in the algebraic integers it stays coprime.

[The converse is not true. If a and b are not coprime in the algebraic
integers it is trivial to find a ring containing the algebraic integers
in which a and b are coprime]>

-William Hughes

.

Relevant Pages

  • Re: JSH: The "Published" paper he dosent what you to know about.
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  • Re: JSH: The "Published" paper he dosent what you to know about.
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  • Re: JSH: The "Published" paper he dosent what you to know about.
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