Re: JSH: what is the object ring ??

In sci.math, tommy1729
<tommy1729@xxxxxxxxx>
wrote
on Mon, 10 Sep 2007 05:26:22 EDT
<3559391.1189416413430.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>:
what the **** is an object ring JSH ??

you have been talking about it for years.

what is it ?

i did not see a definition , not even a bad one ...

common harris , amaze us with your math definition of an object ring.

and i mean a clear statement , not like this:

an object ring proves FLT, it is commutative associative but not
distributiv and it factors integers

tommy1729

The object ring is of course nonexistent, but the
general idea, apparently, is that, with a certain magical
quasi-factorization of a certain parameterized polynomial,
JSH somehow "proves" that 7 has to divide exactly one
of the roots thereof. Therefore, at least one of the
roots comprise members of a new ring, which has the very
interesting property (thus spake JSH, anyway) that it
contains exactly two units, despite the fact that the
proper subring of all roots of polynomials with highest
term having coefficient +1 or -1, or, if one prefers, the
proper subring of all algebraic integers, has an infinitude
of units (one such set being sqrt(n^2+1)+n for any nonzero
integer n, but there's lots of others), and of course any
unit in the subring is also a unit in the superring.

Or something like that. I do not remember the polynomial,
but it doesn't matter; the roots of the quadratic JSH
uses can be easily determined using standard methods,
and neither one is divisible by 7. (JSH might have been
slightly better off staying with his original cubic; at
least there one would have had to apply stonger methods
similar to those first espoused by Cardano, Tartaglia,
and/or dal Ferro. Not that it would have helped much anyway.

http://www.sosmath.com/algebra/factor/fac11/fac11.html

is a nice, concise boildown of the general techniques
that can be used to solve a general cubic.)

Of course the trouble is that one cannot apply the Unique
Factorization theorem to algebraic integers. Basically,
1 = 1 * 1 uniquely in the positive integers, and to an
extent 1 = 1 * 1 = (-1) * (-1) in the integers (unique
up to sign), but
1 = (sqrt(n^2+1) + n ) * (sqrt(n^2+1) - n) for any integer n,
and that's far from a complete enumeration of all
factorizations of the value 1 -- which means 1 cannot be
uniquely factored over the algebraic integers, though one
might make a case that 1 is uniquely factorable up to a
unit -- but there's an awful lot of units to play with. [*]

Also, x^2 = 7 suggests that exactly one of the roots might
be divisible by 7, were they integers -- which of course
they're not; they're +sqrt(7) and -sqrt(7), neither of which
is divisible by 7. And, 7 is factorable into sqrt(7) * sqrt(7)
and cbrt(7) * cbrt(7) * cbrt(7).

There's also the little problem that the definition
of a ring requires among other things a set of items,
two operators, and left and right distributivity using
those operators. A non-distributive ring is therefore
quite impossible.

http://mathworld.wolfram.com/Ring.html

At this point JSH usually goes into a screaming hissy-fit
or some such, but the facts are clear: his theory (if
one can dignify it as such) falls apart like so much wet
tissue paper when squirted at with anything stronger than
warm butterfly spit. ;-)

[*] it turns out that the ring of all algebraic integers, and
the field of all algebraic numbers, are both denumerable, so
there's not that many. One trick involves sorting the
equations by degree and by maximum absolute value of
any coefficient.

--
#191, ewill3@xxxxxxxxxxxxx
fortune: not found

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