Re: Is the polynomial ring interpretation of abstract algebra A[X]

On Jul 1, 10:21 am, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <d49f93ae-4ad7-4a62-a196-b5b1b8eee...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> "Tim BandTech.com" <tttppp...@xxxxxxxxx> writes:
...
> The only substantial disagreement that I see is that you claim that
> for a polynomial of real coefficients that those coefficients aren't
> actually real.

That was not the claim. In a polynomial ring over some base ring A the
"coefficients" are from the ring A. Moreover, the set A is a subset of
A[X], i.e. each element of A is also an element of A[X].

> We are not free to mix products as
> x z
> where x is real and z is complex since these are two sets,k not one.

We are, because R is a subring of C.

> Your own looseness here is exactly part of the problem. As I
> understand it a specification of
> polynomial with real coefficients

You are again confusing the polynomials as a function and as the elements
of a polynomial ring. Meaning that you still do not understand what a
polynomial ring actually is.

I don't see that the functional distinction does anything for the
conflict that I am expressing.
In functional systems there is a domain and a range. Under the ring
nomenclature these are the same, so long as you work in products and
sums. The function does not distinguish the conflict and furthermore
does allow for real valued functions such as sin(x) to be expressed
via the polynomial, with some additional constraints. Expressing that
sin(x) is real does nothing to diminish it as a function. If anything
it strengthens it since the alternative form where X is undefined does
not seem to build anything at all. It is off in never-never land so
long as X is in never-never land.

I wonder if given your statement that R is a subring of C if the same
can be proven of X. X seems so nondescript that such a proof will
likely be flawed. If it is not flawed than at least it grants X some
qualities beyond a symbol. Beneath here I see you are claiming that
X X
is a new element and I see this yet this new element is composed of
two elements. Until we mark this new element
c = X X
then we do not have an element. Upon accepting the element c we then
have the collapse of the polynomial. The claim that it provides a
solid and stable structure to perform number theory on is gone.
Further attempting to distinguish c from a is likewise very difficult.
The polynomial
a0 + a1 X + a2 X X
can happily accept a substition
a' = a0 + a2 X X
to become
a' + a1 X .
Here a' is an element of A[X] which carries A within it. To me because
X has not widened out A[X] there are no new elements beyond A in A[X].
This thinking applies to any set. Simply posing a variable does not
make that variable anything special. Somehow this one slips through
the logic of most people. It is the ray gun of a childhood fantasy
battle, only to be trumped by A[X[Y]]...

All of this garbage goes away by simply taking the ring definition as
pristine. Working in a generic ring A polynomials can be expressed and
could be applied to any specific ring, but upon specifying A as say
the reals then all products and sums become real valued.

- Tim

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