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

On Jul 6, 10:05 am, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1b2ffbed-40df-4da6-84cb-18b9dd559...@xxxxxxxxxxxxxxxxxxxxxxxxxxx> "Tim Golden BandTech.com" <tttppp...@xxxxxxxxx> writes:
 > On Jul 2, 10:29 am, "*** T. Winter" <***.Win...@xxxxxx> wrote:
...
 > > May be.  But in a polynomial ring the ring elements are not functions.
 > > Addition and multiplication is defined between the elemens of that ring,
 > > and those operations make it a ring.  It is called a polynomial ring
 > > because the elements can be seen as polynomials with coefficients of
 > > a base ring.
 >
 > You are widening out the discussion here and I appreciate that. I am
 > free to compose purely from the ring of reals:
 >    x - x^3 / 3! + x^5 / 5! - x^7 / 7! + ...

No, you are not free to compose that.  You should know that what you write
here is not a polynomial but an infinite series, which is a limit in disguise.

 > These are a series of elements
 >    { x,+1!,x,x,x,-3!,x,x,x,x,x,+5!,x,x,x,x,x,x,x,-7!, ... }.
 > The redundancy in x can be eliminated if you wish.

What redundancy?

 > This function is named sin(x).
 > You say:
 >    "the ring elements are not functions."
 > but your distinction is not so strong given that a composition of
 > elements with sum and product operators does form a function.

Yes, but as in A[X], X is not an element of A, it does not form a function.

 > Particularly we can simply consider the element
 >    x
 > and call it a function, though it's not very exciting.

You can call it that, but in the context of A[X], X is *not* a function.

 >                                                        Likewise since
 > you have already stated that
 >    x * x
 > is an element contradicting your statement above with a function of
 > your own claim as one element.

Eh?  This makes absolutely no sense to me.  If X is an element of a ring,
so are X*X and X*X*X.  Otherwise it would not be a ring.

 > These are merely the qualities of the definition of ring which we
 > discuss. The polynomial form happens to be one compositional pattern
 > of the freedoms that exist. Using your previous context we can accept
 > any polynomial expression as an element, though I have argued that the
 > distinction ought to go more as you have put it here so that
 >    x * x
 > is a product of two elements, that element being assignable to say y,
 > but until we form that expression the proper issuance of a new element
 > has not actually been made.

Whether it is or is not a new element does not depend on whether we give
it a separate name or not.  It depends on the structure of the ring and
the ring operations, that is all.

 >                           This is just quibbling and I don't see that
 > much is here to make any provable point on the qualities of X in the
 > polynomial A[X].

But X is ony a name of an element that is not in A itself, nothing more,
nor less.  So we add to A a new element, call it X.  If we want to make
what we have into a ring, we have to define addition and multiplication
between the elements of A and the new element X.  This means also that
we get new elements like X+X, X*X, etc.  That is the way to form a ring..

 > > May be, but we can also define a "functional ring" where the elements are
 > > functions of a base ring, as addition and multiplication are defined, this
 > > also forms a ring.  (But it is better to use only functions that are
 > > defined everywhere on the base ring.)  So with R the base ring, sin(x)
 > > and cos(x) are elements of that functional ring and sin(x)*cos(x) is
 > > another element, as is sin(x)+cos(x).
 >
 > Well, this is fine. We can see these as polynomials too.

No, we can't, because they aren't.

 > >  > I wonder if given your statement that R is a subring of C if the same
 > >  > can be proven of X.
 > >
 > > What about X?  X is not a ring, so it can not be a subring of something,
 > > nor can something be a subring of it.
 >
 > Errr... as I understand it X is being interpreted as
 >    Z is in Q is in R is in C ... is in X .

Clearly you do not understand.  X is not a set, so it is not a ring.  It
is just the name of a new element added to a base ring A.

 > Oh, I see that notation is bad It's more like
 >    Z is in Q is in R is in C ... is in A[X]

It is not either.  Q is *not* in Z[X], because Q is a field and Z[X] is
*not* a filed.  In general, given a ring A, A[X] is *never* a field,
because it does not contain the inverse of X, and so C is *never* in A[X]..

 > There is a little bit of set theory here I need to review.
 > Like, if
 >    C - R
 > is the set of complex numbers with its subset of real numbers removed
 > does this just leave I the imaginary numbers?

No.  2 + i is also in C - R.

 >                                               Or alternatively is the
 > R which is a subset of C just the values
 >    x + 0 i
 > where x is in R?

That is a different question, and that is true.

 > I believe that this is the correct interpretation, which leaves
 >    x + 1.2 i
 > in the set C after R is removed, otherwise we could freely substitute
 > this expression into the products and sums as R, which is clearly
 > going to break things.

Of course removing R from C does not remove x + 1.2 i for any value of x.
Why should you think that would be the case?

 >                        I'm sorry if this digression is not easy to
 > follow. This could be important to the interpretation of A[X] with A
 > removed.

Interestingly, (X) is an ideal in A[X], and that is the same as A[X] with
A removed.

Not really. Just consider the polynomial 1 + X...

-- m
.

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