Re: Question about "polynomial evaluations"

On Jun 17, 2:16 pm, Brian Chandler <imaginator...@xxxxxxxxxxxxx>
wrote:
Brian Chandler wrote:

Stuff addressed to Tim BandTech.com, in which I referred to an example
on these pages from W W Sawyer's little book:  http://imaginatorium.org/private/sawyer.gif

I said:

But here you are simply wrong. In particular, in the example Sawyer
gives of polynomials over a finite field ({0, 1} with mod 2
arithmetic, call it F2), the polynomial x^2+x _always_ evaluates to 0.
So in the field of "values" of these polynomial expressions (there
_are_ only two values), these two things are "equal". But in the ring
of polynomials over F2, these are not the same polynomials.
Mathematics is the study of patterns, and is just as free to study the
pattern of *polynomials* as it is to study the pattern of *polynomial
evaluations* (though this is stunningly uninteresting, since it's
exactly the same as the field underlying the polynomials).

The last bit seems to be simply wrong:

Yes; what I think you were aiming for is what I mentioned elsewhere:
polynomial functions as opposed to polynomials.

in this particular case, define
the "evaluation" e(P) of an arbitrary  polynomial P as follows.

If we substitute a value for x, this must be 0 or 1, and considering
these two values, there are four cases:


Yikes. Okay...


[needs fixed font]

When
x=0  x=1  "Evaluation"
 0    0    0
 1    1    1
 0    1    S (for 'same')
 1    0    D (for 'different')

If we define the obvious operations on the Evaluations, e(a) + e(b) = e
(a+b), e(a) * e(b) = e(a*b), we get (I think) that e() is a ring
homomorphism, and these Evaluations must form a ring.

I think this is what you might be groping for:

Let R be any (commutative, for simplicity) ring, and let R[X] be the
ring of polynomials in X with coefficients in R. Let F(R) be the ring
of functions (note: functions, *not* homomorphisms) R-->R with
pointwise addition and product; that is, if f,g:R-->R are any two
functions from R to R, then (f+g)(x)=f(x)+g(x) and (f*g)(x) = f(x)*g
(x) for each x in R, with the operations on the right hand side being
the operations of R.

There is a natural ring homomorphism eval:R[X]-->F(R) given by
"evaluation": if p(x) in R[X], say p(x) = a_0 + a_1x + ... + a_m*x^m,
then p(x) defines a function eval(p): R-->R by the rule eval(p)(r) = p
(r) = a_0 + a_1*r + ... + a_m*r^m.

The image of eval in F(R) is a subring of F(R); its elements are
called the "polynomial functions" from R to R.

Now, take R = F_2, the field of two elements. The polynomials x^2 and
x are distinct elements of F_2[X]; however, their images eval(x^2) and
eval(x) are equal as elements of F(F_2), since eval(x^2)(0) = 0 = eval
(x)(0) and eval(x^2)(1)=1=eval(x)(1). That is, the map eval:F_2[X] -->
F(F_2) is not one-to-one.

On the other hand, if R is the integers, the rationals, the real
numbers, etc., the map eval *is* one-to-one.

In fact, one can prove that if R is *any* finite field, then *every*
element of F(R) is a polynomial function; in particular, since F(R) is
finite, the map eval cannot be one-to-one.

What is more, for a very large class of rings, in particular for any
field and for Z, one can show that the map eval is one-to-one *if and
only if* the ring is infinite. That is: distinct polynomials
correspond to distinct functions exactly when the underlying ring is
infinite.

In short, the issue you are dealing with is the distinction between
polynomials and polynomial functions. When the map eval is one-to-one,
it is common to identify the two (as the ring R[X] will be isomorphic
to the subring of polynomial functions under the map eval); but when
eval is *not* one-to-one, that's when you get things like distinct
polynomials giving the same polynomial function.

Hope that helps.

--
Arturo Magidin
.

Relevant Pages

  • Re: Possible flaw in the polynomial ring A[X] construction
    ... distinguish it from the product of polynomial functions. ... for rings of finite characteristic, so let us consider the base ring to ... the ring of polynomials over this field is ... with the base ring being the field of two elements. ...
    (sci.math)
  • Re: Possible flaw in the polynomial ring A[X] construction
    ... distinguish it from the product of polynomial functions. ... for rings of finite characteristic, so let us consider the base ring to ... the ring of polynomials over this field is ... with the base ring being the field of two elements. ...
    (sci.math)
  • Re: Possible flaw in the polynomial ring A[X] construction
    ... I construct the polynomial ring with real coefficients Afrom ... likewise collapsed to the reals by strict adherence to the ring ... In the ring of polynomial FUNCTIONS, perhaps, but that is quite ... different from the ring of polynomials. ...
    (sci.math)
  • Re: Possible flaw in the polynomial ring A[X] construction
    ... with further operations such as a quotient ring. ... then the ring Aof polynomials with real coefficients, ... than just heaping piles of construction upon other piles of ... "While the polynomial itself is necessarily infinite in length, ...
    (sci.math)
  • Re: Understanding the quotient ring nomenclature
    ... A ring is defined as containing elements. ... undefined domain and an infinite length of terms. ... infinite series are reals. ... long as the two polynomials being multiplied have finitely many terms, ...
    (sci.math)