Re: Question about "polynomial evaluations"

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

<snipped>

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

Ah, yes. Thanks -- your explanation made a lot of it clearer.

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.

Is this proof trivial? It looks as though you can just "construct" a
polynomial for any function, given an unlimited amount of "space" to
play with. Intuitively, it ought to be possible to make a polynomial
(function) that maps p to q, and everything except p to 0, then you
just add them together. [?]

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

OK, so I could rewrite these as the polynomial functions 0, 1, x, x+1,
but I'll leave the single character identifiers for ASCIIvenience, and
repeat the multiplication table for these polynomial functions:

* 0 1 S D
0 0 0 0 0
1 0 1 S D
S 0 S S 0
D 0 D 0 D

Is this right? I suppose this reminds me how feeble my grasp is on
what a ring homomorphism is. There's a "shape-preserving" map from
polynomials (which certainly don't have zero divisors) to polynomial
functions (which do?) Hmm, so it means that the "zero-
polynomials" (like x^2+x) are an ideal, which doesn't do much except
remind me how feeble my grasp is on what an ideal is.

Brian Chandler
.

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
    ... Because for a polynomial function on the ring GF, for each x, fis ... For every ring A, the ring of formal polynomials over that base ring, ... cannot be the same as the finite ring of polynomial functions. ... the distinction. ...
    (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: A "How would you do this"-type of question. not Java-specific.
    ... a dot will appear on the map to indicate where that ZIP code ... coords and extrapolate those to x,y coords on your graphic map? ... arbitrary encoding -- some large-scale patterns are evident ... evaluating such high-degree polynomials with useful accuracy! ...
    (comp.lang.java.help)
  • Re: Fields and transcendentals
    ... homomorphism fails to be an isomorphism it's because t is ... since Kis formed by the inclusion of all polynomials in t, ... there be some elements of Kthat map to 0. ... and find a polynomial having t as a root. ...
    (sci.math)
Quantcast