functions not polynomials

[Paul Taylor <pt@cs.man.ac.uk>]

Let  f: Z x Z -> Z  be a binary FUNCTION (in the sense of sets)
on the integers, with the property that

 - for each x:Z,  f(x,-) : Z -> Z is a (agrees with a unique)
   POLYNOMIAL, whose coefficients are functions of x; and similarly

 - for each y:Z,  f(-,y) : Z -> Z is also a polynomial.

Then  f(x,y)  was itself a polynomial in two variables.

"You just use Newton's difference method to find the co..."

.... unterexample.

Taking N for simplicity first, consider the binomial coefficient
(x,n) as an nth degree polynomial in x, ie
(x,0) = 1
(x,1) = x
(x,2) = x(x-1)/2
(x,3) = x(x-1)(x-2)/3!
and so on.  Newton's finite difference method provides the coefficients
of these generating polynomials by taking successive differences (the
finite analogue of successive derivatives).

But then    sum_n (x,n)(y,n)
is a function NxN->N that is a polynomial in x for each y and vice versa
but isn't itself a polynomial.

None of the web pages that I found about this method was especially
clear, so I'm not going to recommend any, but they are there if you
want to look for them yourself.

The more general method (as described on these pages, which is why
I didn't think they were that good) takes differences from any sequence
of points, not just 0, 1, 2, ....  Hence, given any enumeration of
a countable field K (this probably works for Z too) we can generalise
both the method of fitting polynomials and the counterexample.


Of course, since I'm a categorist, it was the categorical formulation
that interests me.  Any thoughts on that would be appreciated, despite
the above failure of the simple example.

In fact, I'd quite like to hear from any students of universal algebra
out there who also know how monads encode algebraic theories, and might
be willing to help as sparring partners for my present mad scheme.

Paul Taylor