Re: Integer-Valued Polynomials


Chip Eastham wrote:
Maury Barbato wrote:
Hello,
let D be a domain, K its quotient field, and F an
extension field of K. The ring of integer-valued
polynomials in n indetermintes over D consists of those
polynomials in K[x_1,...,x_n] which give a value in D
for every argument in D^n (standard definition).
What happens if we replace K[x_1,...,x_n] with
F[x_1,...,x_n] in this definition? Do we obtain the same
set of polynomials? What if D=Z, K=Q and F=C?
Thank you very mauch for your ideas.

Hi, Maury:

I don't believe we do pick anything "extra" in polynomials
over the extension field K, by an interpolation argument.

First let us note that one can have a "integer-valued"
polynomial which is not simply integer coefficients:

f(x) = x(x+1)/2

for example. However a polynomial with coefficients in
(for example) the complex field C and integer-valued
for all arguments in Z must agree with a polynomial of
equal degree with coefficients in Q (e.g. constructed
by Newton divided differences) on all arguments Z,
and hence be identical to that polynomial over Q.

Ah, as Jyrki points out, we do need domain D to be
infinite to make this argument work, as otherwise a
multiple of a nonzero polynomial vanishing only all of
D can be constructed.

-- c

.

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