Re: Integer-Valued Polynomials


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.

regards, chip

Two remarks:
(I)Newton method works also for several
variables and for general fields?
(II) Even if it works, it can't be very useful. How could
you apply it, to use later the "Identity Polynomials
Principle"?
But your argument intended maybe to be only intuitive.

However, thank you very much for your help.
My Best Regards,
Maury
.

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