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?

Can't be true in this generality. If D is finite, say
of
cardinality q, then you can add any multiples of
(x_i^q-x_i) to an integer-valued polynomial without
affecting this property.

What if D=Z, K=Q and F=C?

OTOH it seems to me that requiring D to be infinite
is
a sufficient condition as well: write elements of F
in terms of a K-basis B (assume that 1_K is in B).
Then you can write all the polynomials in
F[x_1,...,x_n]
as K[x_1,...,x_n]-linear combinations of constants
from B.

IIRC the following holds:
D infinite => (p(x_1,x_2,...,x_n) with p in
in K[x_1,...,x_n]
vanishes on all of D^n, iff p=0)


Yes, this is surely true. It can be easily proved by
induction on the number of indeterminates, and using the
fact that a polynomial in one indeterminate of degree k
has at the most k roots.

For a polynomial from F[x_1,...,x_n] to be
integer-valued
it must be at least K-valued, so by the above result
the coefficient polynomial p_b of an element b from
B\setminus{1_K}
must be the zero polynomial (in order to give rise to
the
all-zero polynomial function). Therefore only p_1 can
be
non-zero and the claim follows.

Cheers,

Jyrki
.

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