Re: Integer-Valued Polynomials
- From: Jyrki Lahtonen <lahtonen@xxxxxx>
- Date: Tue, 28 Feb 2006 14:55:04 +0200
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 K[x_1,...,x_n]
vanishes on all of D^n, iff p=0)
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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