For simplicity this is about polynomials in just two variables.

Any $f\in\mathbb Q[X,Y]$ can be written as a linear combination of monomials $X^iY^j$ and therefore as a sum of polynomials $p_{ij}\in\mathbb Q[X^iY^j]$ over one variable:

$$\displaystyle f(X,Y)=p_{00}+\sum_{\gcd(i,j)=1}p_{ij}(X^iY^j).$$

My question: is the subring of all integer valued polynomials $f$ over two variables identical with the subring of all sums of integer valued polynomials $p_{ij}$ over one variable as above?

An integer valued polynomial in one variable is a polynomial $p$ with rational coefficients such that $p(\mathbb Z)\subseteq \mathbb Z$. And corresponding for polynomials over several variables.

It might be some abuse of language to call $p_{ij}$ polynomials over one variable, they may rather be polynomials over one variable applied to monomials $X^iY^j$.

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    $\begingroup$ What about $x(x-1)y(y-1)/4$? $\endgroup$ – Fedor Petrov Jan 7 '19 at 20:22
  • $\begingroup$ @Okay, thanks! It's a proper subring then. $\endgroup$ – Lehs Jan 7 '19 at 20:31
  • $\begingroup$ The counterexample given above is the product of binomial coefficient polynomials $\binom{x}{2}\binom{y}{2}$. Integer-valued polynomials in one variable $X$ are linear combinations of binomial coefficient polynomials $\binom{X}{i}$, so do you want to consider polynomials $p_{ij}(\binom{X}{i}\binom{Y}{j})$ instead? $\endgroup$ – KConrad Jan 7 '19 at 20:43
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    $\begingroup$ @Lehs is it a subring? $\endgroup$ – Fedor Petrov Jan 7 '19 at 20:43
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    $\begingroup$ @KConrad the polynomials ${x \choose i}{y\choose j}$ form a basis in the $\mathbb{Z}$-module of integer-valued polynomials in two variables. So you do not even need these $p_{ij}(\cdot)$, only integer coefficients. $\endgroup$ – Fedor Petrov Jan 7 '19 at 20:45

Every polynomial function $f : \mathbb{Z}^2 \to \mathbb{Z}$ is a $\mathbb{Z}$-linear combination of the polynomials $\pmatrix{x \\ i} \pmatrix{y \\ j}$ with $i,j \geq 0$.

Indeed, let $f : \mathbb{Z}^2 \to \mathbb{Z}$ be a polynomial function. By interpolation, we know that $f$ comes from a polynomial in $\mathbb{Q}[x,y]$. Since the polynomials $\pmatrix{x \\ n}$ form a basis of $\mathbb{Q}[x]$, we may write $f(x,y)=\sum_{i,j \geq 0} a_{i,j} \pmatrix{x \\ i} \pmatrix{y \\ j}$ with $a_{i,j} \in \mathbb{Q}$.

Since $f(0,y) = \sum_{j \geq 0} a_{0,j} \pmatrix{y \\ j}$ is integer-valued, we have $a_{0,j} \in \mathbb{Z}$. Moreover, by the recursion formula for binomial coefficients \begin{equation*} f(x+1,y)-f(x,y) = \sum_{i \geq 1, j \geq 0} a_{i,j} \pmatrix{x \\ i-1} \pmatrix{y \\ j} \end{equation*} so that $a_{1,j} \in \mathbb{Z}$, and an easy induction on $i$ gives $a_{i,j} \in \mathbb{Z}$ for every $i,j$.

The same proof works for polynomial functions $f : \mathbb{Z}^n \to \mathbb{Z}$.

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