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 $X^iY^j$.



Integer valued polynomials over several variables
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 $f : \mathbb{Z}^n \to \mathbb{Z}$.


