Integer valued polynomials in two variables The ring of integer valued polynomials, $\{ f \in \mathbb{Q}[x] : f(\mathbb{Z}) \subseteq \mathbb{Z} \}$ is fairly well-known to be generated as Abelian group by the binomial coefficients, $f_k(n) = \binom{n}{k}$ of degree $k$.
What about the corresponding two-variable ring, $\{ f \in \mathbb{Q}[x,y] : f(\mathbb{Z}^2) \subseteq \mathbb{Z} \}$? Does it have a nice generating set?
Multinomial coefficients are usually $\binom{n}{i,j}$ which would still be a single variable polynomial (the $i$ and the $j$ would affect the denominator). Nothing else came to mind.
Also, I guess I probably want the generators to satisfy $f(\omega^2) \subseteq \omega$ where $\mathbb{N}=\{0,1,2,\ldots\} \subset \mathbb{Z}$, which just works out for the binomial coefficients. If that is impossible, that would also be useful knowledge, since theoretically these count things.
 A: We require that $f(x,y)\in\mathbb{Z}$ for $x,y\in\mathbb{Z}$.
Represent $f$ as a polynomial in $y$ with coefficients in $\mathbb{Q}[x]$.
Since $\operatorname{im}f|_{\mathbb{Z}^2}\subseteq\mathbb{Z}$
we know that $f$ can be made to take the form
$$ f(x,y) = \sum_j f_j(x)\binom{y}{j}
\quad\text{with}\quad
f_j(x)\in\mathbb{Q}[x];\quad
f_j(\mathbb{Z})\subseteq\mathbb{Z}$$
where $j$ runs through a finite subset of the nonnegative integers.
Conversely, every such expression fulfills the requirements we have for $f$.
Now since $\operatorname{im}f_j|_{\mathbb{Z}}\subseteq\mathbb{Z}$, 
we also know that $f_j$ can be made to take the form
$$f_j(x) = \sum_i f_{i,j}\binom{x}{i}
\quad\text{with}\quad f_{i,j}\in\mathbb{Z}$$
where $i$ runs through a finite subset of the nonnegative integers.
Conversely, every such expression fulfills the requirements we have for $f_j$.
Taken together, this means that we have the generating set
$$\binom{x}{i}\binom{y}{j} \quad\text{with}\quad i,j\in\{0,1,2,\ldots\}$$
the members of which have pairwise distinct degrees in $x$ or $y$.
Therefore, that generating set is minimal.
