# Integer valued polynomials over several variables

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 $$X^iY^j$$.

• What about $x(x-1)y(y-1)/4$? – Fedor Petrov Jan 7 '19 at 20:22
• @Okay, thanks! It's a proper subring then. – Lehs Jan 7 '19 at 20:31
• 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? – KConrad Jan 7 '19 at 20:43
• @Lehs is it a subring? – Fedor Petrov Jan 7 '19 at 20:43
• @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. – 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 $$f : \mathbb{Z}^n \to \mathbb{Z}$$.