TITLE: Generators for the ideal of entire functions vanishing on $\mathbb Z\times\mathbb Z\subset \mathbb C\times\mathbb C$
QUESTION [9 upvotes]: For $n=1,2$, let $R_n$ be the ring of entire complex functions in $n$ complex variables.  Let $I$ be the ideal $R_2$ of functions $f$ such that $f(\mathbb Z\times\mathbb Z)=\{0\}$. Is $I$ generated by the two functions $(x,y)\mapsto\sin(\pi x)$ and $(x,y)\mapsto\sin(\pi y)$  ?
The reason why I am asking this is the analogous case in $R_1$: If $J$ is the  ideal $R_1$ of functions $f$ such that $f(\mathbb Z)=\{0\}$ then, for $f\in J$, $\frac{f(x)}{\sin(\pi x)}$ seems to be an entire function.  This implies $J=\sin(\pi x)R_1$.

REPLY [3 votes]: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}\def\cO{\mathcal{O}}$I now have an elementary, although to my mind less intuitive argument.
Lemma: Let $f_n(y)$ be any sequence of entire functions $\CC \to \CC$, indexed by $n \in \ZZ$. Then there is an entire function $f : \CC^2 \to \CC$ with $f(n,y) = f_n(y)$ for all $n \in \ZZ$.
Proof: We will take $f(x,y)$ of the form
$$f_0(y) + \sum_{n=1}^{\infty} \frac{\prod_{m=1}^{n-1} (x^2-m^2)}{\prod_{m=1}^{n-1} (n^2-m^2)} \left( a_n(y) \frac{n-x}{2n} + b_n(y) \frac{n+x}{2n} \right) \left( \frac{x^2}{n^2} \right)^{C_n}$$
for some entire functions $a_n(y)$ and $b_n(y)$ and some positive integers $C_n$ which we compute inductively as I will now describe.
Note that, for $p>n$, the product $\prod_{m=1}^{p-1}$ vanishes at $x = \pm n$, so terms beyond the $n$-th term don't effect the value of the sum at $x = \pm n$. Also, at $x = \pm n$, we have $\tfrac{x^2}{n^2} =1$; we have $\tfrac{n-x}{2n}=1$ at $x = -n$ and $0$ at $x=n$, and we have  $\tfrac{n+x}{2n}=1$ at $x = n$ and $0$ at $x=-n$. So, if we have choose $a_m(y)$, $b_m(y)$ and $C_m$ for $m<n$, there is a unique choice of $a_n(y)$ and $b_n(y)$ which will make the sum correct at $x = \pm n$. 
Make that choice for $a_n(y)$ and $b_n(y)$. Then choose $C_n$ large enough that the $n$-th summand is $\leq 2^{-n}$ whenever $|x|$ and $|y| \leq n-1$. (Since $\{ (x,y) \in \CC^2 : |x|, |y| \leq n-1 \}$ is compact and $\left| \tfrac{x^2}{n^2} \right| \leq (1-1/n)^2$ on this region, taking $C_n$ large enough will work.) Thus, the sum will be uniformly convergent on compact subsets, and will thus give an entire function. $\square$
Now, let $h : \CC^2 \to \CC$ be an entire function with $h(\ZZ^2)=0$. For every integer $n$, the function $\tfrac{h(n,y)}{\sin (\pi y)}$ will extend to an entire function $\CC \to \CC$. So, by the lemma, there is some entire function $f(x,y)$ with $f(n,y) = \tfrac{h(n,y)}{\sin (\pi y)}$ for every integer $n$. So $h(x,y) - f(x,y) \sin (\pi x)$ vanishes for all $x \in \ZZ$. Therefore, $\tfrac{h(x,y) - f(x,y) \sin (\pi x)}{\sin (\pi y)}$ will extend to an entire function $g(x,y)$. So $h(x,y) = f(x,y) \sin (\pi x) + g(x,y) \sin(\pi y)$.