Generators for the ideal of entire functions vanishing on $\mathbb Z\times\mathbb Z\subset \mathbb C\times\mathbb C$ 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 $J=\sin(\pi x)R_1$.
 A: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\cO{\mathcal{O}}\def\dbar{\overline{\partial}}$This is true, and is a great example of the power of sheaf theoretic ideas in complex analysis. I will give (a) a high level cohomological proof, explaining that the key point is Cartan's theorem B and (b) a restatement of the argument in terms of the $\dbar$-lemma, without explicitly mentioning sheaf cohomology. (Specifically, I want the form of the $\dbar$-lemma that Wikipedia calls "extended Dolbeault-Grothendieck", with $\epsilon = \infty$.) I can't find an elementary proof.
This has nothing to do with the details of $\CC^2$ and $\mathbb{Z}^2$. Here is the real result:
Theorem Let $[ B_{ji}(z_1, \ldots, z_r)]$ be an $n \times m$ matrix of entire functions and let $[v_j(z_1,\ldots, z_r)]^T$ be a vector of $n$ entire functions. Suppose that, for every point in $\CC^r$, there is a vector $[u_i(z_1, \ldots, z_r)]^T$ of $m$ functions defined near that point, solving the equation $v(z) = B(z) \ v(z)$ near that point. Then there is a vector $u$ of $m$ entire functions solving $v(z)=B(z) \ u(z)$ globally.
Application to your result Take $r=2$, $m=2$, $n=1$ and the matrix $B(x,y) = [\sin (\pi x) \ \sin (\pi y)]$.
Notational convections Capitals letters $A$ and $B$ are matrices of holomorphic functions; their entries are $B_{ji}$ and $A_{ih}$. Lower case roman letters $t$, $u$ and $v$ are vectors of holomorphic functions. Letters with a tilde over them, like $\tilde{t}$ and $\tilde{u}$ are vectors of smooth functions. Letters $h$, $i$, $j$, $k$, $\ell$ are indices.
Sheaf theoretic proof Let $\cO$ be the sheaf of holomorphic functions. The matrix $B$ defines a map $\cO^{\oplus m} \to \cO^{\oplus n}$. Let $\mathcal{K}$ be the kernel of this map and $\mathcal{I}$ the image; both are coherent sheaves of $\cO$-modules. 
Consider the short exact sequence $0 \to \mathcal{K} \to \mathcal{O}^{\oplus m} \to \mathcal{I} \to 0$. By Cartan's theorem B, we have $H^1(\CC^r, \mathcal{K})=0$, and thus we have a short exact sequence $0 \to H^0(\CC^r, \mathcal{K}) \to H^0(\CC^r, \mathcal{O}^{\oplus m}) \to H^0(\CC^r, \mathcal{I}) \to 0$. The vector of entire functions $v$ is an element of $H^0(\CC^r, \cO^{\oplus n})$ and the condition that $v = B u$ is locally solvable means that $v$ actually lies in $H^0(\CC^r, \mathcal{I})$. So there is some $u \in H^0(\CC^r, \mathcal{O}^{\oplus m})$ which maps to $v$, and this is the $u$ we seek. $\square$
In the particular case that $m = 2$, $n=1$ and the functions $B_{11}(z)$ and $B_{12}(z)$ have no common factor, I can rewrite this proof to not mention cohomology. Set $A_{11}(z) = B_{11}(z)$ and $A_{21}(z) = - B_{12}(z)$. Note that, if $B_{11}(z) u_1(z) + B_{12}(z) u_2(z)=0$ then there is a unique holomorphic function $t(z)$ with $u_j(z) = A_{j1}(z) t(z)$. In sheaf language, we have a short exact sequence $0 \to \cO \overset{A}{\longrightarrow} \cO^{\oplus 2} \overset{B}{\longrightarrow} \cO \to 0$, but if you are scared of sheaves, then the statement is perfectly explicit; you could even replace the $A$ and $B$'s with explicit sines if you wanted to.
More elementary proof in this case Cover $\CC^r$ by open sets $\Omega_k$ on which there are holomorphic solutions $u_k$ to $B u_k = v$. Let $\phi_k$ be a partition of unity subordinate to the cover $\Omega_k$ and put $\tilde{u} = \sum \phi_k u_k$. Then $B \tilde{u} = v$, but $\tilde{u}$ is a vector of smooth functions, not holomorphic ones. 
On the open set $\Omega_k$, we have $B (\tilde{u} - u_k)= 0$, so there is a vector of smooth functions $\tilde{t}_k$ with $\tilde{u} - u_k = A \tilde{t}_k$. (You might want to write this out for the original example of the two sine functions.) 
Hitting both sides with the $\dbar$ operator, we have $\dbar\left( \tilde{u} \right) = A \ \dbar\left( \tilde{t}_k \right)$. Here we have used that $A_{ih}(z)$ and $(u_k)_i(z)$ are holomorphic and thus $\dbar(A) = \dbar(u_k) = 0$. Thus, on the open set $\Omega_k \cap \Omega_{\ell}$, we have $A \ \dbar\left( \tilde{t}_k \right) = A \ \dbar\left( \tilde{t}_{\ell} \right)$. But the matrix $A$ is injective, so we have $\dbar\left( \tilde{t}_k \right)=\dbar\left( \tilde{t}_{\ell} \right)$ where both sides are defined. So there is a global $(0,1)$-form $\eta$ which restricts to $\dbar\left( \tilde{t}_k \right)$ on each $\Omega_k$.
The equation $\eta = \dbar\left( \tilde{t}_k \right)$ shows that $\eta$ is $\dbar$-closed. So, by the $\dbar$-Poincare lemma, there is a global smooth function $\tilde{t}$ with $\dbar(\tilde{t}) = \eta$. On each $\Omega_k$, we see that $\dbar(\tilde{t}) = \dbar(\tilde{t}_k)$. 
Put $u = \tilde{u} - A\ \tilde{t}$. Then $B \ u = B \tilde{u} - B A \tilde{t} = v$, since $BA=0$. But also, on each $\Omega_k$, we have  $$\dbar(u) = \dbar(\tilde{u}) - A \ \dbar(\tilde{t})=\dbar(\tilde{u}) - A \ \dbar(\tilde{t}) = A\  \dbar(\tilde{t}_k) - A\ \dbar(\tilde{t}) = A\ \left( \dbar{\tilde{t}} - \dbar{\tilde{t}_k} \right) = 0.$$ 
So $\dbar(u)=0$ and $u$ is holomorphic. We have solved the equation $Au=v$ with a vector of global holomorphic functions as required. $\square$
The key property of $\CC^r$ is that $\CC^r$ is Stein. If we take the standard example of a non-Stein domain, $\Omega = \CC^2 \setminus \{ (0,0) \}$, then note that the function $1$ vanishes at every point of $\Omega$ where $z_1=z_2=0$, but that we cannot find holomorphic functions $u_1(z_1, z_2)$ and $u_2(z_1, z_2)$ on $\Omega$ solving $z_1 u_1 + z_2 u_2 = 1$. This illustrates that we really did need a condition like Stein-ness.
A: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}\def\cO{\mathcal{O}}$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 $h(x,y) = f(x,y) \sin (\pi x) + g(x,y) \sin(\pi y)$.
