I've been recently wondering how to prove the existence of a Hilbert polynomial for finitely generated bigraded modules $M$ over a polynomial ring $R=k[X_0,...,X_n,Y_0,...,Y_m]$ with the usual bigraded structure; concretely, is there a polynomial $P\in \mathbb{Q}[T,S]$ such that $P(a,b)=\text{dim}_k M_{a,b}$ for all sufficiently large $a,b\in \mathbb{Z}$?
I know that the usual proof of this fact for the graded case uses the existence of a finite free resolution of the module. So I've been trying to apply the same technique for the bigraded case: I take a resolution of $M$ of the form $$ 0\rightarrow K \rightarrow L_{n+m+1}\rightarrow ... \rightarrow L_{0}\rightarrow M\rightarrow 0$$ where $L_i$ are finitely generated free modules (now, free in the sense of having a basis of bihomogeneous elements), $K$ is finitely generated and all the maps are compatible with the bigrading and of bidegree (0,0). We can view this resolution as a graded one (take, for instance, $M_r=\sum_{a+b=r} M_{a,b}$ as the $r$-degree piece of $M$), and then Hilbert's Syzygy Theorem asserts that $K$ is also free $\textbf{as graded module}$. I've trying to prove that in fact we can take a basis of $K$ consisting of bihomogeneous elements, but I didn't succeed (in fact I don't even know if what I'm trying to prove is true). So here's the question:
Can we assure that a finite free resolution of this sort exists for bigraded modules (that is, in which every term is freely generated by bihomogeneous elements)? Is there an analogous Hilbert Syzygy Theorem for bigraded modules? And finally (and importantly!), can this be extended to the multigraded case?
Thank you all a lot in advance.