Question on bigraded module.

Question

I have a question on an argument in this survey.

Let $A$ be a commutative Noetherian ring, $A[x_1,...,x_m, y_1,...,y_p]$ be the bigraded algebra over $A$, with deg$x_i=(1,0)$, deg$(y_j)=(d_j,1)$.

Let $\mathcal{M}$ be a finitely generated bigraded module over $A[x_1,...,x_m, y_1,...,y_p]$. Define $\mathcal{M_{n}}:=\bigoplus_{a\ge 0}\mathcal{M}_{(a,n)}$. Then, the author claimed that $\mathcal{M_{n}}$ is a graded module over $A[x_1,...,x_m]$.

Now, the author claimed that if $m=0$, then $\mathcal{M_{n}}$ has only finitely many graded component.

My question is : How do we prove that it has only finitely many graded component ?

Answer 1

For $n \neq 0$, $\mathcal{M}_n$ (the submodule generated by homogeneous things whose "$y$-degree" is $n$) is generated as an $A$-module by products of the form $y_{j_1} \dotsm y_{j_n}$. For a given $n$ there are only finitely many such products, so all but finitely many of the $\mathcal{M}_{a,n}$ are zero. The case $n=0$ is not hard.