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Suppose $M= \bigoplus_{n\in \mathbb Z} M_n$ is a finitely generated graded module over a Noetherian graded commutative ring $A=\bigoplus_{n\in \mathbb Z}A_n$.

If $A$ is positively graded ($A_n=0$ if $n<0$), then each $M_n$ is finitely generated as $A_0$-module (Atiyah, McDonald: Introduction to commutative algebra, beginning of chap. 11).

But what happens, if $A$ is not assumed to be positively graded, like $A= \mathbb{Q}[t,t^{-1}]$ ? Is $M_n\;(n \in \mathbb Z)$ also finitely generated over $A_0$ in this case ?

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A submodule over $A_0$ of $M_n$ can be written as the intersection with $M_n$ of the $A$-module it generates. Now try ascending chains. – Wilberd van der Kallen Sep 18 '12 at 16:05
The proof suggested by Wilberd shows in fact that if $G$ is a commutative group, $R$ is a commutative $G$-graded ring, and $M$ is a noetherian $G$-graded $R$-module, then every component of $M$ is a noetherian $R_0$-module. – Fred Rohrer Sep 18 '12 at 21:03
up vote 8 down vote accepted

The answer is yes. Instead of showing that $M_n$ is finitely generated we may show it has the property that any ascending sequence of $A_0$-submodules stabilizes. If $N$ is an $A_0$-submodule of $M_n$, consider $M_n\cap AN$. It is a sum of the $M_n\cap NA_i$. One sees it is $N$ itself. So $N$ can be recovered from $AN$. Now if $N_1\subset N_2\dots$ is an ascending sequence of $A_0$-submodules of $M_n$, the ascending sequence $AN_1\subset AN_2\dots$ stabilizes, hence so do the $N_i=M_n\cap AN_i$.

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@Wilberd @Ralph: Thanks for your answers. Your proofs are not only more general but even shorter than the one in Atiyah-McDonald! – tj_ Sep 19 '12 at 10:36

As already shown by Wilberd, $M_n$ is finitely generated as $A_0$-module. This can be seen as follows:

As a finitely generated module over the Noetherian ring $A$, $M$ is a Noetherian $A$-module. Hence the submodule generated by $M_n \le M$ is finitely generated by, say, $x_1,...,x_k \in M_n$. Now, if $x \in M_n$ there are $a_1,...,a_k \in A$ such that $x=\sum_{i=1}^ka_ix_i$. Write $a_i = \sum_j a_{ij}$ with $a_{ij}$ homogeneous of degree $j$. Since $x,x_i$ are homogeneous of degree $n$, we find $x= \sum_{i=1}^k a_{i,0}x_i$ with $a_{i,0} \in A_0$. Thus $x_1,...,x_k$ generate $M_n$ over $A_0$.

Remark: It's sufficient to assume that $A$ is left Noetherian (and not necessarily commutative).

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