Let $A$ be a nonnegatively graded algebra and $M$ a nonnegatively graded $A$-module. Then, $A\_{>0}M$ is a graded $A$-submodule of $M$.


(Here, given a nonnegatively graded algebra $A$, we've defined $A\_{>0} := \oplus\_{i > 0} A\_i$.)


My question is as follows. How do I see that $M$ is finitely generated as an $A$-module if and only if $M/A\_{>0}M$ is finitely generated as an $A$-module?


Is this well-known? Can I find a proof of this anywhere? Or could anybody supply a proof?



$M$ is finitely generated as $A$-module iff $M/A_{>0}M$ is finitely generated as $A$ module?
This is essentially Nakayama's lemma (not literally, but the same proof). More generally, the result is that a graded module map $\phi\colon N\to M$ is surjective if and only if the induced map $N\to M/A\_{>0}M$ is surjective. "Only if" is obvious; consider the "if" direction. The induced map being surjective is the same as saying that $M=A\_{>0}M+\operatorname{im}\phi$, so reducing both sides modulo $\operatorname{im}\phi$, we see that $M/\operatorname{im}\phi=A\_{>0}(M/\operatorname{im}\phi)$. This is impossible, since $M/\operatorname{im}\phi$ has an element of minimal degree (since the grading is non-negative) and that's obviously not in $A\_{>0}(M/\operatorname{im}\phi)$.


Note if $M$ is allowed to have elements in arbitrary degrees the result is false: consider $k[x,x^{-1}]$ as a graded module over $k[x]$. 


