Let $$R$$ be a graded Noetherian ring and $$M$$ be f.g. graded $$R$$-module. What I want to show is that

1. Primary decomposition of submodule of $$M$$ can be taken in terms of homogeneous modules. (In fact it is sufficient to show at decomposition of $$0$$.)
2. It is well known that there is a finite filtration $$0=M_0 \subseteq \cdots \subseteq M_n=M$$ such that $$M_i/M_{i-1} \cong R/P_i$$ for some prime ideal $$P_i$$ (in fact, this $$P_i$$ is an associated prime of $$M$$.) Can we take these $$M_i$$ and $$P_j$$ homogeneous ones??

I could verify that in given conditions, every associated primes of $$M$$ is homogeneous.

Thank you for any helps.

Step 1: Let $$M$$ be a f.g. graded non-zero module over a graded Noetherian ring $$R$$. Consider the set $$S:=\{\operatorname{Ann} x:x\in M\text{ is non-zero and homogeneous} \}$$. Note that each ideal of $$S$$ is homogeneous. Use Noetherian property to deduce it has a maximal element $$P=\operatorname{Ann}(x)$$. You can show that $$P$$ is prime and hence $$P$$ is a homogeneous prime ideal. EDIT: Assume $$fg\in \operatorname{Ann}(x)$$ where $$f,g$$ are homogeneous. Say $$f\not\in\operatorname{Ann}(x)$$ Then $$\operatorname{Ann}(x)\subsetneq\operatorname{Ann}(gx)$$ By maximality, this forces $$gx=0\implies g\in\operatorname{Ann}(x)$$
Step 2 Let $$M,R$$ be as above. Consider $$M_0=0$$ $$M_1=Rx$$ where $$x$$ is as above. Note that $$M_1$$ is a homogeneous sub-module. Use Step 1 on $$M/M_1$$ to get $$M_2/M_1$$. Then consider $$0=M_0\hookrightarrow M_1\hookrightarrow M_2\hookrightarrow \dots$$
Note that each $$M_i$$ is homogeneous and $$M_i/M_{i-1}\cong R/P_{i}$$ where $$P_i$$ is a homogeneous prime ideal. This chain terminates since $$M$$ is Noetherian and the final module must be $$M$$ by Step 1.
• There was a pretty serious slip up in the previous version. One has to consider only non zero homogeneous elements else you will always get $R$. Sep 28, 2020 at 3:21