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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.

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1 Answer 1

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Indeed you can. Here's an outline.

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.

I think this answers both your questions.

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  • $\begingroup$ 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 $R$. $\endgroup$
    – user6
    Commented Sep 28, 2020 at 3:21

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