## Question on local cohomology

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

Let $M$ be a positive graded finitely generated module over a positive graded commutative ring $R$. Assume that $R_0$ is a local ring with maximal ideal $m_0$. Let $d$ be the Krull dimension of $M$. In the book "Local cohomology : An algebraic introduction with geometric application" of R.Y. Sharp and Broddman, the author claim that if $d=0$ then the set of associated primes ideal of $M$ is $\lbrace m_0 \oplus R_+ \rbrace$, and therefore there exist a $t\in \mathbb{N}$ such that :$R_{+}^{t}M=0$.

My question is :

1. Why Ass(M)= $\lbrace m_0 \oplus R_+ \rbrace$

2. Why from that we have :$R_{+}^{t}M=0$ for some $t\in \mathbb{N}$ ?

-

## 2 Answers

First, claim 1 as stated is wrong - consider the zero module.

Second, I guess that you talk about a step in the proof of Theorem 15.3.1 in Brodmann-Sharp. If so, then you have more hypotheses than you mentioned. Beside others, $R$ is noetherian and - most important - $M$ is $0$-dimensional. (And $M$ is not "positively graded", a notion that seems not reasonable for graded modules.)

So, what you want to show is that under these hypotheses, $M$ is $R_+$-torsion. For this it suffices to show that $M$ is `$\mathfrak{m}_0+R_+$`-torsion. More general, it suffices to show that a $0$-dimensional finitely generated graded module $M$ over a *local graded ring with *maximal ideal $\mathfrak{m}$ is $\mathfrak{m}$-torsion. And this is indeed the case. Namely, $0$-dimensionality means that the graded ring $R/(0:_RM)$ is $0$-dimensional. Now, $\sqrt{(0:_RM)}$ is the intersection of the graded primes containing $(0:_RM)$. But as $R/(0:_RM)$ is $0$-dimensional, $\mathfrak{m}$ is the only such prime, implying $\sqrt{(0:_RM)}=\mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated (as $R$ is supposed to be noetherian), there exists $t\in\mathbb{N}$ with $\mathfrak{m}^t\subseteq(0:_RM)$, and this implies that $M$ is an $\mathfrak{m}$-torsion module as desired.

-
@Fred Rorher: Why do we have : $M$ is $m_{0}+R_+$ torsion then $M$ is $R_+$ torsion ? Could you please make it more precise ? – Axy Dec 4 at 10:13

1
Dear @Axy, this is because if $\mathfrak{a}$ and $\mathfrak{b}$ are ideals with $\mathfrak{a}\subseteq\mathfrak{b}$, then the $\mathfrak{b}$-torsion functor is a subfunctor of the $\mathfrak{a}$-torsion functor. – Fred Rohrer Dec 4 at 10:18
Sorry Fred Rohrer, but I still do not get it. my question is follow: Since M is $m_0+R_+$ torsion, then there exist $t$ such that $(m_0+R_+)^tM=0$. From this, how can we get $R_{+}^{k}M=0$ for some k – Axy 14 secs ago – Axy Dec 4 at 10:32

1
Dear @Axy, $R_+\subseteq\mathfrak{m}_0+R_+$, hence $R_+^t\subseteq(\mathfrak{m}_0+R_+)^t$, thus $R_+^tM\subseteq(\mathfrak{m}_0+R_+)^tM=0$, and therefore $R_+^tM=0$. – Fred Rohrer Dec 4 at 10:35
Thank @Fred Rohrer for that. I am so stupid. Could you explain for me that in the proof in that book, the authors proved that : $M_{\mathfrak{p}_{0}}=N_{\mathfrak{p}_{0}}$ for each $\mathfrak{p}_0\in \text{Spec}(R_0)$. So why can we reduce the problem for the local case ? I still do not understand the idea of the author. Sorry if you feel my question is stupid. – Axy Dec 4 at 10:44
show 2 more comments

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Let $m = m_0 \oplus R_+$ which is the homogeneous maximal ideal (also it is maximal). Then the first condition implies that Min(M) = Ass(M) = {$m$}. In particular, dim $M =$ dim $R/m = 0$. This says that $M$ is an Artinian module. So there exists $s$ such that $M_s = 0$. So taking $t \ge s$ would do the job.

-

@Young su: What do you mean by Min(M) ?  – Axy Dec 4 at 10:31
I meant the set of minimal primes in Supp(M). But here Supp(M) = {m} as well.  – Youngsu Dec 5 at 3:47
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 67, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.939772367477417, "perplexity_flag": "head"}