# If $M$ is a positively graded finitely generated module of dim 0, then why $R_{+}^{t}M=0$ for some $t\in \mathbb{N}$?

Let $M$ be a positively graded finitely generated module over a positively 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" by M.P.Brodmann and R.Y.Sharp, the authors claim that if $d=0$ then the set of associated prime ideals of $M$ is $\lbrace m_0 \oplus R_+ \rbrace$, and therefore there exists a $t\in \mathbb{N}$ such that $R_{+}^{t}M=0$.

My question is :

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

2. Why does this imply $R_{+}^{t}M=0$ for some $t\in \mathbb{N}$?

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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.

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@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 '12 at 10:13
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 '12 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 '12 at 10:32
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 '12 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 '12 at 10:44

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.

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@Young su: What do you mean by Min(M) ? – Axy Dec 4 '12 at 10:31
I meant the set of minimal primes in Supp(M). But here Supp(M) = {m} as well. – Youngsu Dec 5 '12 at 3:47