Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $t\in \mathbb{N}$ ?

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
@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
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 '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
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 '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
show 2 more comments

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.

share|improve this answer
    
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.