Question on local cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:59:41Z http://mathoverflow.net/feeds/question/115371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115371/question-on-local-cohomology Question on local cohomology Axy 2012-12-04T07:54:18Z 2012-12-04T10:13:48Z <p>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 :$R_{+}^{t}M=0$.</p> <p>My question is : </p> <ol> <li><p>Why Ass(M)= $\lbrace m_0 \oplus R_+ \rbrace$ </p></li> <li><p>Why from that we have :$R_{+}^{t}M=0$ for some $t\in \mathbb{N}$ ?</p></li> </ol> http://mathoverflow.net/questions/115371/question-on-local-cohomology/115379#115379 Answer by Youngsu for Question on local cohomology Youngsu 2012-12-04T09:15:43Z 2012-12-04T09:15:43Z <p>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.</p> http://mathoverflow.net/questions/115371/question-on-local-cohomology/115380#115380 Answer by Fred Rohrer for Question on local cohomology Fred Rohrer 2012-12-04T09:19:16Z 2012-12-04T10:13:48Z <p>First, claim 1 as stated is wrong - consider the zero module.</p> <p>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.)</p> <p>So, what you want to show is that under these hypotheses, $M$ is $R_+$-torsion. For this it suffices to show that $M$ is <code>$\mathfrak{m}_0+R_+$</code>-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.</p>