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Well, I have asked this question on MO, I do not like asking duplicate question on both sites, however, my question might be too elementary for them, so I decided to post it here. Sorry if you feel it is not appropriate.

Let $A$ be a commutative Noetherian ring, $R$ be a standard graded algebra over $A$, $M$ be finitely generated graded $R$-module. Let $R_{+}$ be the irrelevant ideal. The Castelnuovo-Mumford regularity of $M$ or regularity for short is defined as follows: $$\text{reg}(M):=\text{max}\lbrace a(H_{R_+}^{i}(M))+i|\ i\ge 0\rbrace,$$ where $a(H)$ denotes the maximal non-vanishing degree of a graded module $H$. Let $x_1,...,x_s$ be linear forms in $R$. This set of elements is called an $M$-filter regular sequence if $x_{i}\notin \mathfrak{p}$ for any associated prime $\mathfrak{p}\nsupseteq R_{+}$ of $(x_1,...,x_{i-1})M$ for $i=1,...,s$.

Then, people claim that $\text{reg}(M)=\text{max}\lbrace a((x_1,...,x_i)M:R_{+}/(x_1,...,x_{i})M)|\ i=1,...,s\rbrace$.

Could you explain for me why the two above definitions are equivalent ? What is the motivation of filter regular sequence ?

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up vote 4 down vote accepted

For all $k\ge 0$ we set $a_k(M)=a(H_{R_+}^{k}(M))$.

Lemma. Let $z\in R_1$ be an $M$-filter regular element. Then, for all $k\ge 0$, we have $$a_{k+1}(M)+1\le a_k(M/zM)\le\max\{a_k(M),a_{k+1}(M)+1\}.$$

Proof. By the definition of $M$-filter regularity and a similar result for modules to this one Noetherian ring and primary decomposition result we get $(0:_Mz)\subseteq\bigcup_{n\ge 0}(0:_MR_+^n)$. In particular $(0:_Mz)$ is annihilated by $R_+^n$ for some $n\ge 0$ which in terms of local cohomology means that $H_{R_+}^i(0:_Mz)=0$ for all $i\ge 1$. This shows easily that $H_{R_+}^i(M)=H_{R_+}^i(M/(0:_Mz))$ for all $i\ge 1$. From the following short exact sequence $$0\longrightarrow M/(0:_Mz)\stackrel{z\cdot}\longrightarrow M\longrightarrow M/zM\longrightarrow 0$$ we get a long exact sequence for the local cohomology $$H_{R_+}^i(M)_n\longrightarrow H_{R_+}^i(M/zM)_n\longrightarrow H_{R_+}^{i+1}(M)_{n-1}\longrightarrow H_{R_+}^{i+1}(M)_n$$ and this proves the assertion.

Proposition. Let $z_1,...,z_s\in R_1$ be an $M$-filter regular sequence. Then $\max\{a_i(M)+i|\ i=0,\dots,s\}=\text{max}\lbrace a((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)|\ i=0,...,s\rbrace.$

Proof. By using Lemma successively we get $$a_i(M)+i\le a_0(M/(z_1,\dots,z_i)M)\le\max\{a_j(M)+j|\ j=0,\dots,i\}.$$ This implies that $$\max\{a_i(M)+i|\ i=0,\dots,t\}=\max\{a_0(M/(z_1,\dots,z_i)M)|\ i=0,\dots,t\}$$ for all $t\le s$.

Now we identify $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ with $$\bigcup_{n\ge 0}(z_1,...,z_i)M:R_{+}^n/(z_1,...,z_{i})M.$$ Set $a=a_0(M/(z_1,\dots,z_i)M)$. Then we have $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a\subseteq(z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M\subseteq H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ and therefore $a((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)=a$.

Edit. $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a\subseteq(z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M$ since $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ is a graded $R$-module and $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a$ is its top graded part, so $R_+H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a=0$.

$(z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M\subseteq H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ implies $((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)_j\subseteq H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_j$ for all $j$. For $j>a$ the right hand side is $0$, while for $j=a$ is not. This shows that $a((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)=a$.

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The result and its proof are coming from Trung's papers. – user26857 Oct 23 '12 at 11:06

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