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Let $R$ be a polynomial ring over a field $k$,: $k[x_{1},..x_{n}]$, $\mathfrak{m}=(x_0,...,x_{n})$ and $M$ be a finitely generated $R$ module.

In a paper of Kodiyahlam, he define the Castelnuovo-Mumford regularity of $M$ to be the least integer number $m$ such that for every $j$ the $j$-th syzygy of $M$ is generated in degree less or equal $m+j$.Then, he conclude that the Castelnuovo-Mumford regularity of $M/\mathfrak{m}M$ is equal to the maximal degree of generator of $M$.

Could you please so me why the CM regularity of $M/\mathfrak{m}M$ is equal to the maximal degree of generator of $M$ ?

If $I$ and $J$ are two finitely generated ideal of $R$ and $J\subseteq I$ then do we have the CM regularity of $J$ is less or equal the CM regularity of $I$ ?

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The second question has answer "no." –  Charles Staats Oct 8 '12 at 13:35
    
@Charles Staats : Could you please post a counter-example here ? –  Knot Oct 8 '12 at 15:20
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Try resolving $(x,y)$ and $(x^2,y^2)$ over $R=k[x,y]$. The intuition should in fact be the exact opposite: ideals that are "deeper" in the ring should have larger (i.e. worse) regularity. –  Graham Leuschke Oct 8 '12 at 16:32

1 Answer 1

Notice that $M/\mathfrak{m}M$ is an $R$-module of finite length, so $H^i_{\mathfrak{m} }(M/\mathfrak{m}M) = 0$ for all $i>0$ and $H^0_{\mathfrak{m} }(M/\mathfrak{m}M) = M/\mathfrak{m}M$. Recalling that CM regularity can be computed via local cohomology module (see Brodmann-Sharp: local cohomology), we have $$reg (M/\mathfrak{m}M) = \max \{ end (H^i_{\mathfrak{m} }(M/\mathfrak{m}M)) +i | i= 0,...,n \}.$$ Here, consider a graded $R$-module $N = \oplus_iN_i$, we denote $end(N) = \sup \{i | N_i \neq 0\}$. Therefore $reg (M/\mathfrak{m}M) = end ((M/\mathfrak{m}M))$. $end ((M/\mathfrak{m}M))$ is equal to the maximal degree of generator of $M$ by graded Nakayama lemma.

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What is $end$ ? –  Piotr Achinger Oct 9 '12 at 4:17
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What do you mean by $\text{end}(H^{i}_{\mathfrak{m}}(M/\mathfrak{m}M)$ ? –  Knot Oct 9 '12 at 4:19
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@ Knot: you said that you read a paper of Kodiyahlam about CM regularity. You should study basic fact of CM regularity before reading this paper. –  Pham Hung Quy Oct 9 '12 at 5:05
    
Thanks for including the definition! –  Piotr Achinger Oct 9 '12 at 5:07

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