MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $I$ ?

share|cite|improve this question
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
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

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.

share|cite|improve this answer
What is $end$ ? – Piotr Achinger Oct 9 '12 at 4:17
What do you mean by $\text{end}(H^{i}_{\mathfrak{m}}(M/\mathfrak{m}M)$ ? – Knot Oct 9 '12 at 4:19
@ 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

Your Answer


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.