Timeline for Modules over a Gorenstein ring
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 4, 2011 at 12:20 | vote | accept | ashpool | ||
| Sep 8, 2010 at 18:33 | comment | added | Hailong Dao | @kwan: it follows from the fact that M_p is maximal CM (assuming M is maximal CM). So if depth M_p=0, depth R_p must also be 0. | |
| Sep 8, 2010 at 17:40 | comment | added | ashpool | @Hailong: I couldn't find it in Bruns-Herzog but I found a nice proof in Eisenbud which also covers A-regularity implying M-regularity. Thanks! By the way, in your comment when you mentioned Ass(M) is a subset of Ass(A) (Aug 17 5:09), did you mean that the union of primes in Ass(M) is contained in the union of primes in Ass(A)? Your argument doesn't seem to go beyond that. | |
| Sep 7, 2010 at 19:45 | comment | added | Hailong Dao | @kwan: If $R$ is CM then maximal CM modules localize. This should be in Bruns-Herzog. | |
| Sep 7, 2010 at 19:39 | comment | added | ashpool | @Hailong: Sorry, I can't figure out why maximal Cohen-Macaulay modules localize. Is it a general fact or are you assuming $R$ to be Gorenstein? | |
| Sep 6, 2010 at 15:40 | comment | added | Hailong Dao | No, but if $M$ is maximal Cohen-Macaulay, then $M_p$ also maximal Cohen-Macaulay. | |
| Sep 5, 2010 at 23:36 | comment | added | ashpool | Sorry, I'm having hard time following almost everything you say. Why does depth(M_p)=0 imply depth(R_p)=0? Are you implying depth(M)=depth(R) localizes? | |
| Aug 17, 2010 at 5:09 | comment | added | Hailong Dao | @kwan: the set of associated primes of M is a subset of the set of associated primes of R (depth M_p=0 implies depth R_p=0). A R-reg element has to be outside all ass. primes of R, so also outside all ass. primes of M. | |
| Aug 16, 2010 at 21:28 | comment | added | ashpool | Could you give me a hint (or reference) for showing the A-regularity and M-regularity coincide when A and M have the same depth? | |
| Aug 16, 2010 at 16:47 | comment | added | Hailong Dao | @kwan: I actually meant depth(M) =depth(R), that why we need to pass to a high syzygy. | |
| Aug 16, 2010 at 16:20 | comment | added | ashpool | Just to make sure: by "full depth" do you mean depth(M)=dim(M) ? | |
| Aug 16, 2010 at 15:54 | comment | added | Hailong Dao | @kwan: If $M$ has full depth, then a regular element in $R$ is also regular on $M$. | |
| Aug 9, 2010 at 16:36 | comment | added | ashpool | Are you implying that when $R$ and $M$ have the same depth, there exists a sequence in $R$ that is both $R$-regular and $M$-regular? Or is the projective dimension of $M$ still not affected if I kill, say, a sequence that is $R$-regular but not $M$-regular? | |
| Aug 9, 2010 at 4:00 | comment | added | Hailong Dao | I was typing too fast: one can conclude the sequence splits from knowing N is injective by definition (-: | |
| Aug 8, 2010 at 5:34 | comment | added | Yemon Choi | +1 for the last paragraph. | |
| Aug 8, 2010 at 4:49 | history | answered | Hailong Dao | CC BY-SA 2.5 |