# Canonical module

Let $(R,m)$ be a Cohen-Macaulay ring and $M$ a maximal Cohen-Macaulay $R$-module.

Assume there exists canonical module $K_R$. Set $L:=\{f(x)|x\in M, f\in Hom_R(M,K_R)\}$. Then $M$ is $R$-module.

I want to find some conditions are equivalent to (or implies) the following condition $$L\subseteq mK_R$$

• By the way, Firststep, did you mean to write in the third row "Then $L$ is an $R$-module."? (I mean not $M$). Sep 4 '16 at 8:02

It seems to me that what you would like can be rephrased the following way:

Consider the short exact sequence $$0\to mK_R \to K_R \to K_R/mK_R \to 0,$$ and apply the functor $Hom_R(M,\_\_)$ to obtain the exact sequence $$0\to Hom_R(M,mK_R) \to Hom_R(M,K_R) \to Hom_R(M,K_R/mK_R) \to\\ \to Ext^1_R(M,mK_R) \to Ext^1_R(M,K_R)\to Ext^1_R(M,K_R/mK_R)\to \dots$$

Now, what you are asking for is that the first non-trivial morphism in this sequence, $$Hom_R(M,mK_R) \to Hom_R(M,K_R)$$ is surjective (i.e., an isomorphism).

So, the actual condition is that the next morphism is zero $$Hom_R(M,K_R) \to Hom_R(M,K_R/mK_R),$$ in other words that no morphism $M\to K_R/mK_R$ can be lifted to a morphism $M\to K_R$.

An easy way this could happen would be if the latter group were zero, i.e., if $Hom_R(M,K_R/mK_R)=0$, but I don't think this can happen; $K_R/mK_R$ is a $k=R/m$-vector space. So is $M/mM$ and by Nakayama's lemma both of these are non-trivial and hence there are non-trivial morphisms between them.

Another, somewhat more drastic way this happens is if the former group is zero, i.e., if $Hom_R(M,K_R)=0$. This actually can happen, but it probably defeats the point of your question as in that case also $L=0$ and the condition holds trivially.

I know this is of little help, but my feeling is that this shows that it is unlikely that what you are asking for happens in a natural way under non-trivial circumstances.

• Thank you so much. Could you check for me the second morphism is zero in case $R=k[[X,Y]]/(XY)$ and $M=(X)/(XY)$. The trivial case is also good for me.
– TNAn
Sep 3 '16 at 2:28
• I guess, this is covered by curious's answer, so it's OK. Sep 3 '16 at 17:20

The case when $R$ is Gorenstein is simple: $K_R=R$, and $L\subseteq m$ iff there is no surjection $M\to R$ iff $M$ has no summand isomorphic to R. In the example you are interested in, mentioned in your comment to Sándor's answer, $M$ is indecomposable and non-free, so it holds.

• Indecomposable + non-free implies that $M$ has no summand isomorphic to $R$. $M$ (in the case mentioned by Firststep) is a principlal ideal with non-zero annihilator, so not free. I don't understand your second and third questions. Sep 3 '16 at 23:38
• The statement is that $L\subseteq m$ iff $R$ is not a summand of $M$. If $M$ is indecomposable, this is equivalent to $M$ is not free. As for the specific example, $M =(x)$, whose annihilator is $(y)$, non-zero. Sep 4 '16 at 18:16
• I had no intention to treat you as an idiot. I honestly did not understand what the point of confusion was. Now you deleted your questions, it is hard for me to explain. Anyhow, since you did comment to Firststep that his example was covered by my answer, I assumed that you knew I was talking about that example. Sorry for any misunderstanding. Sep 4 '16 at 23:43
• OK, no sweat, I didn't think you did it knowingly, but it was frustrating that you just kept repeating the clear part. I probably overreacted. Sorry about that. I read your answer first, and wrote my first comment to you. I only read Firststep's comment later and I saw that that was covered by your answer, but I did not realize that you were talking about that specific example, because in my timeline they were reversed. I thought you were referring to the original question. (Although if you saw my comment, why did you think I did not understand your argument?). Anyway, no hard feelings. Sep 5 '16 at 0:18