Let $(R,m)$ be a Cohen-Macaulay local ring which possesses the canonical module $K_R$. Then $R$ is said to be an almost Gorenstein local ring, if there is an exact sequence $0 \to R \to K_R \to C \to 0$ of $R$-modules such that $\mu (C) = e^0_m(C).$
The injection $0 \to R \to K_R$ does not always exist. But in some cases it does; A trivial example is Gorenstein rings, in which $ R \cong K_R.$ Here a question arise:

What conditions can be posed on ring so that there is an injection $0 \to R \to K_R$?

Thank you.

  • $\begingroup$ For any ring as above which is a domain, such an inclusion exists, for example. What kind of answer are you expecting? $\endgroup$ – Mohan Mar 16 '16 at 0:08
  • $\begingroup$ @Mohan if you post your comment (with more expression) it is an answer. I want conditions on ring that this happen. $\endgroup$ – user 1 Mar 17 '16 at 14:40

As soon as the dimension of $R$ is at least $1$ and $R$ is locally Gorenstein at the associated primes, then $K$ is isomorphic to an ideal of pure height $1$. In particular $K$ (as an ideal) contains a non-zerodivisor, so allows an injection $R \longrightarrow K$.

If $R$ is Artinian, then $K \cong E_R(k)$ is the injective hull of the residue field, and every element is annihilated by a power of the maximal ideal, so there is no such injection.

  • $\begingroup$ thank you. for "If R is Artinian" case; why " so there is no such injection" $\endgroup$ – user 1 Mar 17 '16 at 7:10
  • $\begingroup$ If $R$ is Artinian, length of $R$ and $K$ are equal and so injection implies isomorphism and so no such injection exists unless $R$ is Gorenstein. $\endgroup$ – Mohan Mar 17 '16 at 13:47
  • $\begingroup$ Or: existence of an injection $R \longrightarrow M$, for any module $M$, is equivalent to the existence of an element $x \in M$ with trivial annihilator. No elements of $K$ have trivial annihilator when $R$ is Artinian. $\endgroup$ – Graham Leuschke Mar 17 '16 at 14:22
  • $\begingroup$ so isnt it better the statement change to R is Artinian, and is not Gorenstein ? $\endgroup$ – user 1 Mar 17 '16 at 14:33
  • $\begingroup$ Well, yes. \ \ \ $\endgroup$ – Graham Leuschke Mar 17 '16 at 14:41

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