when there is an injection $0 \to R \to K_R$?

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.

• For any ring as above which is a domain, such an inclusion exists, for example. What kind of answer are you expecting? – Mohan Mar 16 '16 at 0:08
• @Mohan if you post your comment (with more expression) it is an answer. I want conditions on ring that this happen. – 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.
• 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. – Mohan Mar 17 '16 at 13:47
• 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. – Graham Leuschke Mar 17 '16 at 14:22