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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:

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

Thank you.

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:

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

Thank you.

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.

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user 1
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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:

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

Thank you.