Let $(R, {\frak m})$ be a local ring and set $k=R/{\frak m}$. Over a Gorenstein local ring of dimension $n \geq 1$ we have $E_R(k)={\frak m} \, E_R(k)$. Is the relation true when $\dim R \geq 1$ and $R$ is not necessarily Gorenstein?
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Can you give a source of the Gorenstein case? – MooS Mar 05 '18 at 15:37
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If $R$ is a Gorenstein local ring then $H^n_{\frak m}(R) \cong E_R(k)$. On the other hand for any finitely generated $R-$module $M$ one has $H^n_{\frak m}(M) \cong H^n_{\frak m}(R) \otimes M$. Hence $0=H^n_{{\frak m}}(k) \cong H^n_{\frak m}(R) \otimes k \cong E_R(k) \otimes k.$ – Mohammad Bagheri Mar 05 '18 at 16:09
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Can't you do the same for any local noetherian ring? Instead of the local cohomology of $R$, take the total derived local cohomology of the dualizing complex. – MooS Mar 05 '18 at 17:02
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If $R$ is an integral domain, then $xE_R(k)=E_R(k)$ for any non-zero $x$, since $E_R(k)$ is divisible. So, I suspect it is true in general, though can't think of an argument right now. – Mohan Mar 06 '18 at 02:25
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@Mohan One can at least make this argument work if $R$ satisfies $S_1$, i.e. has a non-zero divisor $x$. Then $xM=M$ for any injective $R$-module $M$ and in particular $\mathfrak mM=M$. Thus it holds for positive dimensional Cohen-Macaulay rings. – MooS Mar 06 '18 at 08:30
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So one should investigate the case $R=k[x,y]/(xy,x^2)$. – MooS Mar 06 '18 at 08:34
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Thanks for all useful comments. – Mohammad Bagheri Mar 07 '18 at 10:57