Let R be a regular local ring with $ \text{dim} R = d $. If $ 0\rightarrow R\rightarrow I_0\rightarrow ...\rightarrow I_d\rightarrow 0 $. Then why for $ 0\leq i\leq d-1 $, the socle of $ I_i $ is zero? And why this is not true in $I_d$?
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$\begingroup$ What are the $I$'s? I assume your sequence above is exact. With no conditions, this is not true. For example take $d=1$, with $0\to R\to I_0\to k$ split exact where $k$ is the residue field. $\endgroup$– MohanCommented Apr 21, 2013 at 14:30
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$\begingroup$ Mohan, I think the question is about an injective resolution. Anyway, voted to close as this is not a real question. $\endgroup$– Hailong DaoCommented Apr 22, 2013 at 3:26
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$\begingroup$ You are right. this is a minimal injective resolution. $\endgroup$– MaxCommented Apr 22, 2013 at 12:01
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I believe that you meant to say that the Bass number with respect to the maximal ideal is $0$ if $i \neq d$ and 1 if $i = d$. This is a consequence of the Gorensteiness of the ring $R$. I am pretty sure Matsumura's commutaive ring thoery book contains this result. In fact the vanishing part comes from the Cohen-Macauayness of the ring $R$. And $\dim_{R/m} Ext^d_R (R/m, R) = 1$ makes the ring of (Cohen-Macaulay) type $1$. This also is well known, a local ring having a canonical module is Gorenstein if and only if it is a Cohen-Macaulay ring of type $1$.
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$\begingroup$ Youngsu, if depth M is $>0$ then the socle is $0$. $\endgroup$ Commented Apr 22, 2013 at 3:27
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$\begingroup$ Yes, you are absolutely right. I meant to say Cohen-Macaulay type. Let me edit this. $\endgroup$– YoungsuCommented Apr 22, 2013 at 3:51