Let $(R,m)$ be a local ring and $E=E(R/m)$ be the injective hull of $R/m$.
Put $R_t=R/m^t$. What is the injective hull of residue field of $R_t$?
I guess it is $(0:_E m^t) =\{x \in E \;:\;xm^t=0\}$.
If it is correct then give the reason.
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2 Answers
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Modules over $R_t$ are the same as modules over $R$ which are annihilated by $m^t$. From this it is easy to see directly from the definition that if $M$ is an injective $R$-module, then $(0:_M m^t)$ is an injective $R_t$-module.
(In general, if $J$ is an ideal in $A$ and $M$ is an injective $A$-module, then $(0:_M J)$ is injective $A/J$-module.)
Thus $(0:_E m^t)$ is an injective $R/m^t$-module. Now you just have to check that its socle is equal to $R/m$. But again, its socle as an $R$-module and its socle as an $R_t$-module are the same, and by assumption its socle as an $R$-module is $R/m$.
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You're right. The reasons are somewhat too long to explain here, but you can take a look at this document from a Summer School on injective modules, especially
theorem 2.11 and corollary 2.12.
