Let $R$ be a commutative Noetherian ring and $I$ be an ideal of $R$. If $p$ is a prime ideal of $R$, is the following statement true:
$$Hom_R(R/I,E_R(R/p))= \begin{cases} E_{R/I}(R/p)~~ if \hspace{.4cm} I\subseteq p\\ 0 \hspace{1.7cm} ~~ if \hspace{.4cm} I \nsubseteq p. \end{cases} $$
It is known that $$Hom_R(R/I,E_R(R/p)) \cong (0 :_{E(R/p)} R/I).$$ However, it is not immediately clear how to show that $(0 :_{E(R/p)} R/I) \cong E_{R/I}(R/p)$. Could you please provide some assistance in proving this statement?
Thank you for your help.
This is straight forward. I assume that you know how to deal with the case $I$ not contained in $p$.
So assume that $I\subset p$. Denote by $J$ the set of all elements in $E(R/p)$ annihilated by $I$. Easy to see that if $J$ is injective over $R/I$, then it is the injective hull of $R/p$ as an $R/I$ module.
So, let $0\to A\stackrel{f}{\to} B$ be an exact sequence of $R/I$ modules and let $g:A\to J$ be a homomorphism. Composing with the inclusion $J\subset E(R/p)$ and using the fact that $E(R/p)$ is injective over $R$, we can extend $f$ to an $R$- module homomorphism $B\to E(R/p)$. But, since $I$ annihilates $B$, the image of $B$ under this extension is contained in $J$ and so we are done.