I have a commutative ring $R$, and a prime ideal $P$ of $R$. I also have a module $E$ such that $R/P$ is a submodule of $E$ and every submodule $F\neq (0)$ of $E$ satisfies $F\cap R/P\neq (0)$. ($E$ is an essential extension of $R/P$).
What I need to show:
Given an element $r\notin P$, the map:
$f:E\to E$ defined by $f(x) = rx$ is an isomorphism.
When I try to show injectivity, I assume that $f(x) = P$. ($P$ is the $0$ element of $E_R(R/P)$.
Then $rx = P$. Now here is my issue: If $x\in R/P$, then I can show that $x = P$ since $r\notin P$. But if $x\notin R/P$, all I know is that it is in $E$. I think I need to get a contradiction somehow.
So if $x\neq 0$, I consider that $Rx\neq (0)$ is a submodule of $E$, and therefore there is a $q\neq 0$ in $R$ such that $qx\in R/P$. Since by assumption $rx = 0$, $r(qx) = q(rx) = q(0) = 0$.
So since $qx \in R/P$, I can write $qx = x_q + P$, but this is where I am stuck (assuming I am even on the right track). I was just blinding applying the hypothesis here and hoping things would come out the way I wanted.
Any tips on how I can show that $x= P$?