$\mathrm{Hom}_R(N,M)$ is an essential extension of $\mathrm{Hom}_R(N,L)$ $M$ is an $R$-module and an essential extension of $L$. $N$ is a finitely generated submodule of $M$. Then $\mathrm{Hom}_R(N,M)$ is an essential extension of $\mathrm{Hom}_R(N,L)$.


I tried to use the definition of essential extension: intersection of every submodule of $\mathrm{Hom}_R(N,M)$ with $\mathrm{Hom}_R (N,L)$ is not trivial.
But I can not. Can you help me?
$L\subset M$ is essential, and $N\subset M$ is finitely generated. Then $\operatorname{Hom}(N,L)\to\operatorname{Hom}(N,M)$ is essential.


Let $f\in\operatorname{Hom}(N,M)$, $f\ne0$.   

Since $N$ is finitely generated and $L$ is essential in $M$ there is $a\in R$ such that $af(N)\subseteq L$ and $af(N)\ne0$. (This can be shown as follows: set $N=Rx_1+\cdots+Rx_n$, and suppose that $f(x_i)\ne 0$ for all $i$. Then there is $a_1\in R$ such that $a_1f(x_1)=y_1\in L\setminus\{0\}$. If $a_1f(x_i)=0$ for all $i\ge2$ then set $a=a_1$. Suppose $a_1f(x_2)\ne0$. Then there is $a_2\in R$ such that $a_2a_1f(x_2)=y_2\in L\setminus\{0\}$. If $a_1a_2f(x_i)=0$ for all $i\ge3$ then set $a=a_1a_2$, and so on.)   

Thus $af\ne0$, $af\in\operatorname{Hom}(N,L)$, and we are done.
this is false SINCE for example hom(M, N) has Z module structure and not R module structure.