Recall that a right module $M\_R$ is called **semiartinian** if every nonzero homomorphic image has nonzero socle. It's well known that the following two statements are equivalent:


1. $M$ is semiartinian.
2. Every nonzero homomorphic image of $M$ has essential socle.


It's clear that $(2)$ implies (1). But how can I prove the converse?



A characterization of semiartinian modules
Argue by contradiction.  
 Assume that $M$ is semiartinian and that $N$ is a nonzero homomorphic image of $M$ whose socle $A\_0$ is not essential. There will exist a nonzero submodule $B\leq N$ that is disjoint from $A\_0$. Enlarge the submodule $A\_0$ to a submodule $A\_1$ that is maximal for the condition $A\_1\cap B = 0$. The module $N/A\_1$ is nonzero, since
$$0\neq B\cong B/(A\_1\cap B)\cong (A\_1+B)/A\_1\leq N/A\_1.$$
$N/A\_1$ is a quotient of $N$, which is isomorphic to a quotient of $M$, so $N/A\_1$ is isomorphic to a quotient of $M$. Since $M$ is semiartinian, the nonzero quotient $N/A\_1$ must have a nonzero simple submodule. This means that $N$ has a submodule $A\_2$ covering $A\_1$ (i.e., $A\_1\lneq A\_2$, and there is no intermediate submodule $A\_1\lneq C\lneq A\_2$). By the maximality condition on $A\_1$ we have $A\_2\cap B\neq 0$ and by the modularity $\textrm{Sub}(N)$ (= the submodule lattice of $N$) we even get that $S:=A\_2\cap B$ covers $0$. Hence $S$ is a simple submodule of $N$ contained in $B$. All simple submodules of $N$ are contained in the socle $A\_0$ which is contained in $A\_1$. We get that $S\subseteq A\_1\cap B=0$. This is a contradiction, since $S\neq 0$.


