# An indecomposable module $M$ satisfying $\text{soc}(M) \neq 0$, can we conclude that $\text{soc}(M) \hookrightarrow M$ is an essential extension?

Let $$R$$ be a ring with unity and $$M$$ be an indecomposable module over $$R$$. If the socle of $$M$$, $$soc(M) \neq 0$$, then can we conclude that $$\text{soc}(M) \hookrightarrow M$$ is an essential extension?

If $$M$$ is not indecomposable, there is a counterexample. Take $$R = \mathbb{Z}$$ and $$M = \mathbb{Q} \oplus \mathbb{Z}/\mathbb{2Z}$$, then $$soc(M) = \mathbb{Z}/\mathbb{2Z} \neq 0$$, but $$\mathbb{Z}/\mathbb{2Z} \hookrightarrow \mathbb{Q} \oplus \mathbb{Z}/\mathbb{2Z}$$ is not an essential extension; i.e., $$\text{soc}(M) \hookrightarrow M$$ is not an essential extension.

Let $$R=M=k[x,y]/(x^2,xy)$$, where $$k$$ is a field. So $$R$$ has a basis $$1,x,y,y^2,y^3,\dots$$.
Then $$\operatorname{soc}(R)$$ is one-dimensional, spanned by $$x$$, and $$\operatorname{soc}(R)\cap Ry=0$$. So $$\operatorname{soc}(R)$$ is not an essential submodule of $$R$$.