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

Question

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.

Answer 1

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 $R$.