Serial ring

Remarks

Example: let $k$ be a field and $R = k[x]/(x^n)$ with $n \ge 1$. Its ideals are the ideals $(x^i)/(x^n)$ for $0 \le i \le n$, and they form a chain. So $R$ is uniserial over itself on both sides, and $R$ is serial.

Non-example: let $R = k[x, y]/(x, y)^2$. This is a local ring, so $R$ has no idempotents other than $0$ and $1$, and $R$ is indecomposable as a module over itself. If it were serial, it would therefore be uniserial. But the ideals $(x)$ and $(y)$ of $R$ are not comparable, so $R$ is not serial.

Equivalently, $R$ is right serial when $R = e_1 R \oplus \cdots \oplus e_m R$ for some orthogonal idempotents $e_1, \dots, e_m$ with each $e_i R$ uniserial, since a direct summand of $R_R$ has the form $eR$ for an idempotent $e$. A finite-dimensional serial algebra over a field is also called a Nakayama algebra.