4. Examples 4.6.2

If $M$ is a finite $A$ -module, for example $Z / 6 Z$ as a $Z$ module, then $M$ is Noetherian and Artinian
$Z$ is noetherian but not Artinan (In particular, every 1;;principal ideal domain $A$ is noetherian and artinian iff $A$ is a field).
$k [x]$ is a PID and not a field, so it is Noetherian but not Artinian
Consider $k [x_{1}, x_{2}, \dots]$ in countably many variables. This ring is not Noetherian by construction. By what we will see next, since $k [x_{1}]$ is a quotient module of $k [x_{1}, x_{2}, \dots]$ which is not Artinian, the original ring is not Artinian.