Question

Let R be a commutative Q-algebra generated by x1, x2, . . . with the relations xixj = 0 for all i, j. Show that R is semiprimary (that is, rad R is nilpotent, and R/rad R is semisimple), but neither artinian nor noetherian.

Answer 1

Clearly, rad $R$ is the ideal generated by the $x_{i}$ 's. We have $(\mathrm{rad} R)^{2} = 0$ , and $R / \mathrm{rad} R \sim Q$ , so $R$ is semiprimary. The strictly ascending chain $(x_{1}) \subseteq (x_{1}, x_{2}) \subseteq (x_{1}, x_{2}, x_{3}) \subseteq \dots$ shows that $R$ is not noetherian, while the strictly descending chain $(x_{1}, x_{2}, \ldots) \subseteq (x_{2}, x_{3}, \ldots) \subseteq (x_{3}, x_{4}, \ldots) \subseteq \ldots$ shows that $R$ is not artinian.