Another non-regular C-M ring

Question

Among other examples of Cohen-macaulay rings which are not regular I am run into $R=F[x]/(x^2)$ with $F$ a field. It is clear that it is not regular, since every regular local ring must be a domain while $x+(x^2)$ is a zero-divisor. My question is that why $R$ is a local Artinian ring? My try is that $(x)+(x^2)$ is a maximal ideal of $R$. I am aware also of the fact that each Artinian ring is Cohen-Macaulay. Thanks in advance!

Answer 1

$F[x]$ is a principal ideal domain, so any ideal above $(x^2)$ must be generated by a divisor of $x^2$. From this you should see the ring has exactly three ideals.

This should also prime you to see why $F[x]/(x^n)$ is local for any positive integer n.

It's Artinian since it's two dimensional over $F$.