Zero dimensional local ring with maximal ideal not principal.

Question

Probably it is well known. I am looking for a zero dimensional local ring with maximal ideal not principal.

Answer 1

Consider the quotient ring $$ R=\frac{K[X,Y]}{(X,Y)^2} $$ where $K$ is a field. It is zero-dimensional because there's a bijection between ideals of $R$ and ideals of $K[X,Y]$ containing $(X,Y)^2$ under which prime ideals correspond to prime ideals, and $(X,Y)$ is the only prime ideal containing $(X,Y)^2$.

As a by-product of this argument, we also see that $R$ is local because it has just one prime ideal, thus maximal, namely ${\frak m}=(\bar X,\bar Y)$.

Finally, if ${\frak m}$ were principal, since $\bar X$ and $\bar Y$ cannot be a common multiple of some non-invertible element of lower degree (there are none!) we would have $$ X-Y\in(X,Y)^2 $$ But this is impossible, because $(X,Y)^2$ contains no polynomial of degree 1.

Answer 2

Another nice example is $\mathcal O_{\mathbf C_p}/p\mathcal O_{\mathbf C_p}$, where $\mathcal O_{\mathbf C_p}$ is the ring of integers of the completion $\mathbf C_p$ of $\overline{\mathbf Q_p}$. It has dimension $0$, but its maximal ideal is not even finitely generated.