Cohen-Macaulay but not regular

Question

In the Wiki page it is claimed that

$K[[t^2,t^3]]$ is a $1$-dimensional Cohen-Macaulay ring which is not regular. Is there anybody who kindly explain to me the above assertion?

Thanks in advance!

Answer 1

To see that $K[[t^2, t^3]]$ is Cohen-Macaulay, we will show the stronger statement that it is Gorenstein. As $K[[t^2,t^3]]$ is a local ring, this is equivalent to showing that the polynomial ring $K[t^2,t^3]$ is Gorenstein (since a noetherian commutative local ring is Gorenstein iff its completion is). But $K[t^2,t^3] \simeq K[x,y]/(x^3-y^2)$ is a plane curve, hence Gorenstein.

However, if $K[[t^2,t^3]]$ were regular, then since it is a local ring, the Auslander-Buchsbaum theorem implies that it is a UFD. In particular, it is integrally closed. But the integral closure of $K[[t^2,t^3]]$ is $K[[t]]$, a contradiction.

Answer 2

The ring extension $K[[X^2,X^3]]\subset K[[X]]$ is integral, so $\dim K[[X^2,X^3]]=1$ and hence it is CM. Moreover, $K[[X^2,X^3]]$ is local, the maximal ideal being $(X^2,X^3)$. If $K[[X^2,X^3]]$ is regular, then its maximal ideal must be principal. Suppose $(X^2,X^3)=(f)$, $f\in K[[X^2,X^3]]$. Then $X^2=fg$, so $f\mid X^2$. If look in $K[[X]]$ get $f=1$, $f=X$, or $f=X^2$ (eventually multiplied by a constant). The first case is not possible, while in the second $f\notin K[[X^2,X^3]]$. It remains $f=X^2$ and then $X^3=X^2g$ for some $g\in K[[X^2,X^3]]$, a contradiction.

Now use a well known theorem which says that a local noetherian ring is regular iff it has finite global dimension.