TITLE: Cohen-Macaulay but not regular
QUESTION [7 upvotes]: 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!

REPLY [10 votes]: 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.