# $2$-dimensional Noetherian integrally closed domains are Cohen-Macaulay

Any 1-dimensional Noetherian domain is Cohen-Macaulay (C-M).

For the $2$-dimensional case, a condition of being integrally closed is necessary to be added for a Noetherian domain to be C-M, which I could not prove it.

Would anybody be so kind as to solve this?

I also search for a non C-M $2$-dimensional Noetherian domain which is not integrally closed.

Thanks for any cooperation!

#### Solutions Collecting From Web of "$2$-dimensional Noetherian integrally closed domains are Cohen-Macaulay"

Let $R$ be a noetherian integral domain of dimension $2$. If $R$ is integrally closed, then $R$ is Cohen-Macaulay.

From Serre’s normality criterion we have that $R$ satisfies $(R_1)$ and $(S_2)$.
$(R_1)$ gives that all the localizations of $R$ at height one primes are regular, and therefore Cohen-Macaulay. (In fact, we don’t need to use $(R_1)$ in order to prove that $R_{\mathfrak p}$ is Cohen-Macaulay for prime ideals $\mathfrak p$ of height one.)
Now let $\mathfrak p$ be a height two prime ideal of $R$. From $(S_2)$ we get that $\operatorname{depth}R_{\mathfrak p}\ge2=\dim R_{\mathfrak p}$, so $R_{\mathfrak p}$ is Cohen-Macaulay.

$k[x^4,x^3y,xy^3,y^4]$ is $2$-dimensional, not Cohen-Macaulay and not integrally closed.