# Two dimensional normal rings are Cohen-Macaulay

Let $R$ be a Noetherian normal ring of Krull dimension two, I would like to show $R$ is Cohen-Macaulay without referring to Serre's normality criterion. There is a similar question here but it is not what I want.

If we localize away from a prime $\mathfrak{p}$, and the resulting dimension is zero or one, we are done. If $\dim(R_\mathfrak{p}) = 2$, my idea is to produce two elements from $\mathfrak{p}R_\mathfrak{p}$ that can be a part of a regular sequence for $R_\mathfrak{p}$ and this will do the trick. The first element $0 \ne x$ can be chosen easily since $R_\mathfrak{p}$ is a domain, but I am stuck on the second piece. This amounts to being able to pick a non-zero divisor from some minimal prime over $xR_\mathfrak{p}$ (I think)... This is as far as I got, and I would appreciate some hints or corrections if I made a mistake.

Here is just the argument of Serre, without mentioning Serre criterion. You have reduced the situation to $R$ local with maximal ideal say $P$. If depth of $R/xR$ for the non-zero $x$ you chose is one, you are done. So, assume the depth is zero. This means, there exists a non-zero element in $R/xR$ annihilated by $P$, which in $R$ means, there exists $y\not\in xR$ and $Py\subset xR$. Try to show that $y/x$ is integral over $R$ and thus $y\in xR$, since $R$ is integrally closed, to reach the desired contradiction.