Is it true that a height one prime ideal in a Cohen-Macaulay integral domain $R$ is principal? Is the corresponding quotient domain Cohen-Macaulay?
My think is that the grade of the prime ideal is also one, and according to being domain, each nonzero element of the ring is a non-zero-divizor whence forming a regular sequence the length of such maximal sequence is one. Now, for each nonzero $x$ and $y$ in the prime ideal, say $P$, $y$ is not a zero divisor in $R/yR$ .... .
Thanks for help!
All one-dimensional domains are CM. Now just take one which is not a PID, like $\mathbb Z[\sqrt{-5}]$. For instance, $P=(2,1+\sqrt{-5})$ is a non-principal prime of height one. In this case the quotient is a field, so it is CM.
Let $R=K[X_1,\dots,X_n]$, $n\ge 4$. It is known that there are prime ideals $P$ of height two in $R$ such that $R/P$ is not CM; see Bruns and Herzog, exercise 2.1.18(b). Let $f\in P$ be an irreducible polynomial. Then $R/(f)$ is a CM integral domain, $P/(f)$ has height one, it's not principal (why?), and $R/P$ is not CM.
