Let $R = \mathbb{R}[x_1 , \ldots , x_n]$ be a polynomial ring and $I \subset R$ be a principal ideal with $I = \langle f \rangle$. I know that $R$ is a CM ring. So my question is that:
The answer to both (1) and (2) is yes. More generally we have the following:
Proposition. If $R$ is any CM domain with principal ideal $I$, then $R/I$ is CM.
Proof. First recall the definition for (non-local) CM rings, from Stacks Project:
Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
and a fact from the Wikipedia page Cohen-Macaulay ring:
($\bullet$) For a non-zero-divisor $u$ in the maximal ideal of a Noetherian local ring $R$, $R$ is Cohen–Macaulay if and only if $R/(u)$ is Cohen–Macaulay.
For $\mathfrak p \in \operatorname{Spec}R$ we know $R_{\mathfrak p}$ is CM and $I_\mathfrak p$ is principal. By ($\bullet$), we find that for all $\mathfrak p \supset I$ the quotient $R_{\mathfrak p}/I_\mathfrak p$ is CM. Such $\mathfrak p$ correspond precisely to the prime ideals $\bar{\mathfrak p}$ of $R/I$. Finally since $(R/I)_{\bar{\mathfrak p}} = R_{\mathfrak p}/I_{\mathfrak p}$, the proposition follows. $\square$
