Let $(R, \mathfrak{m})$ be a regular local ring and let $\mathfrak{a} \subset R$ be an ideal. Let
$$ \mathfrak{b} = \bigcap \{R \cap \mathfrak{a} \cdot R\_\mathfrak{p} \text{ } \colon \mathfrak{p} \in \text{Ass}(R/\mathfrak{a}) \text{ and } \mathfrak{m} \neq \mathfrak{p} \}. $$


Is it true that $R/\mathfrak{b}$ is a Cohen-macaulay ring?



Cohen-Macaulay quotient ring and symbolic power
I don't think so. Take $R=k[[x,y,z,w]]$ and take $\mathfrak{a}=(x,y)\cap (z,w)$. Then $\operatorname{Ass}(R/\mathfrak{a})=\{(x,y), (z,w)\}$, and $R\cap \mathfrak{a}\cdot R\_P=P$, so $\mathfrak{b}=\mathfrak{a}$, but $R/\mathfrak{a}$ is not Cohen-Macaulay since its vanishing locus is two planes that meet at a point. Do you know this theorem: Let $R$ be a Noetherian ring and suppose that $\mathfrak{a}\subset R$ is an ideal with no embedded prime divisors. Then $$\mathfrak{a}=\bigcap\_{P\in\operatorname{Ass}(R/\mathfrak{a})}\left(R\cap \mathfrak{a}R\_P\right)$$
is a primary decomposition of $\mathfrak{a}$.


