Denote by $R = \mathbb{C}\{x\_1, \dots, x\_n\}$ the ring of germs of analytics maps at the origin in $n$ variables and let $f \in R$ such that $Sing(V(f))=V(x\_1, \dots, x\_{n-1})$ as sets. In addition, assume that $V(\partial\_{1}(f), \dots, \partial\_{n-1}(f))=V(x\_1, \dots,x\_{n-1})$ as sets as well. (By "as sets" we mean without considering multiplicity, only as reduced schemes).


We know that $R/\langle \partial\_{1}(f), \dots, \partial\_{n-1}(f) \rangle$ is a Cohen-Macaulay ring, as it is a quotient of a Noetherian local Cohen-Macaulay ring by a regular sequence (see "Pellikaan, R. (1989). Series of isolated singularities" for proof why this sequence is regular.)


But can we conclude that $R/\langle \partial\_{1}(f), \dots, \partial\_{n}(f) \rangle$ is Cohen-Macaulay as well?



Cohen-Macaulyness of Milnor algebra
I am just posting my comment as one answer. Using the local flatness criterion, the ring $R/\langle \partial\_1(f),\dots,\partial\_n(f)\rangle$ is Cohen-Macaulay if and only if it is flat as a module over $\mathbb{C}\{x\_n\}$, i.e., if and only if multiplication by $x\_n$ is injective on the ring. In the answer to a previous question, I wrote an example where multiplication by $x\_n$ is not injective on the ring, i.e., I wrote an example where the image of $x\_n$ is a zero divisor in the ring. Here is the link to the previous question: [Deformation of isolated singularities and non zero divisors](https://mathoverflow.net/questions/421209/deformation-of-isolated-singularities-and-non-zero-divisors)


