Let $(R,\mathfrak m)$ be a local Cohen-Macaulay ring. Let $I\subseteq \mathfrak m$ be an ideal of $R$. If $R/I \cong S/J$ for some regular local ring $S$ and ideal $J$ of $S$ such that $ht(J)=\mu(J)$, then is it true that $ht(I)=\mu(I)$ ?
2026-03-25 06:07:12.1774418832
When $R/I \cong S/J$, where $R$ is Cohen-Macaulay, $S$ is regular local and $ht(J)=\mu(J)$
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No. Try $R=k[[x,y,z]]/(x^2-yz)$ and $I=(x,y)$.