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Let $R$ be a commutative Noetherian ring with unit and $I$ an unmixed ideal of $R$. Let $x\in R$ be an $R/I$-regular element. Can we conclude that $x+I$ is an unmixed ideal?

Background:

A proper ideal $I$ in a Noetherian ring $R$ is said to be unmixed if the heights of its prime divisors are all equal, i.e., $\operatorname{height} I=\operatorname{height}\mathfrak{p}$ for all $\mathfrak{p}\in \operatorname{Ass} I$.

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Stella
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1 Answers1

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Let $R$ be a noetherian integral domain and $I=(0)$. If $\dim R=2$ and $R$ is not Cohen-Macaulay, then there is $x\in R$, $x\ne 0$, such that $xR$ is not unmixed. (For more details look here.)

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user26857
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    Just to be explicit: if $R = k[s^4, s^3t, st^3, t^4]$, then $(s^3t)$ has height $1$, but the maximal ideal $\mathfrak{m} :=(s^4,s^3t,st^3,t^4)$ is an associated prime of $s^3t$ of height $2$ (notice $\mathfrak{m} \cdot (s^5t^3) \subseteq (s^3t)$) – zcn Mar 13 '14 at 01:08