Let $R=k[x_1,\ldots,x_n]$ be a standard graded polynomial over field $k$ and $I$ an unmixed homogeneous ideal of $R$. Let $x\in R$ be an $R/I$-regular element. Can we conclude that $x+I$ is an unmixed ideal?
Height unmixed ideal and a non-zero divisor
Background:
A proper ideal $I$ in a Noetherian ring $R$ is said to be height unmixed if the heights of its prime divisors are all equal. i.e., $\operatorname{height} I=\operatorname{height}\mathfrak{p}$ for all $\mathfrak{p}\in \operatorname{Ass} I$.
No. For instance, take $R = k[x^4,x^3y, xy^3, y^4]$. This is a $2$-dimensional domain which is not Cohen-Macaulay at the origin. In fact the depth of $R$ at the origin is $1$. Since $R$ is a domain, the ideal $0$ is unmixed. Let $f$ be a homogeneous non zero element in $R$. Then $R/(f)$ is $1$-dimensional with depth $0$. Therefore, the ideal $(0)$ in $R/(f)$ is not unmixed.
The condition is equivalent to $R/I$ satisfies Serre's condition $S_2$.