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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 $\mathfrak{p}\in \operatorname{Ass} I$.

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Stella
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1 Answers1

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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 $S_2$.

Youngsu
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  • @user121097: Let $S = k[a,b,c,d]$ and $\phi:S \to R$. Let $I = \ker \phi$. – Youngsu Mar 26 '14 at 21:09
  • Why the condition is equivalent to $R/I$ satisfies Serre's condition $S_2$. – Stella Apr 08 '14 at 14:03
  • @Stella: Sure, let me try.

    Claim $I$ unmixed and $R/I$ satisfies $S_2$. Then for any $x$ which is a non zerodivisor on $R/I$, then $I,x$ is unmixed:

     – Youngsu Apr 09 '14 at 15:47
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    Since $R/I$ satisfies $S_2$, $R/(I,x)$ satisfies $S_1$. Therefore, it suffices to show that for $P \in Min((I,x))$, height $P$ = height $(I,x) = $ height $I + 1$. Consider the shortest irredundant primary decomposition of $I$, $Q_1 \cap Q_s$. Since $I$ is unmixed, height $Q_i =$ height $I$. We see that $I \subseteq P$ and $P$ prime. Therefore, $Q_i \subseteq P$ for some $i$. Let $p = \surd{Q_i}$. Since height $P/ (p,x) = 0$, height $P/p = 1$. Since $R_P$ is regular local domain, this implies height $P = $ height $p + 1$. Recall that height $p = $ height $I$. – Youngsu Apr 09 '14 at 15:48
  • @ Yongsu: In what circumstances $R/(I,x)$ satisfy $S_2$? If $R/I$ be Cohen-Macaulay then $R/(I+x)$ satisfies $S_2$? – Stella Apr 11 '14 at 08:58
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    @Stella: A Cohen-Macaulay (CM for short) ring satisfies $S_n$ for all $n$. If $R$ is a CM and $x$ a non zerodivisor on $R$, then $R/(x)$ is CM. I don't really know "what circumstances" should mean other than normal or CM. – Youngsu Apr 11 '14 at 17:06
  • Why $\operatorname{height}P/p=1$?please explain more. – Stella Jun 21 '14 at 05:31
  • @Stella: Consider the ring $R_P/p_P$. In this ring $(x)_P$ is primary to the maximal ideal $P_P$. Hence $\dim R_p/p_P = 1$. – Youngsu Jun 24 '14 at 18:56