Skip to main content
Notice removed Draw attention by CommunityBot
Bounty Ended with Francesco's answer chosen by CommunityBot
Tweeted twitter.com/StackMath/status/1188334636692320256
added 98 characters in body
Source Link
user521337
  • 3.7k
  • 1
  • 10
  • 24

Let $(R, \mathfrak m)$ be a local Cohen-Macaulay ring. How to show that $R$ has minimal multiplicity (i.e. $e(R)=\mu (\mathfrak m)- \dim R +1$ ) if and only if $\mathfrak m^2=(\overline x)\mathfrak m$ for some regular sequence $\overline x$ in $R$ ?

If this is not true in general is it at least true if we assume $R/\mathfrak m$ is infinite ?

Let $(R, \mathfrak m)$ be a local Cohen-Macaulay ring. How to show that $R$ has minimal multiplicity (i.e. $e(R)=\mu (\mathfrak m)- \dim R +1$ ) if and only if $\mathfrak m^2=(\overline x)\mathfrak m$ for some regular sequence $\overline x$ in $R$ ?

Let $(R, \mathfrak m)$ be a local Cohen-Macaulay ring. How to show that $R$ has minimal multiplicity (i.e. $e(R)=\mu (\mathfrak m)- \dim R +1$ ) if and only if $\mathfrak m^2=(\overline x)\mathfrak m$ for some regular sequence $\overline x$ in $R$ ?

If this is not true in general is it at least true if we assume $R/\mathfrak m$ is infinite ?

Notice added Draw attention by user521337
Bounty Started worth 50 reputation by user521337
Source Link
user521337
  • 3.7k
  • 1
  • 10
  • 24

Local Cohen-Macaulay ring of minimal multiplicity

Let $(R, \mathfrak m)$ be a local Cohen-Macaulay ring. How to show that $R$ has minimal multiplicity (i.e. $e(R)=\mu (\mathfrak m)- \dim R +1$ ) if and only if $\mathfrak m^2=(\overline x)\mathfrak m$ for some regular sequence $\overline x$ in $R$ ?