Hello All,is This conclusion true? if $(R,m)$ be a local ring & $ Min Ass R=Ass R$ then we Can conclude that $Min Ass \hat{R}=Ass \hat{R}$. ( $\hat{R}$ is $m$-adic completion of $R$) $MinAss$ means minimal primes in $Ass(R)$. " $Min Ass R = Ass R$ " means R has no embedded prime ideals.in fact, if every associated prime ideal of $R$ is minimal then every associated prime ideal of $\hat{R}$ is minimal.

Hello All,is This conclusion true?

If $(R,m)$ is a local ring and $ Min Ass R=Ass R$ then can we conclude that $Min Ass \hat{R}=Ass \hat{R}$? ($\hat{R}$ is $m$-adic completion of $R$)

$MinAss$ means minimal primes in $Ass(R)$. "$Min Ass R = Ass R$" means that $R$ has no embedded prime ideals. In fact, if every associated prime ideal of $R$ is minimal then every associated prime ideal of $\hat{R}$ is minimal?

Hello All,is This conclusion true? if $(R,m)$ be a local ring & $ Min Ass R=Ass R$ then we Can conclude that $Min Ass \hat{R}=Ass \hat{R}$. ( $\hat{R}$ is $m$-adic completion of $R$) $MinAss$ means minimal primes in $Ass(R)$. " $Min Ass R = Ass R$ " means R has no embedded prime ideals

Hello All,is This conclusion true? if $(R,m)$ be a local ring & $ Min Ass R=Ass R$ then we Can conclude that $Min Ass \hat{R}=Ass \hat{R}$. ( $\hat{R}$ is $m$-adic completion of $R$) $MinAss$ means minimal primes in $Ass(R)$. " $Min Ass R = Ass R$ " means R has no embedded prime ideals.in fact, if every associated prime ideal of $R$ is minimal then every associated prime ideal of $\hat{R}$ is minimal.

Hello All,is This conclusion true? if $(R,m)$ be a local ring & $ Min Ass R=Ass R$ then we Can conclude that $Min Ass \hat{R}=Ass \hat{R}$. ( $\hat{R}$ is $m$-adic completion of $R$)

Hello All,is This conclusion true? if $(R,m)$ be a local ring & $ Min Ass R=Ass R$ then we Can conclude that $Min Ass \hat{R}=Ass \hat{R}$. ( $\hat{R}$ is $m$-adic completion of $R$) $MinAss$ means minimal primes in $Ass(R)$. " $Min Ass R = Ass R$ " means R has no embedded prime ideals