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Let $(R,m)$ be a regular local ring having an ideal $I$ such that $I$ is a subset of $m^2$. If $I$ possesses a non-zerodivisor, I want to show that $R/I$ can not be regular.

My try is just that $m$ could be generated by an $R$- sequence with length $d=ht(m)$, also if $a$ is the non-zerodivisor then $gr(m)=gr(m/(a))+1$. Thanks!

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1 Answer 1

up vote 1 down vote accepted

I have to repeat what I said in a previous answer:

From Bruns and Herzog, Proposition 2.2.4, we learn that if $R$ is a regular local ring, then $R/I$ is local regular if and only if $I$ is generated by a subset of a regular system of parameters.

But a regular system of parameters can't contained in $\mathfrak m^2$ since it is a basis of $\mathfrak m/\mathfrak m^2$ over $R/\mathfrak m$.

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