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A well-known theorem in commutative algebra states the fact that if $R$ is a Cohen-Macaulay ring, and $a_1,...,a_r$ is an $R$-sequence, then $R/I$ is Cohen-Macaulay, where $I=(a_1,...,a_r)$.

Now,

Is it true that for any positive integer $n$ the ring $R/I^n$ is Cohen-Macaulay?

I know, somehow, that we should resort to induction and use the short exact sequence $0\to I^n/I^{n+1} \to R/I^{n+1}\to R/I^n\to 0$ but I could not continue. Is there an elementary proof (or counterexample) for that?

Thanks for cooperation!

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    $\begingroup$ If $I$ is an ideal generated by a regular sequence, then $I^m/I^{m+1}$ is isomorphic to $S^m(I/I^2)$ and in particular, a free module over $R/I$. Then the result follows. $\endgroup$ – Mohan Nov 14 '15 at 5:30
  • $\begingroup$ what is $S^m(I/I^2)$? $\endgroup$ – user 1 Nov 14 '15 at 7:01
  • $\begingroup$ @Mohan Thanks for the suggestion! Could you please unravel your answer more to me? $\endgroup$ – karparvar Nov 14 '15 at 16:01
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Suppose $(R,\mathfrak m)$ is local. We want to show that $\operatorname{depth}R/I^n=\operatorname{depth}R/I$. By Theorem 1.18 from Bruns and Herzog we get $(R/I)[X_1,\dots,X_r]\simeq\operatorname{gr}_I(R)$. It follows that $I^{k}/I^{k+1}\simeq(R/I)^{m_k}$ for all $k\ge0$. Now one can show inductively that $\operatorname{Ext}_R^i(R/\mathfrak m,R/I^n)=0$ for $i<\operatorname{depth}R/I$, and $\operatorname{Ext}_R^i(R/\mathfrak m,R/I^n)\ne0$ for $i=\operatorname{depth}R/I$.

Since $\dim R/I^n=\dim R/I$ we are done.

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