# Quotient of ideal generated by regular sequence is a perfect module

I have recently started reading Bruns-Herzog's 'Cohen Macaulay rings' and this is problem 1.4.27 in it.

We say that a module $$M$$ over a Noetherian ring $$R$$ is perfect if the projective dimension of $$M$$ is equal to the grade of $$M$$.

I am required to show that if an ideal $$I$$ is generated by a regular sequence of length $$n$$ in $$R$$, then $$R/I^m$$ is perfect for all natural numbers $$m$$.

I could see that the projective dimension of $$R/I$$ must be $$n$$ and it looks like I should somehow be able to use the Auslander-Buchsbaum formula but I have no clue how to proceed beyond this. Bruns-Herzog mention that the following result (Theorem 1.1.8) will be useful but I am not sure why:

Let $$R$$ be a ring, $$M$$ an $$R$$-module and suppose $$I$$ is an ideal generated by an $$M$$ regular sequence $$a_1, \cdots ,a_n$$ . Then the map $$(M/IM)[X_1,....,X_n] \rightarrow gr_I(M): X_j \mapsto \overline a_j \in I/I^2$$ must be an isomorphism, where $$gr_I(M) := \oplus_{r \geq 0} I^rM/I^{r+1}M$$

I would really appreciate any hint or solution.

By Theorem 1.1.8, $$I^m/I^{m+1}$$ is isomorphic as an $$R/I$$-module to the $$m$$th graded component of $$R/I[X_1, \ldots, X_n]$$, so $$I^m/I^{m+1}$$ is a free $$R/I$$-module of finite rank. Now consider the exact sequence $$0 \to I^m/I^{m+1} \to R/I^{m+1} \to R/I^m \to 0$$ of $$R$$-modules. Now apply induction on $$n$$ to show that $$R/I$$ is perfect. Then apply induction on $$m$$ to show that the projective dimension on $$R/I^m$$ (as an $$R$$-module) is $$n$$ for all $$m$$. Note that $$\mathrm{grade}(I^m) = \mathrm{grade}(I)$$ for all $$m$$.