Quotient of ideal generated by regular sequence is a perfect module

Question

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.

Answer 1

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 $m$.