# A detail in the proof of Auslander-Buchsbaum Theorem

I'm trying to understand the proof of a theorem (Auslander-Buchsbaum) which says that given a local ring $(R,m)$, where $m$ is the maximal ideal, and a finitely generated non-zero $R$-module $M$ such that its projective dimension $\operatorname{pd}_R M$ is finite then the following formula holds: $$\operatorname{pd}_R M + \operatorname{depth} M= \operatorname{depth} R.$$

I remind you that the quantity $\operatorname{depth} M$ in general is defined to be the length of a maximal regular sequence $\bar{x}=x_1,\dots,x_n$ on $M$ in $m$ (i.e. $\bar{x}$ is the maximal sequence of elements in $m$ such that (1) $M/(\bar{x})M\neq 0$ and (2) $x_j$ is a non-zero divisor in $M/(x_1,\dots,x_{j-1})M$).

In the proof of this theorem (proven by induction) it is used the fact that the projective dimension $\operatorname{pd}_R M$ is equal to $\operatorname{pd}_{R/xR}M/xM$, where $x$ is not a zerodivisor on $M$ nor on $R$, but I cannot understand why this should be true... Can you help me?

You can again use induction to prove this fact. Base $\operatorname{pd}_R M=0$ is obvious. If $\operatorname{pd}_R M=n>0$, let $F$ be a finitely generated module s.t. we have an epimorphism $F \to M$, denote kernel of this map by $K$, so we have a short exact sequence $$0 \to K \to F \to M \to 0.$$ But $\operatorname{Tor}_1^R(M, R/xR)=\{a \in M : xa=0\}=0$, thus the sequence over $R/xR$ $$0 \to K/xK \to F/xF \to M/xM \to 0$$ is exact.
$\operatorname{pd}_R K=n-1$, and by induction $\operatorname{pd}_{R/xR} {K/xK}=n-1$, module $F/xF$ is free over $R/xR$ thus $\operatorname{pd}_{R/xR} {M/xM}=n$.
• For using the inductionhypothesis $x$ should not be a zerodivisor of $K$. Why is this true? – user501184 Feb 12 '18 at 13:50