auslander buchsbaum base case free module

Projective dimension zero means that $M$ is projective. The finite-dimensional residue space $M\mathrm{/}\mathfrak{m}M$ has a basis of, say, $r$ elements; lift them to $x_1,\dots ,x_r\in M$. Their span $N$ equals $M$: the finite module $Q=M\mathrm{/}N$ satisfies $Q=\mathfrak{m}Q$, and if $q_1,\dots ,q_s$ generate $Q$, write $q_i={\sum }_j{a}_{ij}{q}_j$ with $a_{ij}\in \mathfrak{m}$. The adjugate identity gives $\det (I-A)q_i=0$ for all $i$, while $\det (I-A)\in 1+\mathfrak{m}$ is a unit by [F3]. Hence $Q=0$.

The resulting surjection $R^r\twoheadrightarrow M$ splits because $M$ is projective. Its kernel $K$ is finite, being the image of the complementary idempotent on $R^r$. Reducing the split sum $R^r\cong M\oplus K$ modulo $\mathfrak{m}$, the chosen residue basis makes $k^r\to M\mathrm{/}\mathfrak{m}M$ an isomorphism, so $K\mathrm{/}\mathfrak{m}K=0$. Applying the finite determinant argument of step 1.1 to $K$ gives $K=0$. Thus $M\cong R^r$, and $M\mathrm{\ne }0$ forces $r\ge 1$.

For any finite sequence $\mathbf{x}$ in $\mathfrak{m}$ and each prefix ${\mathbf{x}}_{<i}$, the quotient $R^r\mathrm{/}({\mathbf{x}}_{<i})R^r$ is $(R\mathrm{/}({\mathbf{x}}_{<i}){)}^r$. Since $r\ge 1$, this quotient is nonzero exactly when $R\mathrm{/}({\mathbf{x}}_{<i})$ is nonzero, and multiplication by $x_i$ on it is injective exactly when multiplication by $x_i$ on the ring quotient is injective. The final quotients are likewise nonzero together. Thus precisely the same finite sequences in $\mathfrak{m}$ are regular on $M\cong R^r$ and on $R$. Their possible lengths, hence their depths, are equal, including depth zero.