auslander buchsbaum syzygy projective dimension

If $K=0$, the given presentation identifies $M$ with the free module $F_0$, contradicting $n>0$. Thus $K\mathrm{\ne }0$. Fix a length-$n$ projective resolution of $M$ supplied by [F1]. Starting with the given $F_0\to M$, form its first $n$ projective presentations by taking finite free surjections onto the successive kernels. Each kernel is finite by [F3]. Only finitely many choices are made, so no axiom of choice is needed. Write $K_n$ for the final kernel; when $n=1$ it is $K$.

Compare these $n$ presentations with the fixed length-$n$ resolution using [F2]. At the first stage the two kernels become isomorphic after adding the other presentation's projective term. Present these stabilized kernels at the next stage by adding identity presentations on those projective summands, and apply [F2] again. Induction through the finite $n$ stages gives $K_n\oplus A\cong P_n\oplus B$, where $P_n$ is the terminal projective of the length-$n$ resolution and $A,B$ are finite direct sums of its projective terms and the constructed finite free terms. Hence $K_n$ is a direct summand of a projective module and is projective. Truncating the constructed sequence therefore gives a length-$(n-1)$ projective resolution of $K$. Thus ${\mathrm{pd}}_{R}K\le n-1$.

If $n\ge 2$ and ${\mathrm{pd}}_{R}K\le n-2$, splicing such a shorter resolution with $0\to K\to F_0\to M\to 0$ gives a resolution of $M$ of length at most $n-1$, contradicting [F1] and ${\mathrm{pd}}_{R}M=n$. For $n=1$, step 2.1 already makes $K$ projective, so its projective dimension is zero. In every case $K\mathrm{\ne }0$ and ${\mathrm{pd}}_{R}K=n-1$.