Positive-degree exact free complexes split over a depth-zero local ring

If $F_j=0$ for every $j>0$, the complex is just $F_0$ in degree zero and the decomposition holds. Otherwise let $i>0$ be the highest index with $F_i\mathrm{\ne }0$. Positive-degree exactness makes $d_i:F_i\to F_{i-1}$ injective. If every matrix coefficient of $d_i$ lay in $\mathfrak{m}$, take a basis vector $v\in F_i$. The nonzero vector $av$ would map to zero, since $\mathfrak{m}a=0$, contradicting injectivity. Thus $d_i$ has a unit matrix entry.

By [F1], split off one contractible identity pair. The remaining complex is still positively exact because the split pair has zero homology. Its sum of ranks in positive degrees is strictly smaller. Repeating the argument of step 1.1 therefore stops after finitely many splits and leaves only a free degree-zero term. This is the asserted decomposition.

In the decomposition, let $c_i$ be the number of identity pairs whose nonzero differential has degree $i$. Then $n_i=c_i+c_{i+1}$ for $i>0$, with $c_{e+1}=0$. Descending from $i=e$ gives $c_i=n_i-n_{i+1}+\cdots +(-1{)}^{e-i}{n}_e=r_i$. The map $d_i$ is identity on $R^{c_i}$ and zero on the complementary summand, so its largest nonzero minor size is $r_i$, and an $r_i$-minor is $1$. No choice principle beyond finite bases is used.