Exact free complexes have the expected ranks and determinantal regular sequences

If $e=0$, there are no differential conditions. Assume $e\ge 1$. For every $\mathfrak{q}\in \mathrm{Ass}(R)$, localize the complex at $\mathfrak{q}$; exactness persists, and [F2] makes $R_{\mathfrak{q}}$ depth zero. By [F1] the localized complex has expected ranks $r_i$ and $(I_i{)}_{\mathfrak{q}}=R_{\mathfrak{q}}$, while every $(r_i+1)$-minor vanishes there.

By the injectivity in [F2], each $(r_i+1)$-minor vanishes already in $R$. Since $(I_i{)}_{\mathfrak{q}}=R_{\mathfrak{q}}$ at every associated prime, $I_i$ is not contained in any such prime. An associated prime exists when $R\mathrm{\ne }0$, so $I_i$ is nonzero and at least one $r_i$-minor is nonzero; hence the rank is exactly $r_i$. If every $I_i=R$, the regular-sequence alternative is immediate.

Otherwise set $J=I_1\cdots I_e⊊R$. Each $I_i$ avoids every associated prime by step 2.1; since those primes are prime, their product $J$ also avoids them. By [F3], choose a nonzerodivisor $x\in J\subseteq \mathfrak{m}$. In particular $x\in I_i$ for each $i$. Reduce the complex modulo $x$. By [F4] it is exact in degrees at least two. After omitting its degree-zero term and shifting indices down by one, the complex $0\to (R\mathrm{/}x{)}^{n_e}\to \cdots \to (R\mathrm{/}x{)}^{n_1}$ is positively exact and has length $e-1$.

Apply the same assertion inductively to that shorter complex over the Noetherian local ring $R\mathrm{/}x$. For its differential inherited from $d_i$ with $i\ge 2$, the alternating expected rank is still $r_i$. Its ideal of $r_i$-minors is $I_i\mathrm{/}(x)$ because $x\in I_i$. Therefore for each $i\ge 2$ either $I_i\mathrm{/}(x)=R\mathrm{/}x$ or this ideal contains a regular sequence of length $i-1$. If it is the unit ideal, $I_i=R$ since $x\in I_i$. Otherwise lift the sequence to elements of $I_i$; prepending the nonzerodivisor $x\in I_i$ gives an $R$-regular sequence of length $i$. For $i=1$, either $I_1=R$ or the single element $x$ supplies the required regular sequence.

The base case and length reduction prove the determinantal alternative for every $e$. The rank conclusion was established in step 2.1 independently of the induction. AC enters through associated-prime and regular-element existence in [F2]–[F3].