Finite nilpotent transfer and exact cohomology dimension bounds

Since each factor $\delta$ increases the $\mathfrak{m}$-adic order, $(\delta h{)}^N=(h\delta {)}^N=0$ for sufficiently large $N$. Both inverse series in the Statement are therefore finite. Multiplying them out gives $Q=\delta -\delta hQ=\delta -Qh\delta$. Since $D^2=0$, $d\delta +\delta d+{\delta }^2=0$. To obtain the needed cancellation identity, multiply $dQ+Qd+QipQ$ on the left by $1+\delta h$ and on the right by $1+h\delta$. Replacing $(1+\delta h)Q=\delta =Q(1+h\delta )$, the result is $(1+\delta h)d\delta +\delta d(1+h\delta )+\delta ip\delta =d\delta +\delta d+\delta (hd+dh+ip)\delta =d\delta +\delta d+{\delta }^2=0$. The multipliers are invertible, so $dQ+Qd+QipQ=0$.

Apply $p$ on the left and $i$ on the right to the last identity to get $(pQi{)}^2=0$, using $pd=di=0$. Substituting $i^{\mathrm{\prime }}=i-hQi$ into $D{i}^{\mathrm{\prime }}-i^{\mathrm{\prime }}{d}_H$, using $dh=1-ip-hd$ and $\delta -\delta hQ=Q$, leaves $h(dQ+QipQ)i=0$. Substitution in $p^{\mathrm{\prime }}D-d_H{p}^{\mathrm{\prime }}$ gives the analogous vanishing using $Q-\delta +Qh\delta =0$. Finally $p^{\mathrm{\prime }}{i}^{\mathrm{\prime }}=pi-phQi-pQhi+pQ{h}^{2}Qi=1$ by the side conditions.

Expand $D{h}^{\mathrm{\prime }}+h^{\mathrm{\prime }}D-1+i^{\mathrm{\prime }}{p}^{\mathrm{\prime }}$ using the formulas in the Statement. Use $dh+hd=1-ip$ to cancel the constant terms, then $\delta hQ=Qh\delta =\delta -Q$ to cancel the terms linear in $\delta$, and group what remains as $h(dQ+Qd+QipQ)h$, which is zero by step1.1. Thus $D{h}^{\mathrm{\prime }}+h^{\mathrm{\prime }}D=1-i^{\mathrm{\prime }}{p}^{\mathrm{\prime }}$. Together with step2.1 and [F1], these are the claimed chain homotopy equivalence identities. All expressions are finite algebraic identities.

The cohomology of the finite complex $H\otimes A$ is that of $C\otimes A$ by step3.1. In degree $q$, elementary kernel-image dimensions give $\dim H^q(H\otimes A,d_H)=\dim (H^q\otimes A)-\dim \mathrm{im}{d}_H^q-\dim \mathrm{im}{d}_H^{q-1}$, which is the displayed formula with $\dim (H^q\otimes A)=\mathrm{\ell }\dim H^q$. Equality in every degree forces every adjacent image to be zero, hence $d_H=0$, and the converse is immediate. Since $Q$ lies in $\mathfrak{m}$, all transfer maps reduce to the original $i,p,h$. Every finite formula commutes with base change; when $d_H=0$, the transferred complex and its base change both have zero differential, so their cohomologies are the free modules $H^q\otimes A$ and $H^q\otimes B$, and the natural base-change map is an isomorphism. This does not assert that base change commutes with the cohomology of an arbitrary nonzero transferred differential.