Additivity of the abstract residue over intersecting subspaces

(Stability of $A+B$.) For $f\in K$ one has $fA<A$ and $fB<B$, hence $f(A+B)=fA+fB<A+B$ by the finite-sums rule of [F1]; therefore $A+B$ satisfies the hypothesis of Existence and uniqueness of the abstract residue map res_V: Omega^1_{K/k} -> k and ${\mathrm{res}}_{A+B}$ is defined.

(Stability of $A\cap B$.) Let $f\in K$, choose finite-dimensional $W_1,W_2\subseteq V$ with $fA\subseteq A+W_1$ and $fB\subseteq B+W_2$, and set $W:=W_1+W_2$. For $x\in fA\cap fB$, write $x=a+w_1=b+w_2$ with $a\in A$, $b\in B$, and $w_i\in W_i$. Then $a=b+(w_2-w_1)\in U:=A\cap (B+W)$. The map $U\mathrm{/}(A\cap B)\to (B+W)\mathrm{/}B$ sending $u+(A\cap B)$ to $u+B$ is injective, and $(B+W)\mathrm{/}B$ is finite-dimensional because it is a quotient of $W$. Hence $U\mathrm{/}(A\cap B)$ is finite-dimensional; choose a finite-dimensional subspace $X\subseteq U$ whose image spans that quotient, so $U\subseteq (A\cap B)+X$. It follows that $x=a+w_1\in (A\cap B)+X+W_1$, a finite-dimensional enlargement. Thus $fA\cap fB<A\cap B$, and since $f(A\cap B)\subseteq fA\cap fB$, also $f(A\cap B)<A\cap B$. Therefore $A\cap B$ is stable and ${\mathrm{res}}_{A\cap B}$ is defined.

(Four compatible projections.) Choose a complement $X$ of $A\cap B$ in $A$, a complement $Y$ of $A\cap B$ in $B$, and a complement $Z$ of $A+B$ in $V$, so that $V=(A\cap B)\oplus X\oplus Y\oplus Z$ with $A=(A\cap B)\oplus X$ and $B=(A\cap B)\oplus Y$; let ${\pi }_{A\cap B},{\pi }_X,{\pi }_Y,{\pi }_Z$ be the projections onto the four summands along the complementary summand, and put ${\pi }_A={\pi }_{A\cap B}+{\pi }_X$, ${\pi }_B={\pi }_{A\cap B}+{\pi }_Y$ and ${\pi }_{A+B}={\pi }_{A\cap B}+{\pi }_X+{\pi }_Y$. Each is a $k$-linear projection of $V$ onto $A$, $B$, $A+B$, $A\cap B$ respectively, and adding the expressions for ${\pi }_A$ and ${\pi }_B$ gives the identity ${\pi }_A+{\pi }_B={\pi }_{A+B}+{\pi }_{A\cap B}$.

(Residues via the projections.) Fix the stable subspace $C\in \{A,B,A+B,A\cap B\}$ with projection ${\pi }_C$ of step 1.3. Since ${\pi }_{C}f$ has image in $C$, it lies in $E_1(C)$; and for $c\in C$ one has $fc\in fC<C$, say $fc=c^{\mathrm{\prime }}+w$ with $c^{\mathrm{\prime }}\in C$ and $w$ in a fixed finite-dimensional space, whence $({\pi }_{C}f-f)(c)={\pi }_C(w)-w$ lies in the finite-dimensional space ${\pi }_C(W)+W$, so ${\pi }_{C}f\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(C))$. Therefore [F2] applies with $f_1={\pi }_{C}f$ and $g_1=g$ and gives ${\mathrm{res}}_C(f\mathrm{d}g)={\mathrm{Tr}}_V([{\pi }_{C}f,g])$ for all $f,g\in K$; denote ${\gamma }_C:=[{\pi }_{C}f,g]\in E_0(C)$.

(Finite-potent subspaces for the two differences.) Put $S:=A+B$ and $D:=A\cap B$. For the nested pair $A\subseteq S$, let $F_{A,S}:={\mathrm{span}}_k\{{\gamma }_A,{\gamma }_S\}$. The image bound ${\gamma }_C(V)\subseteq C+gC$ and $gC<C$ gives ${\gamma }_A(V),{\gamma }_S(V)\subseteq S+W_0$ for a common finite-dimensional $W_0$. The compatible projections of step 1.3 satisfy ${\pi }_A(S)\subseteq A$; using $fS<S$, $gS<S$ and $gA<A$ in the formula ${\gamma }_A=[{\pi }_{A}f,g]$ gives ${\gamma }_A(S)\subseteq A+W_1$ for some finite-dimensional $W_1$, while ${\gamma }_S(S)$ is finite-dimensional because ${\gamma }_S\in E_0(S)$. Also ${\gamma }_A(A)$ is finite-dimensional because ${\gamma }_A\in E_0(A)$. Taking common finite-dimensional error spaces for the two generators and their images, every product of three elements of $F_{A,S}$ maps $V$ into a fixed finite-dimensional space: successively its image lies in $S+W_0$, then in $A$ plus a finite-dimensional space, then in a finite-dimensional space. Thus $F_{A,S}$ is finite potent.

(The key identity and the second finite-potent subspace.) With ${\gamma }_C$ as in step 2.1 and using that $f$ and $g$ commute as endomorphisms of $V$, bilinearity of the commutator gives ${\gamma }_A-{\gamma }_{A+B}=[({\pi }_A-{\pi }_{A+B})f,g]$ and ${\gamma }_{A\cap B}-{\gamma }_B=[({\pi }_{A\cap B}-{\pi }_B)f,g]$, and the projection identity of step 1.3 gives ${\pi }_A-{\pi }_{A+B}={\pi }_{A\cap B}-{\pi }_B$; hence ${\gamma }_A-{\gamma }_{A+B}={\gamma }_{A\cap B}-{\gamma }_B$ as $k$-endomorphisms of $V$. To obtain trace linearity for the second difference, put $S:=A+B$ and $D:=A\cap B$, and for $D\subseteq B\subseteq S$ let $F_{D,B}:={\mathrm{span}}_k\{{\gamma }_D,{\gamma }_B\}$. Both generators map $V$ into $S$ plus a finite-dimensional space; compatibility gives ${\pi }_B(S)\subseteq B$, hence ${\gamma }_B(S)\subseteq B$ plus a finite-dimensional space, and ${\gamma }_D(S)\subseteq D$ plus a finite-dimensional space from its image bound. Further, ${\gamma }_B(B)$ and ${\gamma }_D(D)$ are finite-dimensional because ${\gamma }_B\in E_0(B)$ and ${\gamma }_D\in E_0(D)$, while ${\gamma }_D(B)\subseteq D$ plus a finite-dimensional space from the same image bound. With common finite-dimensional error spaces for the two generators and their images, every product of four elements of $F_{D,B}$ maps $V$ into a fixed finite-dimensional space, so $F_{D,B}$ is finite potent. Thus [F3] gives trace linearity on $F_{D,B}$, and step 3.1 gives trace linearity on $F_{A,S}$; consequently ${\mathrm{Tr}}_V({\gamma }_A)-{\mathrm{Tr}}_V({\gamma }_S)={\mathrm{Tr}}_V({\gamma }_A-{\gamma }_S)$ and ${\mathrm{Tr}}_V({\gamma }_D)-{\mathrm{Tr}}_V({\gamma }_B)={\mathrm{Tr}}_V({\gamma }_D-{\gamma }_B)$.

(Additivity on generators.) Step 4.1 gives the operator identity and the trace-linear formula for the second difference; step 3.1 supplies the trace-linear formula for the first difference. Therefore the two trace differences agree. By step 2.1, these traces are the four abstract residues on $f\mathrm{d}g$; hence ${\mathrm{res}}_A(f\mathrm{d}g)-{\mathrm{res}}_{A+B}(f\mathrm{d}g)={\mathrm{res}}_{A\cap B}(f\mathrm{d}g)-{\mathrm{res}}_B(f\mathrm{d}g)$ for all $f,g\in K$. Since the forms $f\mathrm{d}g$ generate ${\mathrm{\Omega }}_{K\mathrm{/}k}^1$ and both sums of residues are $k$-linear, this proves ${\mathrm{res}}_A+{\mathrm{res}}_B={\mathrm{res}}_{A+B}+{\mathrm{res}}_{A\cap B}$, the formula $(R_5)$. The choices of complements in step 1.3 are the only use of the Axiom of Choice [F4].