The abstract residue under a finite free extension of the coefficient algebra

(Rank-zero case.) If $n=0$, then $K^{\mathrm{\prime }}=0$, $V^{\mathrm{\prime }}=A^{\mathrm{\prime }}=0$, and the basis is empty. The only $f\in K^{\mathrm{\prime }}$ is $0$, ${\mathrm{\Omega }}_{K^{\mathrm{\prime }}\mathrm{/}k}^1=0$, multiplication by $0$ on the rank-zero free $K$-module has trace $0$, and both sides of the residue formula are $0$; stability and basis independence are immediate. Thus assume $n>0$ for the remaining steps.

(Stability of $A^{\mathrm{\prime }}$.) For $f^{\mathrm{\prime }}\in K^{\mathrm{\prime }}$, write $f^{\mathrm{\prime }}{x}_j={\sum }_i{x}_i{f}_{ij}$ with $f_{ij}\in K$ as in [F2]. For each of the finitely many pairs $(i,j)$ choose a finite-dimensional $W_{ij}\subseteq V$ with $f_{ij}A\subseteq A+W_{ij}$, using $f_{ij}A<A$. Then for $a_j\in A$, $f^{\mathrm{\prime }}({\sum }_j{x}_j\otimes a_j)={\sum }_{i,j}{x}_i\otimes f_{ij}{a}_j\in A^{\mathrm{\prime }}+W^{\mathrm{\prime }},W^{\mathrm{\prime }}:={\sum }_{i,j}{x}_i\otimes W_{ij}\mathrm{.}$ The space $W^{\mathrm{\prime }}$ is finite-dimensional over $k$: it is a finite sum of images of finite-dimensional spaces $W_{ij}$ under the $k$-linear maps $v\mapsto x_i\otimes v$. Thus $f^{\mathrm{\prime }}{A}^{\mathrm{\prime }}<A^{\mathrm{\prime }}$, so $A^{\mathrm{\prime }}$ is stable and ${\mathrm{res}}^{V^{\mathrm{\prime }}}$ is defined. The witnesses are chosen separately for the finitely many matrix entries; no $K$-module structure on a witness space is needed.

(Basis independence.) If $x_1^{\mathrm{\prime }},\dots ,x_n^{\mathrm{\prime }}$ is another $K$-basis of $K^{\mathrm{\prime }}$, write $x_j^{\mathrm{\prime }}={\sum }_l{a}_{lj}{x}_l$ with $a_{lj}\in K$; the same computation as in step 1.2 applied to the inverse change of basis shows $A^{\mathrm{\prime }\mathrm{\prime }}={\sum }_j{x}_j^{\mathrm{\prime }}\otimes A<A^{\mathrm{\prime }}$ and $A^{\mathrm{\prime }}<A^{\mathrm{\prime }\mathrm{\prime }}$, so $A^{\mathrm{\prime }\mathrm{\prime }}\sim A^{\mathrm{\prime }}$; by [F1] the spaces $E,E_1,E_2,E_0$ and hence the residue depend only on the commensurability class, so the construction of $A^{\mathrm{\prime }}$ is basis-independent up to commensurability.

(Matrix description and finite potency.) By [F2] a $k$-endomorphism $\gamma$ of $V^{\mathrm{\prime }}$ is exactly a matrix $({\gamma }_{ij})$ of $k$-linear endomorphisms of $V$ acting by $\gamma (x_j\otimes v)={\sum }_i{x}_i\otimes {\gamma }_{ij}(v)$. Let $F\subseteq {\mathrm{End}}_k(V)$ be finite potent, and choose a positive exponent $e$ from its finite-potent-family condition [F4]; if the available exponent is $0$, then $V$ is finite-dimensional and exponent $1$ also works. Let $F^{\mathrm{\prime }}$ be the $k$-subspace of ${\mathrm{End}}_k(V^{\mathrm{\prime }})$ of matrices with every entry in $F$. For any fixed $e$ matrices in $F^{\mathrm{\prime }}$, each entry of their product is a finite sum, over intermediate matrix indices, of length-$e$ products of elements of $F$. Every such product has finite-dimensional image by the exponent condition on $F$, and the finite sum of those images is finite-dimensional. Since there are only finitely many input and output blocks, the whole matrix product has finite-dimensional image. Thus $F^{\mathrm{\prime }}$ is finite potent with exponent $e$, independently of the matrix size $n$; no common space $F^{e}V$ or uniform dimension bound is asserted.

(A projection onto $A^{\mathrm{\prime }}$.) By [F7], choose a basis of $A$ and extend it to a basis of $V$; the map equal to the identity on the first basis and zero on the added basis vectors is a $k$-linear projection $\pi \text{\hspace{0.17em}⁣}:V\to A$. Define ${\pi }^{\mathrm{\prime }}({\sum }_i{x}_i\otimes v_i)={\sum }_i{x}_i\otimes \pi (v_i)$; it is a $k$-linear projection of $V^{\mathrm{\prime }}$ onto $A^{\mathrm{\prime }}$. For $f^{\mathrm{\prime }}\in K^{\mathrm{\prime }}$, step 1.2 gives a fixed finite-dimensional $W^{\mathrm{\prime }}\subseteq V^{\mathrm{\prime }}$ with $f^{\mathrm{\prime }}{A}^{\mathrm{\prime }}\subseteq A^{\mathrm{\prime }}+W^{\mathrm{\prime }}$. If $a^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ and $f^{\mathrm{\prime }}{a}^{\mathrm{\prime }}=b^{\mathrm{\prime }}+w^{\mathrm{\prime }}$ with $b^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ and $w^{\mathrm{\prime }}\in W^{\mathrm{\prime }}$, then $({\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}-f^{\mathrm{\prime }})(a^{\mathrm{\prime }})={\pi }^{\mathrm{\prime }}(w^{\mathrm{\prime }})-w^{\mathrm{\prime }}$. Thus the image of $({\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}-f^{\mathrm{\prime }}){\mathrm{\mid }}_{A^{\mathrm{\prime }}}$ lies in the fixed finite-dimensional space $({\pi }^{\mathrm{\prime }}-1)(W^{\mathrm{\prime }})$, so ${\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}\equiv f^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(A^{\mathrm{\prime }}))$; also ${\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}\in E_1(A^{\mathrm{\prime }})$ because its image lies in $A^{\mathrm{\prime }}$. By [F5], ${\mathrm{res}}^{V^{\mathrm{\prime }}}(f\mathrm{d}g)={\mathrm{Tr}}_{V^{\mathrm{\prime }}}([{\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }},g^{\mathrm{\prime }}])$ for all $f\in K^{\mathrm{\prime }}$ and $g\in K$, where $g^{\mathrm{\prime }}$ acts on $V^{\mathrm{\prime }}$ with matrix $g{\delta }_{ij}$.

(Trace formula.) Let $\gamma \in F^{\mathrm{\prime }}$ for a finite potent $F\subseteq {\mathrm{End}}_k(V)$ and write $\gamma =D+U+L$ as its diagonal, strictly upper-triangular, and strictly lower-triangular block parts. All three matrices lie in the common finite potent subspace $F^{\mathrm{\prime }}$ of step 2.2. Since $U^n=L^n=0$, (T3) gives ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}(U)={\mathrm{Tr}}_{V^{\mathrm{\prime }}}(L)=0$, and (T4) applied within $F^{\mathrm{\prime }}$ gives ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}(\gamma )={\mathrm{Tr}}_{V^{\mathrm{\prime }}}(D)+{\mathrm{Tr}}_{V^{\mathrm{\prime }}}(U)+{\mathrm{Tr}}_{V^{\mathrm{\prime }}}(L)$. Each summand $V_i=x_i\otimes V$ is $D$-stable; repeated use of (T2) on this finite direct sum gives ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}(D)={\sum }_i{\mathrm{Tr}}_{V_i}(D{\mathrm{\mid }}_{V_i})$. For each $i$, the isomorphism ${\iota }_i\text{\hspace{0.17em}⁣}:V\to V_i$, $v\mapsto x_i\otimes v$, identifies $D{\mathrm{\mid }}_{V_i}$ with ${\gamma }_{ii}$. Apply (T5) with $\phi ={\iota }_i$ and $\psi ={\gamma }_{ii}{\iota }_i^{-1}$; since $\psi \phi ={\gamma }_{ii}$ is finite potent, this gives ${\mathrm{Tr}}_{V_i}(D{\mathrm{\mid }}_{V_i})={\mathrm{Tr}}_V({\gamma }_{ii})$. Therefore ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}(\gamma )={\sum }_i{\mathrm{Tr}}_V({\gamma }_{ii})$ for every matrix with entries in one finite potent subspace $F$.

(Matrix of the commutator.) For $f={\sum }_j{c}_j{x}_j\in K^{\mathrm{\prime }}$ the matrix $(f_{ij})$ of multiplication by $f$ on $V^{\mathrm{\prime }}$ is as in [F2], and ${\pi }^{\mathrm{\prime }}$ has block-diagonal matrix $\pi {\delta }_{ij}$; hence ${\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}$ has matrix $(\pi f_{ij})$ and, since $g$ is central in $K^{\mathrm{\prime }}$, the commutator $[{\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }},g^{\mathrm{\prime }}]$ has matrix $([\pi f_{ij},g]{)}_{ij}$. Each entry $[\pi f_{ij},g]$ lies in $E_0(A)$: indeed $\pi f_{ij}\in E_1(A)$ and $\pi f_{ij}\equiv f_{ij}(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(A))$ as in step 2.3, while $g\equiv g(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(A))$, so the commutator lies in $E_1(A)$ and is congruent to $[f_{ij},g]=0$ modulo $E_2(A)$.

(Evaluation of the trace.) The matrix entries of $[{\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }},g^{\mathrm{\prime }}]$ all lie in the finite potent subspace $E_0(A)$, so the trace formula of step 3.1 applies and gives ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}([{\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }},g^{\mathrm{\prime }}])={\sum }_i{\mathrm{Tr}}_V([\pi f_{ii},g])$, the sum being finite over the $n$ diagonal blocks.

(Each diagonal block is a residue.) For each $i$ the endomorphism $\pi f_{ii}$ of $V$ lies in $E_1(A)$ and satisfies $\pi f_{ii}\equiv f_{ii}(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(A))$ by step 2.3 applied to the stable subspace $A$; hence [F5] identifies ${\mathrm{Tr}}_V([\pi f_{ii},g])={\mathrm{res}}^V(f_{ii}\mathrm{d}g)$.

(The trace of the coefficient.) By [F3] one has ${\sum }_i{f}_{ii}={\mathrm{Tr}}_{K^{\mathrm{\prime }}\mathrm{/}K}(f)$ for the matrix $(f_{ij})$ of multiplication by $f$ on the free $K$-module $K^{\mathrm{\prime }}$; substituting step 5.2 into step 5.1 and using the $k$-linearity of ${\mathrm{res}}^V$ ([F5]) in the coefficient, ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}([{\pi }^{\mathrm{\prime }}{f}^{\mathrm{\prime }},g])={\sum }_i{\mathrm{res}}^V(f_{ii}\mathrm{d}g)={\mathrm{res}}^V({\mathrm{Tr}}_{K^{\mathrm{\prime }}\mathrm{/}K}(f)\mathrm{d}g)$, which together with step 2.3 is the asserted identity ${\mathrm{res}}^{V^{\mathrm{\prime }}}(f\mathrm{d}g)={\mathrm{res}}^V({\mathrm{Tr}}_{K^{\mathrm{\prime }}\mathrm{/}K}(f)\mathrm{d}g)$. The choices involved (the basis $x_i$, the projection $\pi$) are the only uses of the Axiom of Choice [F6] beyond those inherited from the linear algebra suppliers; the commensurability class of $A^{\mathrm{\prime }}$ was shown in step 2.1 to be independent of the basis.