Basic properties of the abstract residue: restriction, commensurability, vanishing, logarithmic residues

(Setup) By [F2] fix a $k$-linear projection $\pi \text{\hspace{0.17em}⁣}:V\to A$ with $\pi {\mathrm{\mid }}_A={\mathrm{i}\mathrm{d}}_A$; then $\pi \circ \pi =\pi$, $\pi V=A$, so $\pi \in E_1$ because $A<A$, and $(\pi -{\mathrm{i}\mathrm{d}}_V)(A)=0$ is finite-dimensional, so $\pi \equiv {\mathrm{i}\mathrm{d}}_V(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$.

(Claim 4: the two quotients) Let $g\in K^{\times }$ and $h\in K$ with $hA\subseteq A$, and put $N:=A\cap gA$. The standing hypothesis $fA<A$ applied to $g$ gives $gA<A$, and applied to $g^{-1}$ it gives $g^{-1}A<A$, hence $A=g(g^{-1}A)<gA$ by the stability of $<$ under the $k$-linear map $g$; therefore $(A+gA)\mathrm{/}gA\cong A\mathrm{/}N$ and $(A+gA)\mathrm{/}A\cong gA\mathrm{/}N$ are finite-dimensional, and $N$ is of finite codimension in both $A$ and $gA$. Moreover $hA\subseteq A$ and $hgA=ghA\subseteq gA$, so $m_h$ preserves $A$, $gA$ and $N$ and induces $k$-linear endomorphisms ${\alpha }_A$ of $A\mathrm{/}N$ and ${\alpha }_{gA}$ of $gA\mathrm{/}N$.

(Standard lifts) For $f\in K$ put $f_{\mathrm{♯}}:=\pi \circ f\in {\mathrm{End}}_k(V)$; then $f_{\mathrm{♯}}V=\pi (fV)\subseteq A$, so $f_{\mathrm{♯}}\in E_1$ by [F3], and $(f_{\mathrm{♯}}-f)(A)=(\pi -{\mathrm{i}\mathrm{d}}_V)(fA)\subseteq (\pi -{\mathrm{i}\mathrm{d}}_V)(W)$ for a finite-dimensional $W$ with $fA\subseteq A+W$, so $f_{\mathrm{♯}}\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$; hence for all $f,g\in K$ the pair $(f_{\mathrm{♯}},g_{\mathrm{♯}})$ is admissible in [F6] and ${\mathrm{res}}_V(f\mathrm{d}g)={\mathrm{Tr}}_V([f_{\mathrm{♯}},g_{\mathrm{♯}}])$.

(Claim 4: the lifts and $\theta$) Put $f:=h{g}^{-1}\in K$ and choose the lifts $f_1:=\pi \circ f$ and $g_1:=g$; then $f_{1}V=\pi (fV)\subseteq A$, so $f_1\in E_1$ by [F3], and $(f_1-f)A=(\pi -{\mathrm{i}\mathrm{d}}_V)(fA)$ is finite-dimensional because the standing hypothesis $fA<A$ gives $fA\subseteq A+W$ for a finite-dimensional $W$, so $f_1\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ while $g_1=g$ trivially; hence the pair $(f_1,g_1)$ is admissible in [F6] and ${\mathrm{res}}_V(f\mathrm{d}g)={\mathrm{Tr}}_V(\theta )$ for $\theta :=[f_1,g_1]=\pi m_{fg}-g\pi m_f$. Since $fg=h$ and multiplication by $g$ and $h$ commutes, $g\pi m_f=g\pi m_{g^{-1}}{m}_h=(g\pi g^{-1})m_h={\pi }_g{m}_h$, where ${\pi }_g:=g\pi g^{-1}$ is a $k$-linear projection of $V$ onto $gA$; thus $\theta =(\pi -{\pi }_g)m_h$. No commutation of either projection with multiplication is used. (ii) if $\theta \in E_0$ satisfies $\theta V\subseteq A$, then $\phi :=\theta \text{\hspace{0.17em}⁣}:V\to A$ and the inclusion $\psi \text{\hspace{0.17em}⁣}:A\to V$ are $k$-linear with $\psi \phi =\theta$ finite potent, so (T5) gives ${\mathrm{Tr}}_V(\theta )={\mathrm{Tr}}_A(\theta {\mathrm{\mid }}_A)$, and if moreover $\theta (A)=0$ then ${\mathrm{Tr}}_A(\theta {\mathrm{\mid }}_A)=0$ by [F8]; (iii) if $\theta \in E_0$ and $V\mathrm{\prime }\subseteq V$ satisfies $\theta V\subseteq V\mathrm{\prime }$ and $\theta V\mathrm{\prime }\subseteq V\mathrm{\prime }$, then $\theta$ viewed as a map $V\to V\mathrm{\prime }$ and the inclusion $V\mathrm{\prime }\to V$ are $k$-linear with composite $\theta$, so (T5) gives ${\mathrm{Tr}}_V(\theta )={\mathrm{Tr}}_{V\mathrm{\prime }}(\theta {\mathrm{\mid }}_{V\mathrm{\prime }})$; (iv) if $f_1,g_1\in E_1$ satisfy $f_1\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ and $g_1\equiv g(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ for commuting elements $f,g\in K$, then $[f_1,g_1]\in E_1$ because $E_1$ is a two-sided ideal, while $[f_1,g_1]\equiv [f,g]=0(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ because $E_2$ is a two-sided ideal, so $[f_1,g_1]\in E_1\cap E_2=E_0$.

(Framework) From steps 1.1 and 2.1 we record: (i) for every $f\in K$ and $n\ge 1$ one has $(f_{\mathrm{♯}}{)}^n\in E_1$ and $(f_{\mathrm{♯}}{)}^n\equiv f^n(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$, because $E_1$ and $E_2$ are two-sided ideals; for $n=0$, the identity is a lift of $1$ in $E$ and can be paired with an $E_1$ lift; (ii) if $\theta \in E_0$ satisfies $\theta V\subseteq A$, then $\phi :=\theta \text{\hspace{0.17em}⁣}:V\to A$ and the inclusion $\psi \text{\hspace{0.17em}⁣}:A\to V$ are $k$-linear with $\psi \phi =\theta$ finite potent, so (T5) gives ${\mathrm{Tr}}_V(\theta )={\mathrm{Tr}}_A(\theta {\mathrm{\mid }}_A)$, and if moreover $\theta (A)=0$ then ${\mathrm{Tr}}_A(\theta {\mathrm{\mid }}_A)=0$ by [F8]; (iii) if $\theta \in E_0$ and $V^{\mathrm{\prime }}\subseteq V$ satisfies $\theta V\subseteq V^{\mathrm{\prime }}$ and $\theta V^{\mathrm{\prime }}\subseteq V^{\mathrm{\prime }}$, then $\theta$ viewed as a map $V\to V^{\mathrm{\prime }}$ and the inclusion $V^{\mathrm{\prime }}\to V$ are $k$-linear with composite $\theta$, so (T5) gives ${\mathrm{Tr}}_V(\theta )={\mathrm{Tr}}_{V^{\mathrm{\prime }}}(\theta {\mathrm{\mid }}_{V^{\mathrm{\prime }}})$; (iv) if $f_1,g_1\in E_1$ satisfy $f_1\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ and $g_1\equiv g(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ for commuting elements $f,g\in K$, then $[f_1,g_1]\in E_1$ because $E_1$ is a two-sided ideal, while $[f_1,g_1]\equiv [f,g]=0(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2)$ because $E_2$ is a two-sided ideal, so $[f_1,g_1]\in E_1\cap E_2=E_0$.

(Claim 1: restriction, setup) Let $V^{\mathrm{\prime }}$ be a $k$-subspace with $A\subseteq V^{\mathrm{\prime }}\subseteq V$ and $K{V}^{\mathrm{\prime }}\subseteq V^{\mathrm{\prime }}$; then the restrictions $f_{\mathrm{♯}}{\mathrm{\mid }}_{V^{\mathrm{\prime }}},g_{\mathrm{♯}}{\mathrm{\mid }}_{V^{\mathrm{\prime }}}$ are $k$-linear maps $V^{\mathrm{\prime }}\to A\subseteq V^{\mathrm{\prime }}$ with images in $A$, so they lie in $E_1(V^{\mathrm{\prime }})$, they satisfy $f_{\mathrm{♯}}{\mathrm{\mid }}_{V^{\mathrm{\prime }}}-f{\mathrm{\mid }}_{V^{\mathrm{\prime }}}=(f_{\mathrm{♯}}-f){\mathrm{\mid }}_{V^{\mathrm{\prime }}}$ with $(f_{\mathrm{♯}}-f)(A)$ finite-dimensional, so $f_{\mathrm{♯}}{\mathrm{\mid }}_{V^{\mathrm{\prime }}}\equiv f{\mathrm{\mid }}_{V^{\mathrm{\prime }}}(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(V^{\mathrm{\prime }}))$, and $(f_{\mathrm{♯}}{\mathrm{\mid }}_{V^{\mathrm{\prime }}},g_{\mathrm{♯}}{\mathrm{\mid }}_{V^{\mathrm{\prime }}})$ is admissible for the pair $(V^{\mathrm{\prime }},A)$, giving ${\mathrm{res}}_{V^{\mathrm{\prime }}}(f\mathrm{d}g)={\mathrm{Tr}}_{V^{\mathrm{\prime }}}(\theta {\mathrm{\mid }}_{V^{\mathrm{\prime }}})$ for $\theta :=[f_{\mathrm{♯}},g_{\mathrm{♯}}]$.

(Claim 4: the induced map on $B\mathrm{/}N$) Let $B:=A+gA$ and $N:=A\cap gA$. The endomorphism $\theta =(\pi -{\pi }_g)m_h$ of step 2.2 maps $V$ into $B$, since $\pi m_h(V)\subseteq A$ and ${\pi }_g{m}_h(V)\subseteq gA$. It preserves $B$, and it kills $N$: multiplication by $h$ preserves $N$ by step 1.2, while both $\pi$ and ${\pi }_g$ restrict to the identity on $N\subseteq A\cap gA$. Thus $B$ and $N$ are $\theta$-stable and $\theta$ induces an endomorphism $\bar{\theta }$ of the finite-dimensional quotient $B\mathrm{/}N$. By (T2), first for $B\subseteq V$ and then for $N\subseteq B$, the trace on $V$ is the trace on $B$ plus the trace of the zero induced map on $V\mathrm{/}B$, and the trace on $B$ is the trace on $N$ plus ${\mathrm{Tr}}_{B\mathrm{/}N}(\bar{\theta })$. Both zero terms vanish, so ${\mathrm{Tr}}_V(\theta )={\mathrm{tr}}_{B\mathrm{/}N}(\bar{\theta })$.

(Claim 1: commensurability) Let $A^{\mathrm{\prime }}\sim A$ with $f{A}^{\mathrm{\prime }}<A^{\mathrm{\prime }}$ for all $f\in K$; by [F3] the spaces $E(A^{\mathrm{\prime }})=E(A)$, $E_1(A^{\mathrm{\prime }})=E_1(A)$, $E_2(A^{\mathrm{\prime }})=E_2(A)$ coincide, so the class of admissible lifts in the defining formula of [F6] is the same for the pairs $(V,A^{\mathrm{\prime }})$ and $(V,A)$, and by step 2.1 the pair $(f_{\mathrm{♯}},g_{\mathrm{♯}})$ is admissible in both cases with value ${\mathrm{Tr}}_V([f_{\mathrm{♯}},g_{\mathrm{♯}}])$; hence ${\mathrm{res}}_{A^{\mathrm{\prime }}}(f\mathrm{d}g)={\mathrm{res}}_A(f\mathrm{d}g)$ on all generators and therefore, both maps being $k$-linear, ${\mathrm{res}}_{A^{\mathrm{\prime }}}={\mathrm{res}}_A$.

(Claim 1: vanishing) The hypothesis of claim (1) is that $V\mathrm{/}A$ is finite-dimensional. For any $\theta \in {\mathrm{End}}_k(V)$, the image $(\theta V+A)\mathrm{/}A$ of $\theta V$ under the quotient map $V\to V\mathrm{/}A$ is therefore a subspace of the finite-dimensional space $V\mathrm{/}A$, hence finite-dimensional, since every subspace of a finite-dimensional vector space is finite-dimensional. Thus $\theta V<A$ and $E_1={\mathrm{End}}_k(V)$; taking the lifts $f_1:=f$ and $g_1:=g$ of $f$ and $g$ (endomorphisms as elements of the image of $K$ in $E$) gives ${\mathrm{res}}_V(f\mathrm{d}g)={\mathrm{Tr}}_V([f,g])={\mathrm{Tr}}_V(0)=0$ by [F6], since elements of $K$ commute and ${\mathrm{Tr}}_V(0)=0\cdot {\mathrm{Tr}}_V(0)=0$ by the $k$-linearity of the trace on the finite potent space $E_0$; hence ${\mathrm{res}}_V=0$.

(Claim 2: continuity, full Tate condition) Suppose $fA+fgA+f{g}^{2}A\subseteq A$. This gives $fA\subseteq A$, $fgA\subseteq A$ and $f{g}^{2}A\subseteq A$. Use the admissible lifts $f_1:=\pi f$ and $g_1:=g$ from step 2.1, and put $\theta :=[f_1,g]$. Then $\theta (V)\subseteq A+gA$ because $\pi fg(V)\subseteq A$ and $g\pi f(V)\subseteq gA$. For $a\in A$, the equalities $\pi fga=fga$ and $g\pi fa=gfa=fga$ show $\theta a=0$, using $fgA\subseteq A$ and $fA\subseteq A$. For $ga\in gA$, the equalities $\pi f{g}^{2}a=f{g}^{2}a$ and $g\pi fga=gfga=f{g}^{2}a$ show $\theta (ga)=0$, using $f{g}^{2}A\subseteq A$ and $fgA\subseteq A$. Therefore $\theta$ vanishes on $A+gA$, so ${\theta }^2=0$; its finite-potent trace is zero by [F8], and the defining formula gives ${\mathrm{res}}_V(f\mathrm{d}g)=0$.

(Claim 3: nonnegative powers) If $n\ge 1$, both $(f_{\mathrm{♯}}{)}^n$ and $f_{\mathrm{♯}}$ are admissible $E_1$ lifts of $f^n$ and $f$ by step 3.1(i), so ${\mathrm{res}}_V(f^n\mathrm{d}f)={\mathrm{Tr}}_V([(f_{\mathrm{♯}}{)}^n,f_{\mathrm{♯}}])=0$ because these powers commute. If $n=0$, use $1\in E$ as a lift of $f^0=1$ and the $E_1$ lift $f_{\mathrm{♯}}$ of $f$; this pair is admissible and its commutator is zero, so ${\mathrm{res}}_V(\mathrm{d}f)=0$.

(Claim 1: restriction concluded) For the pair $(V^{\mathrm{\prime }},A)$ of step 3.2 the commutator $\theta =[f_{\mathrm{♯}},g_{\mathrm{♯}}]$ lies in $E_0(V)$ by step 3.1(iv) and satisfies $\theta V\subseteq \pi (fV)+\pi (gV)\subseteq A\subseteq V^{\mathrm{\prime }}$, so $\theta$ is an endomorphism of $V$ with image in $V^{\mathrm{\prime }}$ and $\theta V^{\mathrm{\prime }}\subseteq A\subseteq V^{\mathrm{\prime }}$; by step 3.1(iii) ${\mathrm{Tr}}_V(\theta )={\mathrm{Tr}}_{V^{\mathrm{\prime }}}(\theta {\mathrm{\mid }}_{V^{\mathrm{\prime }}})$, whence ${\mathrm{res}}_V(f\mathrm{d}g)={\mathrm{res}}_{V^{\mathrm{\prime }}}(f\mathrm{d}g)$ on all generators, and both maps being $k$-linear, ${\mathrm{res}}_V={\mathrm{res}}_{V^{\mathrm{\prime }}}$.

(Claim 4: the two traces) The classes of $A$ and of $gA$ modulo $N$ have zero intersection and span $B\mathrm{/}N$, so $B\mathrm{/}N=A\mathrm{/}(A\cap gA)\oplus gA\mathrm{/}(A\cap gA)$ is a direct sum. The map $\pi m_h$ sends $B$ into $A$ and preserves $N$ (its restriction to $N$ is $m_h{\mathrm{\mid }}_N$, since $hN\subseteq N$); therefore it induces an endomorphism of $B\mathrm{/}N$ with image in $A\mathrm{/}N$. Its restriction to $A\mathrm{/}N$ is ${\alpha }_A$ because $ha\in A$ for $a\in A$. Relative to the displayed direct sum, this induced map has image in the first summand, so its trace is $\mathrm{Tr}({\alpha }_A)$. Similarly, ${\pi }_g{m}_h$ sends $B$ into $gA$, preserves $N$ (and restricts to $m_h{\mathrm{\mid }}_N$ there), and induces an endomorphism of $B\mathrm{/}N$ with image in $gA\mathrm{/}N$ whose restriction to that summand is ${\alpha }_{gA}$; its trace is $\mathrm{Tr}({\alpha }_{gA})$. Hence ${\mathrm{Tr}}_{B\mathrm{/}N}(\bar{\theta })=\mathrm{Tr}({\alpha }_A)-\mathrm{Tr}({\alpha }_{gA})$, that is, ${\mathrm{res}}_V(h{g}^{-1}\mathrm{d}g)={\mathrm{Tr}}_{A\mathrm{/}(A\cap gA)}(m_h)-{\mathrm{Tr}}_{gA\mathrm{/}(A\cap gA)}(m_h)$ by steps 2.2 and 3.3.

(Claim 2 concluded) For $\theta$ as in step 4.3 with $\theta (V)\subseteq A$ and $\theta (A)=0$, step 3.1(ii) gives ${\mathrm{Tr}}_V(\theta )=0$, hence ${\mathrm{res}}_V(f\mathrm{d}g)=0$ for all $f,g$ with $fA\subseteq A$ and $gA\subseteq A$, which is equivalent to $fA+gA+fgA\subseteq A$; if $A$ is a $K$-submodule of $V$ then $fA\subseteq A$ and $gA\subseteq A$ for all $f,g\in K$, so ${\mathrm{res}}_V$ vanishes on the generators $f\mathrm{d}g$ and therefore on ${\mathrm{\Omega }}_{K\mathrm{/}k}^1$.

(Claim 3: negative powers) Let $f\in K^{\times }$ and $n\le -2$, and put $F:=f^{-1}\in K$ and $m:=-n-2\ge 0$; from $\mathrm{d}(fF)=0$ and the Leibniz rule one gets $\mathrm{d}f=-f^2\mathrm{d}F$, hence $f^n\mathrm{d}f=-f^{n+2}\mathrm{d}F=-F^m\mathrm{d}F$, and the $k$-linearity of ${\mathrm{res}}_V$ together with step 4.4 applied to $F$ gives ${\mathrm{res}}_V(f^n\mathrm{d}f)=-{\mathrm{res}}_V(F^m\mathrm{d}F)=0$.

(Claim 3: the case $n=0$) Taking $n=0$ in step 4.4 gives ${\mathrm{res}}_V(\mathrm{d}f)=0$ for every $f\in K$, since $\mathrm{d}f=f^0\mathrm{d}f$; this is the "in particular" clause of claim (3).

(Claim 4: the special case) If $gA\subseteq A$ then $N=A\cap gA=gA$, so in the formula of step 4.6 the second quotient $gA\mathrm{/}N$ is the zero space with zero trace, while the first quotient is $A\mathrm{/}gA$ and, for $h=1$, the induced endomorphism ${\alpha }_A$ is induced by the identity, that is, ${\alpha }_A={\mathrm{i}\mathrm{d}}_{A\mathrm{/}gA}$; hence ${\mathrm{res}}_V(g^{-1}\mathrm{d}g)={\mathrm{Tr}}_{A\mathrm{/}gA}(\mathrm{i}\mathrm{d})={\dim }_k(A\mathrm{/}gA)$ by [F7].

Claims (1)-(4) are established: (1) in steps 4.1, 4.5 and 4.2; (2) in steps 4.3, 5.1 and 5.3; (3) in steps 4.4, 5.2 and 5.3; and (4) in steps 1.2, 2.2, 3.3, 4.6 and 5.4, where the general two-term formula of part (4) specialises to the case $gA\subseteq A$ with $h=1$; the Axiom of Choice entered only through the choices of the projection in step 1.1 and of the projection $\pi$ used in steps 3.3 and 4.6 (via [F2]).