The global residue theorem on a smooth proper curve over a perfect field

Reduce to generators and dispose of $g=0$. If $g=0$, then $dg=0$, so $fdg=0$ and every local residue and the abstract residue are zero by their $k$-linearity [F1, F2]. Now take $g\mathrm{\ne }0$. By [F1] the local residues are defined at every closed point; by [F2] the adelic residue on $(V_X,A_X)$ is $k$-linear and defined because [F5] gives $K\subseteq V_X$ and $h{A}_X<A_X$ for every $h\in K$. The elements $fdg$ span ${\mathrm{\Omega }}_{K\mathrm{/}k}^1$ over $k$ by [F2], so it suffices to prove finite support and sum zero for arbitrary $f,g\in K$ with $g\mathrm{\ne }0$.

The adelic residue vanishes. By [F3] applied to the two lattices $A_X$ and the diagonal copy of $K$ (both stable by [F5]) we have ${\mathrm{res}}_{A_X}+{\mathrm{res}}_K={\mathrm{res}}_{K+A_X}+{\mathrm{res}}_{K\cap A_X}$. Each term on the right vanishes: ${\mathrm{res}}_{K+A_X}=0$ because $V_X\mathrm{/}(K+A_X)$ is finite-dimensional by [F5] and the residue vanishes on lattices of finite codimension by [F4]; ${\mathrm{res}}_{K\cap A_X}=0$ because $K\cap A_X=H^0(C,{\mathcal{O}}_C)=k$ by [F5] and [F6], so $K\cap A_X$ is finite-dimensional over $k$ and commensurable with the zero subspace, whose residue is zero by [F4] since zero is a $K$-submodule; and ${\mathrm{res}}_K=0$ because $K$ is a $K$-submodule of $V_X$, again by [F4]. Hence ${\mathrm{res}}_{A_X}(fdg)=0$.

Establish stability of the tail and the finite block decomposition. Choose a finite set $S$ outside which $f$, $g$ and $g^{-1}$ are integral, as in [F5], and put $T=C\setminus S$, $V_T=\{(x_p{)}_{p\in T}:x_p\in A_p\text{ for all but finitely many }p\}$, and $A_T={\prod }_{p\in T}{A}_p$. This is a $K$-module pair: for fixed $h\in K$, its finite pole set and the finite exceptional set for an element of $V_T$ show that their product is integral outside a finite set. More precisely, let $P_h=\{p\in T:{\mathrm{ord}}_p(h)<0\}$, which is finite by the finite-pole calculation in [F5]; if $h=0$ this set is empty. For $p\in P_h$, put $m_p=-{\mathrm{ord}}_p(h)$. The local calculation in [F5] gives ${\dim }_k((h{A}_p+A_p)\mathrm{/}{A}_p)=m_p[\kappa (p):k]<\mathrm{\infty }$, and this quotient is zero for $p\mathrm{\notin }{P}_h$. The coordinate map embeds $(h{A}_T+A_T)\mathrm{/}{A}_T$ into the finite direct sum ${⨁}_{p\in P_h}(h{A}_p+A_p)\mathrm{/}{A}_p$: its kernel is zero because a class whose every coordinate is zero is represented by an element of $A_T$. Thus $h{A}_T<A_T$ for every $h\in K$. The same local calculation proves $h{A}_p<A_p$ for each $p\in S$, so every block pair is stable. By the definitions of the restricted product and product lattices, there are finite direct-sum decompositions $V_X=\left({⨁}_{p\in S}{K}_p\right)\oplus V_T,A_X=\left({⨁}_{p\in S}{A}_p\right)\oplus A_T,$ and the diagonal $K$-action preserves every block.

Split the abstract residue by admissible block lifts and trace additivity. Index the finitely many blocks in step 1.3 by $i$. By [F7] choose a $k$-linear projection ${\pi }_i:V_i\to A_i$ for each block. For $h=f,g$, define $h_i={\pi }_i\circ m_h$, where $m_h$ is multiplication by $h$ on that block. The image of $h_i$ lies in $A_i$, so $h_i\in E_1(A_i)$. Since $h{A}_i<A_i$, choose a finite-dimensional $W_{h,i}$ with $h{A}_i\subseteq A_i+W_{h,i}$. Because ${\pi }_i$ is the identity on $A_i$, $(h_i-m_h)(A_i)\subseteq ({\pi }_i-\mathrm{i}\mathrm{d})(W_{h,i})$, a finite-dimensional space; hence $h_i\equiv m_h(\mathrm{m}\mathrm{o}\mathrm{d}{E}_2(A_i))$. The block sums $\tilde{f}={⨁}_i{f}_i$ and $\tilde{g}={⨁}_i{g}_i$ are therefore admissible lifts for the pair $(V_X,A_X)$: their images lie in $A_X$, and their discrepancies on $A_X$ have finite-dimensional image since there are only finitely many blocks. Their commutator is block diagonal, $\mathrm{\Theta }=[\tilde{f},\tilde{g}]={⨁}_i{\theta }_i$, with ${\theta }_i=[f_i,g_i]$ finite potent by [F2]. For each block choose a positive exponent $N_i$ with ${\theta }_i^{N_i}(V_i)$ finite-dimensional and let $N={\max }_i{N}_i$. Then ${\mathrm{\Theta }}^N(V_X)={⨁}_i{\theta }_i^N(V_i)$ is finite-dimensional, so $\mathrm{\Theta }$ is finite potent. Each block is $\mathrm{\Theta }$-invariant; applying (T2) from [F8] successively to the partial sums of these finitely many blocks gives ${\mathrm{Tr}}_{V_X}(\mathrm{\Theta })={\sum }_{p\in S}{\mathrm{Tr}}_{K_p}({\theta }_p)+{\mathrm{Tr}}_{V_T}({\theta }_T)={\sum }_{p\in S}{\mathrm{res}}_{A_p}(fdg)+{\mathrm{res}}_{A_T}(fdg)\mathrm{.}$ The second equality is the defining commutator formula [F2] on each stable block. By [F6], each point term is the local coefficient-trace residue, so this proves the claimed finite-block decomposition.

Only finitely many local residues are nonzero. If $p\mathrm{\notin }S$, then $f,g\in {\mathcal{O}}_{C,p}$ by construction. By [F9], $fdg$ is regular at $p$, so its coefficient in the completed basis $dt$ has no negative powers; the coefficient-trace formula [F1] gives ${\mathrm{res}}_p(fdg)=0$. Thus the support of the local residues is contained in the finite set $S$.

The tail vanishes and the finite sum is zero. At every $p\in T$, both $f$ and $g$ are integral by the choice of $S$, so pointwise multiplication gives $f{A}_T\subseteq A_T$ and $g{A}_T\subseteq A_T$. It follows also that $fg{A}_T\subseteq A_T$ and $f{g}^2{A}_T\subseteq A_T$, hence the full condition $f{A}_T+fg{A}_T+f{g}^2{A}_T\subseteq A_T$ in [F4] holds (and its stable-$f,g$ special case applies). Therefore ${\mathrm{res}}_{A_T}(fdg)=0$. Combining this with steps 1.2 and 2.1 gives ${\sum }_{p\in S}{\mathrm{res}}_p(fdg)={\mathrm{res}}_{A_X}(fdg)=0$.

Conclude. By step 2.2 the sum over all closed points is the finite sum over $S$, which vanishes by step 3.1. This proves finite support and sum zero for every $fdg$ with $g\mathrm{\ne }0$, while step 1.1 handled $g=0$. Since these generators span ${\mathrm{\Omega }}_{K\mathrm{/}k}^1$ and every local residue and the adelic residue are $k$-linear [F1, F2], the two claims hold for every $\omega \in {\mathrm{\Omega }}_{K\mathrm{/}k}^1$.