The abstract residue computes the coefficient-trace residue at every closed point

(The monomial residues.) In the pair $(V,A)=(k((t)),k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}])$ one has $\mathrm{res}(t^n\mathrm{d}t)=0$ for every integer $n\ge 0$ and every integer $n\le -2$ by the power-residue property [F3], applied to $f=t$; and $\mathrm{res}(t^{-1}\mathrm{d}t)={\dim }_k(k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]\mathrm{/}tk[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}])=1$ by the unit property [F3]. Hence $\mathrm{res}(t^n\mathrm{d}t)={\delta }_{n,-1}$ in every characteristic, and by [F2] also $\mathrm{res}(h_1\mathrm{d}{h}_2)=0$ whenever $h_1,h_2\in k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$.

(The extension setup.) The local pair at $p$ is the pair over $k$ with $K^{\mathrm{\prime }}=\kappa (p)((t))$, $V^{\mathrm{\prime }}=K^{\mathrm{\prime }}=\kappa (p)((t))$ and $A^{\mathrm{\prime }}=\kappa (p)[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$; its residue is ${\mathrm{res}}_p$ on ${\mathrm{\Omega }}_{\kappa (p)((t))\mathrm{/}k}^1$, and the completion of the curve at $p$ identifies $\mathrm{d}t$ with the basis differential and $k(C)$ with a subfield of $\kappa (p)((t))$ by [F4]. The field $\kappa (p)$ is finite separable over $k$ by perfectness of $k$ ([F6]), so $\kappa (p)((t))=\kappa (p){\otimes }_{k}k((t))$ is a free $k((t))$-module of finite rank $d=[\kappa (p):k]$ with basis any $k$-basis $x_1,\dots ,x_d$ of $\kappa (p)$.

(Part (1) for finite Laurent polynomials.) Let $f={\sum }_n{a}_n{t}^n$ and $g={\sum }_m{b}_m{t}^m$ be finite sums. The universal derivation satisfies $\mathrm{d}(t^m)=m{t}^{m-1}\mathrm{d}t$, hence $f\mathrm{d}g={\sum }_{n,m}m{a}_n{b}_m{t}^{n+m-1}\mathrm{d}t$, a finite sum. By $k$-linearity of $\mathrm{res}$ and step 1.1 only the terms with $n+m-1=-1$ survive, each with residue its displayed coefficient. Thus $\mathrm{res}(f\mathrm{d}g)={\sum }_{n+m=0}m{a}_n{b}_m$, the coefficient of $t^{-1}$ in $fDg$.

(Applying the extension formula.) Take $K=k((t))$, $V=K$, $A=k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$, $K^{\mathrm{\prime }}=\kappa (p)((t))$ and $V^{\mathrm{\prime }}=K^{\mathrm{\prime }}$ as in step 1.2, with the $K$-basis $x_1,\dots ,x_d$ of $K^{\mathrm{\prime }}$ given by a chosen $k$-basis of $\kappa (p)$. If $f\in K$ has order $r$, then $fA=t^{r}A$; for $r\ge 0$ this is contained in $A$, while for $r<0$ the quotient $fA\mathrm{/}A$ has finite $k$-dimension $-r$. Thus $fA<A$ for every $f\in K$. The associated subspace $A^{\mathrm{\prime }}={\sum }_i{x}_i\otimes A$ identifies with $\kappa (p){\otimes }_{k}k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]=\kappa (p)[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$, the lattice of the pair in step 1.2. Hence [F5] applies and gives ${\mathrm{res}}_{V^{\mathrm{\prime }}}(f\mathrm{d}g)={\mathrm{res}}_V({\mathrm{Tr}}_{K^{\mathrm{\prime }}\mathrm{/}K}(f)\mathrm{d}g)$ for every $f\in K^{\mathrm{\prime }}$ and $g\in K$.

(Part (1) in general.) Let $f,g\in k((t))$ have lower exponents $-M,-N$, respectively. Choose $R\ge \max (M,N)$ and truncate both: $f=f_0+r_f$, $g=g_0+r_g$, with $f_0,g_0$ finite Laurent polynomials and $r_f=t^{R+1}u$, $r_g=t^{R+1}v$ for $u,v\in k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$. First, $r_f\mathrm{d}{g}_0$ is a finite sum of terms $m{b}_m(r_f{t}^{m-1})\mathrm{d}t$; each coefficient is regular since $m\ge -N$ and $R\ge N$, so its abstract residue is zero by [F2], applied to $h\mathrm{d}t=h\mathrm{d}t$ with $h,t\in k[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$. Next, for each monomial $a_n{t}^n$ of $f_0$, where $n\ge -M$, use $\mathrm{d}{r}_g=(R+1)t^{R}v\mathrm{d}t+t^{R+1}\mathrm{d}v$. Both $(R+1)a_n{t}^{n+R}v$ and $a_n{t}^{n+R+1}$ are regular, so the two terms of $a_n{t}^n\mathrm{d}{r}_g$ have zero residue by [F2], applied respectively to a regular coefficient times $\mathrm{d}t$ and to $a_n{t}^{n+R+1}\mathrm{d}v$ with both factors regular. Finally $r_f,r_g$ are regular, so [F2] gives $\mathrm{res}(r_f\mathrm{d}{r}_g)=0$. By bilinearity, $\mathrm{res}(f\mathrm{d}g)=\mathrm{res}(f_0\mathrm{d}{g}_0)=[t^{-1}](f_{0}D{g}_0)$ by step 2.1. The same truncation does not change the formal coefficient: $r_{f}D{g}_0$, $f_{0}D{r}_g$, and $r_{f}D{r}_g$ all have order at least $0$, since $R\ge M,N$. Therefore $[t^{-1}](f_{0}D{g}_0)=[t^{-1}](fDg)$, proving the coefficient formula for arbitrary Laurent series without identifying ${\mathrm{\Omega }}_{K\mathrm{/}k}^1$ with $K\mathrm{d}t$.

(Coefficientwise trace.) For $f={\sum }_n{a}_n{t}^n\in \kappa (p)((t))$, multiplication by $f$ on the finite-dimensional $k((t))$-vector space $K^{\mathrm{\prime }}$ has matrix ${\sum }_n{t}^n{M}_{a_n}$ in the fixed basis $x_1,\dots ,x_d$, where $M_{a_n}$ is the matrix of multiplication by $a_n$ on $\kappa (p)$ over $k$. Since the matrix has finitely many diagonal entries, its trace is ${\sum }_n{\mathrm{tr}}_k(M_{a_n})t^n={\sum }_n{\mathrm{Tr}}_{\kappa (p)\mathrm{/}k}(a_n)t^n$. This is the coefficientwise trace formula in [F6]; the basis is $\{x_i\}$, not a family indexed by powers of $t$.

(Conclusion.) Let $\omega =f\mathrm{d}t$ with $f\in k(C)\subseteq \kappa (p)((t))$, expanded as ${\sum }_n{a}_n{t}^n$, and take $g=t\in k((t))$. By steps 1.2 and 2.2, ${\mathrm{res}}_p(\omega )={\mathrm{res}}^V({\mathrm{Tr}}_{K^{\mathrm{\prime }}\mathrm{/}K}(f)\mathrm{d}t)$; by step 3.2 and part (1) of step 3.1 applied in the base field, the right side is the coefficient of $t^{-1}$ in ${\mathrm{Tr}}_{K^{\mathrm{\prime }}\mathrm{/}K}(f)$, namely ${\mathrm{Tr}}_{\kappa (p)\mathrm{/}k}(a_{-1})$; this is exactly the coefficient-trace residue of [F4], so the two definitions agree at $p$, and by perfectness of $k$ this holds at every closed point of $C$. The Axiom of Choice [F7] is inherited from the residue suppliers.