Annihilators of regular sections under the local residue pairing

The pairing is well defined. Trivialize $\mathcal{L}$ near $p$ and pass to the completion as in [F2]. The completed local lattice is $\kappa (p)\mathrm{⟦}t\mathrm{⟧}$, its completed rational fiber is $\kappa (p)((t))$, and the completed stalk of the dual twist is $\kappa (p)\mathrm{⟦}t\mathrm{⟧}dt$ by [F3]; these describe completions of the actual local objects, not equalities with them. If a representative of $c$ is changed by an element of ${\mathcal{L}}_p$, its product with the given regular section $s$ changes by an element of ${\omega }_{C,p}$, so the residue is unchanged because regular differentials have zero residue by [F4]. This gives the pairing displayed in the statement.

Expand in Laurent series. Write $c={\sum }_{n<0}{c}_n{t}^n$ and $s={\sum }_{m\ge 0}{s}_m{t}^{m}dt$ with coefficients in $\kappa (p)$, the first sum having only finitely many nonzero terms and the second a possibly infinite power series in the completed stalk $\kappa (p)\mathrm{⟦}t\mathrm{⟧}dt$. Then $cs={\sum }_{n<0,m\ge 0}{c}_n{s}_m{t}^{n+m}dt$ has coefficient of $t^{-1}$ equal to ${\sum }_{n<0}{c}_n{s}_{-n-1}$, a finite sum, so by the coefficient-trace formula of [F4], ${\mathrm{res}}_p(cs)={\mathrm{Tr}}_{\kappa (p)\mathrm{/}k}\left({\sum }_{n<0}{c}_n{s}_{-n-1}\right)={\sum }_{n<0}{\mathrm{Tr}}_{\kappa (p)\mathrm{/}k}(c_n{s}_{-n-1})$.

Compute the first annihilator. If $c\in {\mathcal{L}}_p$, then its product with every regular $s\in ({\omega }_C\otimes {\mathcal{L}}^{-1}{)}_p$ is regular, so its residue is zero by [F4]. Conversely suppose $c\mathrm{\notin }{\mathcal{L}}_p$, and let $n<0$ be the smallest exponent with $c_n\mathrm{\ne }0$ in its finite negative Laurent part. By nondegeneracy in [F5], choose $a\in \kappa (p)$ with ${\mathrm{Tr}}_{\kappa (p)\mathrm{/}k}(c_{n}a)\mathrm{\ne }0$. Lift $a$ to $\tilde{a}\in {\mathcal{O}}_{C,p}$ using the residue map ${\mathcal{O}}_{C,p}\to \kappa (p)$, and take the regular section $s=t^{-n-1}\tilde{a}dt$; its exponent is nonnegative. In the completed product, the coefficient of $t^{-1}dt$ is exactly $c_{n}a$, because $n$ is the smallest exponent of $c$ and the constant coefficient of $\tilde{a}$ is $a$. Thus ${\mathrm{res}}_p(cs)=\mathrm{Tr}(c_{n}a)\mathrm{\ne }0$. Hence the annihilator of the regular stalk inside ${\mathcal{L}}_{\eta }$ is exactly ${\mathcal{L}}_p$, and the induced pairing is nondegenerate on the left.

Compute the dual annihilator. If $s$ is regular, then its product with every $c\in {\mathcal{L}}_p$ is regular, so the residue is zero by [F4]. Conversely let a rational section $s$ have completed expansion $s={\sum }_{m\ge m_0}{s}_m{t}^{m}dt$ with a nonzero negative coefficient, and choose its smallest exponent $m_0<0$. By [F5] choose $a\in \kappa (p)$ with ${\mathrm{Tr}}_{\kappa (p)\mathrm{/}k}(a{s}_{m_0})\mathrm{\ne }0$. Set $q=-m_0-1\ge 0$, lift $a$ to $\tilde{a}\in {\mathcal{O}}_{C,p}$, and take $c=t^q\tilde{a}\in {\mathcal{L}}_p$. The coefficient of $t^{-1}dt$ in $cs$ is exactly $a{s}_{m_0}$: any positive-degree term of $\tilde{a}$ would pair with a coefficient of $s$ below $m_0$, which is zero. Hence ${\mathrm{res}}_p(cs)=\mathrm{Tr}(a{s}_{m_0})\mathrm{\ne }0$. Therefore the annihilator of ${\mathcal{L}}_p$ inside the rational dual-twist fiber consists exactly of its regular stalk.

Identify the local model. Steps 2.1 and 2.2 use only finite Laurent coefficients: the most negative coefficient of a principal part, or the first negative coefficient of a rational differential, is detected by a regular test section whose residue coefficient is chosen using the nondegenerate trace form of $\kappa (p)\mathrm{/}k$. These finite tests prove the asserted annihilators and make no assertion about the full algebraic duals of the completed power-series or Laurent-series spaces. The argument divides by no integer, so it applies in every characteristic.