A uniformizer differential generates the module of differentials

(Set-up.) The curve $C$ is geometrically integral of chain dimension one and $C\to \mathrm{Spec}k$ is smooth [F1]; the local ring $A={\mathcal{O}}_{C,p}$ is a Noetherian discrete valuation ring with maximal ideal ${\mathfrak{m}}_p=(t)$ [F4]; and the residue field $\kappa (p)$ is finite separable over $k$ by hypothesis.

(The stalk is finite free.) By [F2] the stalk ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1$ is a finite free $A$-module, say of rank $r\ge 0$, and reduction modulo ${\mathfrak{m}}_p$ identifies ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1{\otimes }_A\kappa (p)$ with $({\mathrm{\Omega }}_{C\mathrm{/}k,p}^1)\mathrm{/}{\mathfrak{m}}_p{\mathrm{\Omega }}_{C\mathrm{/}k,p}^1\cong \kappa (p{)}^{\oplus r}$.

(The cotangent isomorphism.) By [F3] the assignment $[x]\mapsto \mathrm{d}x\otimes 1$ is an isomorphism ${\mathfrak{m}}_p\mathrm{/}{\mathfrak{m}}_p^2\stackrel{\sim }{\to }{\mathrm{\Omega }}_{C\mathrm{/}k,p}^1{\otimes }_A\kappa (p)$, and it carries the class $[t]$ to $\mathrm{d}t\otimes 1$.

(The uniformizer spans the cotangent space.) Since ${\mathfrak{m}}_p=(t)$ by [F4], every element of ${\mathfrak{m}}_p$ is $at$ for some $a\in A$, and ${\mathfrak{m}}_p^2=(t^2)$, so the $\kappa (p)$-vector space ${\mathfrak{m}}_p\mathrm{/}{\mathfrak{m}}_p^2=(t)\mathrm{/}(t^2)$ is one dimensional with basis the class $[t]$.

(Rank one and the basis.) The isomorphism of step 2.2 carries the basis $[t]$ of the one-dimensional space ${\mathfrak{m}}_p\mathrm{/}{\mathfrak{m}}_p^2$ of step 2.3 to $\mathrm{d}t\otimes 1$, so $\mathrm{d}t\otimes 1$ is a $\kappa (p)$-basis of ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1{\otimes }_A\kappa (p)$; comparing with step 2.1 gives $r=1$, so the free module ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1$ has rank one and assertion 1 holds. Write ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1=Ae$ and $\mathrm{d}t=ae$ with $a\in A$; the reduction $\bar{a}\in \kappa (p)$ is the coefficient of $\mathrm{d}t\otimes 1$ in the basis $e\otimes 1$, hence $\bar{a}\mathrm{\ne }0$, so $a$ is a unit of the local ring $A$ and $\mathrm{d}t$ is also a basis, which is assertion 2.

(The generic point.) Let $\eta$ be the generic point of $C$ and $k(C)={\mathcal{O}}_{C,\eta }$ its function field, so that $k(C)$ is the fraction field of the domain $A$. Localizing the free rank-one module ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1=A\mathrm{d}t$ at the zero prime and applying the localization isomorphism of [F5] gives ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1{\otimes }_{A}k(C)\cong {\mathrm{\Omega }}_{k(C)\mathrm{/}k}^1=k(C)\mathrm{d}t$, so $\mathrm{d}t$ is a $k(C)$-basis and every rational differential is uniquely $a\mathrm{d}t$ with $a\in k(C)$; this is assertion 3, and it agrees with the free rank-one statement of [F5].

(Completion.) The $A$-module ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1$ is finitely generated and $A$ is Noetherian by [F4], so [F6] applies with $I={\mathfrak{m}}_p$ and gives an isomorphism ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1{\otimes }_A\widehat{A}\stackrel{\sim }{\to }\widehat{{\mathrm{\Omega }}_{C\mathrm{/}k,p}^1}$ carrying $\mathrm{d}t\otimes 1$ to the image of $\mathrm{d}t$; since ${\mathrm{\Omega }}_{C\mathrm{/}k,p}^1=A\mathrm{d}t$ is free of rank one, its completion is $\widehat{A}(\mathrm{d}t\otimes 1)$, a free rank-one $\widehat{A}$-module with basis the image of $\mathrm{d}t$. This is assertion 4.

(Perfect residue fields and conclusion.) If $k$ is perfect then every algebraic extension of $k$ is separable by [F7], and every closed point $p$ of the finite-type $k$-scheme $C$ has finite residue field $\kappa (p)$ by [F7], so the hypotheses on $\kappa (p)$ hold at every closed point of a smooth integral curve over a perfect field; assertions 1, 2, 3 and 4 are established, and choice [F8] is inherited from the smooth-locally-free theorem [F2] in step 2.1, the DVR theorem [F4] in step 1.1, and the completion theorem [F6] in step 4.2, together with their suppliers. No additional choice is made locally.