Are Weil differentials on a Function Field same as the differentials on the corresponding Riemann surface?

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I was reading "Number Theory on Function Fields" by Michael Rosen, and there is a notion of "Weil differentials" on a Function Field. It intuitively seems to me that a function field should correspond to a non-singular algebraic curve. Even if that is not the case, a non-singular algebraic curve $f(x, y)$ over $\mathbb{C}$ would definitely give us a function field corresponding to $\mathbb{C}[x,y]/(f(x,y))$. Corresponding to $f(x,y)$, we'll also have a Riemann surface. Is the notion of Weil differential on this Function field the same as the notion of differential forms on the Riemann surface which is corresponding to the curve $f(x,y) = 0$?

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Question: "Is the notion of Weil differential on this Function field the same as the notion of differential forms on the Riemann surface which is corresponding to the curve $f(x,y)=0$?"

Answer: Let $K$ be the complex number field and $A:=K[x,y]/(f)$ with $C:=Spec(A)$, and assume $A$ is a domain with quotient field $K(A)$. There is an isomorphism

$$\Omega^1_{K(A)/K} \cong S^{-1}\Omega^1_{A/K}:=(\Omega^1_{A/K})_{\eta}$$

where $\eta:=(0) \in A$ is the generic point and $S:=A-(0)$. There is a canonical map

$$i:\Omega^1_{A/K} \rightarrow \Omega^1_{K(A)/K},$$

but $i$ is not an isomorphism in general. An "algebraic differential form" is an element $\omega\in \Omega^1_{A/K}$ and a "meromorphic algebraic differential" is an element of $\omega_m \in \Omega^1_{K(A)/K}.$

question: "Even if that is not the case, a non-singular algebraic curve $f(x,y)$ over $C$ would definitely give us a function field corresponding to $C[x,y]/(f(x,y))$."

To speak about differential forms on the associated Riemann surface, you must construct the corresponding complex manifold. A meromorphic differential form on the associated Riemann surface is defined using holomorphic functions. A meromorphic algebraic differential form is defined using rational functions and these do not correspond 1-1 to each other.