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Let $k$ a field and $F$ a finite extension of $k(x)$. Let the rational 1-forms $$Fdx = \{ f dx, f \in F\}=\{ f dg, f,g \in F\} $$

(obeying to the rules of $F$-modules, of $k$-linearity and $d1=0$, $d(gh) = gdh+hdg$, plus the algebraic ones : for example with $F = k(x)[y]/(y^2-x^3-x)$ then the algebraic rule is $0 = d(y^2-x^3-x) = (-3x^2-1)dx + 2ydy$ so that $\frac{dy}{dx} = \frac{3x^2+1}{2y} \in F$ and $FdF = Fdx+F dy = Fdx$)

Let $$\ell : F^* \to F dx, \qquad \ell(u) = \frac{du}{u}$$ It is an homomorphism with kernel $K^*$ and for any non-constant $g \in F^*$ then $u \mapsto \frac{\ell(u)}{dg}$ is a logarithm $F^\times/K^\times \to F$ (where $K = \overline{k} \cap F$)

I don't think I can interpret those things in term of Zariski topology.

Is there a notion of continuity or algebraicity that fits to $\ell $ ? Same question with the map to divisors $\text{div} : F^* \to \text{Div}(F)$ and $Fdx \to \text{Div}(F)$. How should I think to those kind of maps in the context of algebraic varieties ?

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    $\begingroup$ Hi, the formula $u \mapsto \frac{\mathrm{d}u}{u}$ is functorial for morphisms of schemes, so for any $S$-scheme there is a morphism of $S$-group schemes $dlog : \mathbb{G}_{m,S} \to \Omega_{S}^{1}$. See for example Hartshorne, "Algebraic Geometry", III, Exercise 7.4 (c). $\endgroup$ – Minseon Shin Jul 9 at 23:49
  • $\begingroup$ @MinseonShin Hartshorne is too high for me. Are you saying that if $\phi : X\to Y$ is a non-constant morphism of affine curves defined over $k$ then $\phi^* : k(Y) \to k(X)$ is a field embedding and we have an embedding $\Phi^* :k(Y)dk(Y) \to k(X)dk(X),\Phi^* (fdg) = \phi^*(f)d\phi^*(g)$ and $\ell_{k(X)} \circ \phi^* = \Phi^* \circ \ell_{k(Y)}$ ? Or does functoriality add something else ? $\endgroup$ – reuns Jul 10 at 20:39
  • $\begingroup$ Yes, but it's more natural to replace the target of your morphisms $\ell_{X}$ by $\Omega_{X}^{1}$, the module of differentials. $\endgroup$ – Minseon Shin Jul 11 at 3:49

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