# Continuity of the “logarithm” $u \mapsto \frac{du}{udx}$ on function fields

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 ?

• 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). – Minseon Shin Jul 9 at 23:49
• @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 ? – reuns Jul 10 at 20:39
• Yes, but it's more natural to replace the target of your morphisms $\ell_{X}$ by $\Omega_{X}^{1}$, the module of differentials. – Minseon Shin Jul 11 at 3:49