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$\def\sO{\mathcal{O}}$In Serre's Algebraic Groups and Class Fields, Ch. II, no. 7, we find the definitions of (i) the meromorphic differentials on an algebraic variety $X$ (this is denoted $D_k(F)$ below; Serre calls them rational differentials) and (ii) assuming $\dim X=1$, the order of a meromorphic differential form $\omega\in D_k(F)$ at a simple (i.e., non-singular) point $P\in X$ (denoted $v_P(\omega)$ below):

Recall briefly the general notion of a differential on an algebraic variety $X$: First of all, if $F$ is a commutative algebra over a field $k$, we have the module of $k$-differentials of $F$, written $D_k(F)$; it is an $F$-module, endowed with a $k$-linear map $$ d: F \rightarrow D_k(F), $$ satisfying the usual condition $d(x y)=x \cdot d y+y \cdot d x$. The $d x$ for $x \in F$ generate $D_k(F)$ and $D_k(F)$ is the "universal" module with these properties. For more details, see [11], exposé 13 (Cartier).

These remarks apply in particular to the local rings $\mathcal{O}_P$ and to the field of rational functions $F=k(X)$ of an algebraic variety $X$ (of any dimension $r$). Reducing to the affine case, one immediately checks that the $\underline{\Omega}_P=D_k\left(\mathcal{O}_P\right)$ form a coherent algebraic sheaf on $X$; furthermore $$ D_k(F)=D_k\left(\mathcal{O}_P\right) \otimes_{ \mathcal{O}_P} F . $$ If $P$ is a simple point of $X$ and if $t_1, \ldots, t_r$ form a regular system of parameters at $P$, the $d t_i$ form a basis of $D_k\left(\mathcal{O}_P\right)$; this can be seen, for example, by applying thm. 5 of exposé 17 of the Seminar cited above. Thus the sheaf of $\underline{\Omega}_P$ is locally free over the open set of simple points of $X$ (it thus corresponds to a vector bundle which is nothing other than the dual of the tangent space).

Now if we come back to the case of a curve satisfying the conditions of no. 1, we see that, in this case, $D_k(F)$ is a vector space of dimension $1$ over $F=k(X)$ and that the sheaf $\overline{\Omega}$ of the $\underline{\Omega}_P$ is a subsheaf of the constant sheaf $D_k(F)$. If $t$ is a local uniformizer at $P$, the differential $dt$ of $t$ is a basis of the $\sO_P$-module $\underline{\Omega}_P$ and it is also a basis of the vector $F$-vector space $D_k(F)$. Thus if $\omega\in D_k(F)$, we can write $\omega=fdt$, with $f\in F$. Then supposing $\omega\neq 0$, we put $$ v_P(\omega)=v_P(f). $$ one sees immediately that this definition is indeed invariant, i.e., independent of the choice of $dt$.

I don't understand the very last sentence. How does one prove invariance? Suppose $s\in\sO_P$ is another uniformizer, so that $s=ut$, where $u$ is a unit of $\sO_P$, and that for $\omega\in D_k(F)$, we have $fdt=\omega=gds$, where $f,g\in F$. Why is then $v_P(f)=v_P(g)$? One has $ds=udt+tdu$, but I don't know how to exploit this.

Elías Guisado Villalgordo
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  • For reference: it is possible to extend the definition of the order of a meromorphic function to work as well at a singular point of a curve (maybe somehow that of a meromorphic differential too?). More generally, one can define the order of a Cartier divisor on a locally Noetherian scheme at a closed irreducible subset of codimension 1, cf. Görtz, Wedhorn, Algebraic Geometry I, 2nd ed., Ch. 11, (11.13) (the definition is equation (11.13.3)). The Stacks Project defines the order of an invertible meromorphic function at a closed irreducible subset of codimension 1, tags 02RJ and 0EMH (3). – Elías Guisado Villalgordo Oct 26 '23 at 08:48

1 Answers1

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As $\Omega_P$ is generated by $dt$ as an $\mathcal{O}_P$-module, we can write $du=fdt$ for some $f\in\mathcal{O}_P$. Then $ds=(u+tf)dt$, and so the valuation of $u+tf$ is zero, as is the valuation of 1.

KReiser
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  • $\def\sO{\mathcal{O}}$One can only infer $0=v_P(1)=v_P(u+tf)$ after showing well-definedness, right? Suppose $fdt=\omega=gds$ and write $s=ut$, where $u$ is a unit in $\sO_P$; and write $du=hdt$, where $h\in\sO_P$. Then we have $ds=(u+th)dt$ and $fdt=gds=g(u+th)dt$; whence $v_P(f)=v_P(g)+v_P(u+th)$. Hence, we just have to show that $0=v_P(u+th)$. But why so? – Elías Guisado Villalgordo Sep 29 '23 at 12:49
  • Okay, I just understood it: $u+th$ must be a unit in $\mathcal{O}_P$ since otherwise $u=(u+th)-th$ would be a non-unit. – Elías Guisado Villalgordo Sep 29 '23 at 12:52
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    @ElíasGuisadoVillalgordo just use the properties of valuations: $v(u)=0$, $v(th)=v(t)+v(h)\geq 1$, so $v(u+th)=0$ since the definition of a valuation includes $v(a+b)\geq\min(v(a),v(b))$ with equality when the two valuations are different. – KReiser Sep 29 '23 at 13:42
  • @ElíasGuisadoVillalgordo is there anything else you're looking for here? – KReiser Oct 06 '23 at 15:41
  • Nope. Just accepted the answer, thanks . – Elías Guisado Villalgordo Oct 06 '23 at 15:43