In these rough prep-notes, we are working towards the proof of the Riemann-Hurwitz genus formula. “0-geometry” means we want to use only fields and their discrete valuations so that we can port some of this later to number fields.
Before we have seen that any field $K$ of transcendence degree $1$ over $k$ with $K \cap \overline{k} = k$ is really the function field $K = k(C)$ of a smooth projective curve $C$ defined over $k$.
A geometric point $P \in C$ is a discrete valuation ring $\mathcal{O}_P$ in the extended field $K^e = K \otimes \overline{k} = \overline{k}(C)$.
Aim: To determine the genus of $C$ from $K^e$ and the discrete valuation rings $\mathcal{O}_P$.
Divisors: For $f \in K^e$ and $P \in C$ we denote the valuation of $f$ in the discrete valuation ring $\mathcal{O}_P$ by $\operatorname{ord}_P(f)$ (that is, $f = u t^{\operatorname{ord}_P(f)}$ for $t$ is a uniformizer and $u$ a unit in $\mathcal{O}_P$).
We claim that there are only finitely many $P \in C$ such that $\operatorname{ord}_P(f) \not= 0$ and that $\sum_{P \in C} \operatorname{ord}_P(f) = 0$.
We can assume that $f \notin \overline{k}$ and so the subring $\overline{k}[f] \subset K^e$ is a polynomial ring. Let $R$ be the integral closure of $\overline{k}[f]$ in $K^e$ (which is a finite field extension of $\overline{k}(f)$ say of dimension $r$). Then $R$ is a Dedekind domain, projective of rank $r$ over $\overline{k}[f]$ and there are maximal ideals $\mathcal{P}_i$ in $R$ such that
$$(f) = \mathcal{P}_1^{e_1} \cdots \mathcal{P}_s^{e_s}$$
Because the localization of $R$ at $\mathcal{P}_i$ is a discrete valuation ring with residue field $\overline{k}$, each $\mathcal{P}_i$ defines a point $P_i \in C$ and we have $\sum_i e_i = r$.
Similarly, let $S$ be the integral closure of the polynomial algebra $\overline{k}[\frac{1}{f}]$ in $K^e$, then there are maximal ideals $\mathcal{Q}_j$ (corresponding to points $Q_j \in C$) such that
$$\left(\frac{1}{f}\right) = \mathcal{Q}_1^{f_1} \cdots \mathcal{Q}_t^{f_t}$$
and $\sum_j f_j = r$. But then the divisor of $f$ satisfies the claims
$$\operatorname{div}(f) = \sum_{P \in C} \operatorname{ord}_P(f) [P] = \sum_{i=1}^s e_i [P_i] – \sum_{j=1}^t f_j [Q_j]$$
Differentials forms: Consider the $K^e$-vectorspace $\Omega_C$ spanned by all ‘differential forms’ $\mathrm{d}f$ where $f \in K^e$, subject to the usual rules:
- $\mathrm{d}(f+g)=\mathrm{d}f+\mathrm{d}g$ for all $f,g \in K^e$.
- $\mathrm{d}(fg) = f\,\mathrm{d}g + g\,\mathrm{d}f$ for all $f,g \in K^e$.
- $\mathrm{d}a = 0$ for all $a \in \overline{k}$.
We claim that $\Omega_C$ has dimension one. More precisely, if $x \in K^e$ is transcendental over $\overline{k}$ such that $K^e$ is a finite separable field extension of the subfield $\overline{k}(x)$, then $\Omega_C = K^e \mathrm{d}x$.
The proof is a computation. Let $g \in K^e$ have a minimal polynomial over $\overline{k}(x)$ of the form
$$G(Y) = Y^n + f_1 Y^{n-1} + \cdots + f_{n-1} Y + f_n$$
with all $f_i \in \overline{k}(x)$. Now consider these two polynomials in $\overline{k}(x)[Y]$ :
$G_1(Y) = n Y^{n-1} + (n-1) f_1 Y^{n-2} + \cdots + f_{n-1}$, and
$G_2(Y) = Y^n + \frac{\partial f_{1}}{\partial x} Y^{n-1} + \cdots + \frac{\partial f_{n-1}}{\partial x} Y + \frac{\partial f_n}{\partial x}$.
By the above equations among differential forms we get
$$0 = \mathrm{d} G(g) = G_2(g)\,\mathrm{d}x + G_1(g)\, \mathrm{d}g$$
Because $G_1(g) \not= 0$ by separability, it follows that $\mathrm{d}g \in K^e \mathrm{d}x$. Done!
Genus: In particular, if $t$ is a uniformizing parameter of the discrete valuation ring $\mathcal{O}_P$, then for any differential form $\omega \in \Omega_C$ there is a unique $f \in K^e$ such that $\omega = f\,\mathrm{d}t$. We define $\operatorname{ord}_P(\omega) =\operatorname{ord}_P(f)$. Clearly, this number depends only on $\omega$ (and $P$), but not on the choice of uniformizer (check!).
Slightly more involved is the claim that $\operatorname{ord}_P(\omega) \not= 0$ for finitely many $P \in C$. Here’s the idea:
Take $x \in K^e$ such that $K^e$ is a finite separable extension of $\overline{k}(x)$ of dimension $r$, write $\omega = f\,\mathrm{d}x$ and consider the corresponding cover $x\colon C \rightarrow \mathbb{P}^1$. As before, there are at most $r$ points of $C$ lying over a point $Q \in \mathbb{P}^1$.
Now, write $K^e=\overline{k}(x)(\alpha)$ and let $D \in \overline{k}(x)$ be the discriminant of the minimal polynomial of $\alpha$ over $\overline{k}(x)$. Then, away from the finite number of poles and zeroes of $D$, there are precisely $r$ points of $C$ lying over any point $Q \in \mathbb{P}^1$.
So, removing a finite number of points from $C$, in the remaining $P \in C$ we have $f(P) \not= 0,\infty$, $x(P) \not= \infty$ and $x-x(P)$ is a uniformizer of $\mathcal{O}_P$. But in such points we have $\operatorname{ord}_P(\omega) =\operatorname{ord}_P(f\,\mathrm{d}(x-x(P))) = 0$.
The number $\sum_{P \in C} \operatorname{ord}_P(\omega)$ is thus well-defined and we claim that it doesn’t depend on the choice of differential form. For, any other form can be written as $\omega’ = f \omega$ for some $f \in K^e$ and then we have
$$\sum_{P \in C} \operatorname{ord}_P(\omega’) = \sum_{P \in C} (\operatorname{ord}_P(f) +\operatorname{ord}_P(\omega))$$
and we know already that $\sum_{P \in C} \operatorname{ord}_P(f)=0$. The genus $g_C$ of the curve $C$ is then determined from that number by $2g_C – 2 = \sum_{P \in C}\operatorname{ord}_P(\omega)$.
Example: Take the projective line $\mathbb{P}^1$ corresponding to the purely transcendental field $\overline{k}(x)$ and consider $\omega = \mathrm{d}x$. In a point $\alpha \not= \infty$ we know that $x-\alpha$ is a uniformizer, so
$$\operatorname{ord}_{\alpha}(\omega) = \operatorname{ord}_{\alpha}(\mathrm{d}x) = \operatorname{ord}_{\alpha}(\mathrm{d}(x-\alpha)) = 0$$
In $\infty$ the uniformizer is $\frac{1}{x}$, whence
$$\operatorname{ord}_{\infty}(\omega) =\operatorname{ord}_{\infty}(\mathrm{d}x) =\operatorname{ord}_{\infty}\left(-x^2\,\mathrm{d}\left(\frac{1}{x}\right)\right) = -2$$
Thus, $\sum_{P \in \mathbb{P}^1} \operatorname{ord}_P(\omega) = -2$ and so the genus of the projective line $g_{\mathbb{P}^1} = 0$.
