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0-geometry: Curves

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This entry is part 1 of 4 in the series prep notes

In this series we collect our rough prep-notes for the lectures ahead. We focus on the main ideas and give precise references. More complete course-notes may follow afterwards.

Fix a perfect field $k$ (say a finite field) with algebraic closure $\overline{k}$ and absolute Galois group $G = \operatorname{Gal}(\overline{k}/k)$.

Aim: to study smooth projective $k$-curves via their function fields. This will allow us later to associate ‘curves’ to number fields.

We need two categories:

$\mathsf{Curves}/k$

The objects are smooth projective algebraic curves defined over $k$ (that is, a smooth closed subvariety $C$ of dimension one of some projective space $\mathbb{P}^n(\overline{k})$ defined by a set of homogeneous polynomials all of their coefficients belonging to $k$). We will call such objects curves defined over $k$.

The morphisms will be surjective algebraic maps $C \to C’$ defined over $k$ (that is, all coordinate functions have their coefficients in $k$). Remember that any non-constant rational map between two curves is automatically surjective. We will call such morphisms covers.

$\mathsf{1Fields}/k$

The objects are field extensions $K$ of $k$ of transcendence degree one with $k$ as their ‘field of constants’. That is, $K \cap \overline{k} = k$.

The morphisms will be field inclusions $K \hookrightarrow K’$ fixing $k$.

Main result: These categories are (anti)-equivalent to each other.

Details are in section I.6 of Robin Hartshorne’s Algebraic Geometry when $k = \overline{k}$ and modifications for the general case are in section II.2 of Joseph Silverman’s The Arithmetic of Elliptic Curves.

Sketch of proof: The direction from curves to fields is straightforward.

The contravariant functor $\mathsf{Curves}/k \longrightarrow \mathsf{1Fields}/k$ assigns to a curve $C$ its function field $k(C)$ (the field consisting of all rational functions $f:~C \rightarrow \overline{k}$ defined over $k$).

This functor associates to a cover $\phi:~C \mapsto C’$ the field-inclusion $\phi^{\ast}:~k(C’) \rightarrow k(C)$ obtained by composition (that is, $\phi^{\ast}(f) = f \circ \phi~:~C \rightarrow \overline{k}$ for all $f \in k(C’)$).

Conversely, the contravariant functor $\mathsf{1Fields}/k \longrightarrow \mathsf{Curves}/k$ assigns to a field $K$ of transcendence degree one

  • the geometric points $C(\overline{k})$ of the curve $C$, which is the set of all discrete valuations rings in $K \otimes \overline{k}$ with residue field $\overline{k}$. The Galois group $G$ acts on this set, and,
  • the schematic points of $C$ are the $G$-orbits of this action. Equivalently, these are the discrete valuation rings of $K$ with residue field a finite field extension $L$ of $k$. The degree of such a scheme-point is the size of the $G$-orbit (or the $k$-dimension of the residue field $L$ of the discrete valuation ring).

Example: Under this equivalence, the purely transcendental field $k(x)$ corresponds to the projective line $\mathbb{P}^1$ over $k$. Its geometric points $\mathbb{P}^1(\overline{k})$ are the points

$$\{ [\alpha : 1]~:~\alpha \in \overline{k} \} \cup \{ \infty = [1 : 0] \}$$

The discrete valuation ring of $\overline{k}(x)$ corresponding to $[\alpha : 1]$ has uniformizing parameter $x-\alpha$ and the one corresponding to $\infty$ has uniformising parameter $\tfrac{1}{x}$.

The Galois group fixes $\infty$ and acts on the point $[\alpha : 1]$ as it does on $\alpha \in \overline{k}$. Hence, the schematic points of $\mathbb{P}^1$ are $\infty$ together with all irreducible monic polynomials in $k[x]$.

Under the equivalence, the set of all non-constant maps $C \mapsto \mathbb{P}^1$ corresponds to the set of all $k$-field morphisms $k(x) \hookrightarrow k(C)$ and as these are determined by the image of $x$ they are determined by $f \in k(C)$. The cover corresponding to $f$

$$C(\overline{k}) \mapsto \mathbb{P}^1(\overline{k}) \quad \text{maps} \quad P \mapsto [f(P) : 1]$$

if $f$ is regular in $P$ and to $\infty$ otherwise.

About lieven le bruyn

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0-geometry: Genus

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This entry is part 2 of 4 in the series prep notes

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 $g_{\mathbb{P}^1} = 0$.

About lieven le bruyn

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0-geometry: Hurwitz

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This entry is part 3 of 4 in the series prep notes

The genus formula of Bernhard Riemann (left) and Adolf Hurwitz (right) asserts that if $\phi: C_1 \rightarrow C_2$ is a separable cover of curves, we have an inequality relating their genera:

$$2 g_{C_1} – 2 \geq \deg(\phi)(2 g_{C_2} -2) + \sum_{P \in C_1} (e_{\phi}(P)-1)$$

Let’s first understand all these terms. The cover $\phi: C_1 \rightarrow C_2$ is separable if the induced field-extension $\phi^{\ast}(\overline{k}(C_2)) \subset \overline{k}(C_1)$ is finite and separable. The dimension $[\overline{k}(C_1) : \phi^{\ast}(\overline{k}(C_2))]$ is called the degree of $\phi$. By using a discriminant argument as in these notes we know that for all but finitely many points $Q \in C_2$ there are exactly $\deg(\phi)$ points of $C_1$ lying over it.

In general, let $P \in C_1$ with corresponding discrete valuation ring $\mathcal{O}_P$ in $\overline{k}(C_1)$, then $\mathcal{O}_P \cap \phi^*(\overline{k}(C_2))$ is a discrete valuation ring in $\overline{k}(C_2) \simeq \phi^*(\overline{k}(C_2))$ and thus of the form $\mathcal{O}_Q$ for some $Q \in C_2$. Naturally we have $\phi(P)=Q$.

If $R$ is the integral closure of $\mathcal{O}_Q$ in $\overline{k}(C_1)$, then $R$ is a semi-local Dedekind domain and a PID. If $t_Q$ is a uniformizer of $\mathcal{O}_Q$ we have

$$(t_Q) = P_1^{e_1} \cdots P_r^{e_r}$$

where the $P_i$ are the maximal ideals of $R$ which corresponds to points $P_i \in C_1$. The integer $e_i$ is called the ramification index of $\phi$ in $P_i$ and will be denoted $e_{\phi}(P_i)$. Clearly we have that $\deg(\phi) = \sum_i e_{\phi}(P_i)$.

Further, for almost all $P \in C_1$ we will have $e_{\phi}(P)=1$. With these notations we can now begin the

Proof of the Riemann-Hurwitz inequality: Because $\phi$ is separable, we have an inclusion

$$\phi^*~:~\Omega_{C_2} \hookrightarrow \Omega_{C_1} \qquad \phi^*(f\,\mathrm{d}x)=\phi^*(f)\,\mathrm{d} \phi^*(x)$$

Take a point $Q \in C_2$ with uniformizer $t_Q \in \mathcal{O}_Q$ and write $\omega = f d t_Q \in \Omega_{C_2}$. For the finitely many $P_i \in C_1$ lying over $Q$ we have (as before) that
$\phi^*(t_Q) = u t_{P_i}^{e_i}$ with $e_i = e_{\phi}(P_i)$ and $u$ a unit in the discrete valuation ring $\mathcal{O}_{P_i}$. But then,

$$\phi^*(\omega) = \phi^*(f)\,\mathrm{d} \phi^*(t_Q) = \phi^*(f)\,\mathrm{d}(u t_{P_i}^{e_i}) = \phi^*(f)\left(e_i u t_{P_i}^{e_i-1} + \frac{\mathrm{d}u}{\mathrm{d}t_{P_i}} t_{P_i}^{e_i}\right)\,\mathrm{d}t_{P_i}$$

The valuation $\operatorname{ord}_{P_i}$ of $e_i u t_{P_i}^{e_i-1}$ is $e_i-1$ (unless $e_i=0$ in $k$, that is $char(k) | e_i$) whereas the valuation of $(\tfrac{\mathrm{d}u}{\mathrm{d}t_{P_i}}) t_{P_i}^{e_i}$ is $\geq e_i$. But then,

$\operatorname{ord}_{P_i}(\phi^* \omega) \geq \operatorname{ord}_{P_i}(\phi^*(f)) + e_i -1 = \operatorname{ord}_Q(f) e_i + e_i – 1$
$= \operatorname{ord}_Q(\omega) e_{\phi}(P_i) + e_{\phi}(P_i) -1$

Summing these inequalities over all $P \in C_1$ we get for the degree of the divisor

$\deg(\operatorname{div}(\phi^*(\omega))) \geq \sum_{P \in C_1} (e_{\phi}(P) \operatorname{ord}_{\phi(P)}(\omega)+e_{\phi}(P) -1)$
$=\sum_{Q \in C_2} (\sum_{P \in \phi^{-1}(Q)} (e_{\phi}(P) \operatorname{ord}_Q(\omega) + e_{\phi}(P) -1))$

and because we already know that $\deg(\phi) = \sum_{P \in \phi^{-1}(Q)} e_{\phi}(P)$ for all $Q \in C_2$, this is equal to

$=(\sum_{Q \in C_2} \deg(\phi) \operatorname{ord}_Q(\omega))+(\sum_{P \in C_1} (e_{\phi}(P)-1))$
$=(\deg(\phi))(\deg(\operatorname{div}(\omega))) + \sum_{P \in C_1} (e_{\phi}(P)-1)$

Plugging in the relation between the genus and the degree of the divisor of a nonzero differential form, we have here the Riemann-Hurwitz inequality!

About lieven le bruyn

Also blogs at NeverEndingBooks.

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