Algebraic Function Fields of Genus 0

\( \def\mcal#1{\mathcal{#1}} \def\Z{\mathbb{Z}} \def\P{\mathbb{P}} \DeclareMathOperator{\Div}{Div} \def\R{\mathbb{R}} \)In this post I want to discuss the exercises from our Algebraic Curves course I worked on (and struggled with) last week.

So, first of all: unfortunately, this post will not be entirely self-containing. There are quite a few statements upon which the statement I write about is build upon — some of which I will explain in more detail, while others will only be stated briefly or even only referenced to. However, except for the characterization which this post is about, there will be no “rigorous” proofs. For a more detailed explanation and for proofs of these statements consult Stichtenoth’s book “Algebraic Function Fields and Codes” (which features the same [or at least a similar] notation as this posting).

Some (Fundamental) Preliminaries

The goal of this post is to give a characterization of algebraic function fields of genus 0 — and in order to state and prove this, I’ll introduce some foundational concepts from the theory of algebraic curves. Let $f$ be a bivariate polynomial with coefficients in a field $K$, $f \in K[X,Y]\setminus K$. Then $f$ is called algebraic curve, and all points $(x,y)\in K^2$ with $f(x,y) = 0$ are called $K$-rational points. The set of $K$-rational points of an algebraic curve $f$ is denoted as $C_f(K)$. Furthermore, $C_f$ stands for the set of $\overline{K}$-rational points (i.e. the rational points over the algebraic closure of $K$). A point $P = (\alpha, \beta) \in C_f$ on the curve is called regular point if one of the first-order formal partial derivative of $f$ with respect to $x$ or $y$ does not evaluate to $0$ at $P$.

The fields we are primarily interested in are so-called algebraic function fields, these are defined as field extensions $L/K$ (i.e. $L$ and $K$ are both fields with $L \supseteq K$), such that there is an element $t \in L$ which is transcendental (think of a “variable”) over $K$ with $[L : K(t)] < \infty$. In particular, in this setting $L$ is said to be an algebraic function field over $K$. The connection of these function fields to the $K$-rational points from above is as follows: for irreducible $f \in K[X,Y]$ we can show that $K(C_f)$, the field of regular functions on $C_f$ defined over $K$ (which is defined as the quotient field of the factor ring $K[X,Y]/(f)$), is a function field over $K$ — and vice versa, a function field $L/K$ with transcendental element $t\in L$ where $L/K(t)$ is separable is isomorphic to an appropriate field of rational functions $K(C_f)$.

For a function field $L/K$, we define the field of constants as $\tilde K := \lbrace\alpha \in L \mid \alpha \text{ is algebraic over } K\rbrace$. For the sake of simplicity, we assume that all function fields encountered below are function fields over their field of constants, $L/\tilde K$.

A special case of function fields are so-called rational function fields: a function field $L/K$ is said to be rational if we have $L = K(t)$ for a transcendental $t\in L$. Later on, when we come to the characterization of this special class of function fields of genus 0, rational function fields play a special role.

The next basic concept we need is the notion of valuation rings: let $\mathcal{O}$ with $K\subseteq \mathcal{O} \subset L$ be a ring. Then $\mathcal{O}$ is said to be a valuation ring if for all $x\in L\setminus \lbrace 0\rbrace$ we have $x\in \mathcal{O}$ or $x^{-1} \in \mathcal{O}$. It is easy to show that $\mathcal{O}$ is a local ring, meaning that there is a unique proper maximal ideal $\mathcal{P} \trianglelefteq \mcal{O}$. This so-called valuation ideal is given by all non-units in $\mcal{O}$, that is $\mcal{P} = \lbrace x\in \mcal{O} \mid x = 0 \text{ or } x^{-1} \not\in \mcal{O}\rbrace$ — and it also is a principal ideal. Furthermore, note that for every valuation ring there is a so-called discrete valuation, which is a map $v : L \to \Z \cup \lbrace\infty\rbrace$ that fulfills $v(\alpha) = \infty$ if and only if $\alpha = 0$, $v(\alpha \beta) = v(\alpha) + v(\beta)$, and $v(\alpha + \beta) \geq \min(v(\alpha), v(\beta))$.

For example, when considering rational function fields $L/K$ where $L = K(t)$ a class of valuations can be constructed easily: fix an irreducible element $p\in K[t]$. Then, for $x\in L^\times$ we can write $x$ uniquely as $x = p^r \frac{f}{g}$ for $r\in \Z$, $f, g \in K[t]$ with $(f,g)=1$ and $p \nmid fg$. The function defined as $v_P(x) := r$ and $v_P(0) = \infty$ then is a discrete valuation; the respective place is the principal ideal $(p)$.

These valuation rings play an important role in the context of algebraic curves: it can be shown that in some sense, they correspond to rational points on the curve. The following theorem illustrates that:

Theorem (regular points and valuation rings): Let $f \in K[X,Y]$ be irreducible, $P = (\alpha, \beta) \in C_f$ a rational point on $f$. Then the following statements hold:

  1. If $K(\alpha, \beta)$ is separable over $K$ and if $K(C_f)_P$ (the localization of $K(C_f)$ at $P$, i.e. $\lbrace  \frac{g + (f)}{h + (f)} \mid g, h\in K[X,Y],\ h(\alpha, \beta) \neq 0\rbrace$) is a discrete valuation ring of the function field $K(C_f)/K$, then $P$ is a regular point of $C_f$.
  2. If $P$ is a regular point of $C_f$, then $K(C_f)_P$ is a discrete valuation ring.

Due to this correspondence, it is not entirely irrational to consider the set of all valuation ideals (as these are unique and their respective valuation ring can be constructed from them) — this is called the set of places. We denote the set of places of an algebraic function field $L/K$ as $\mathbb{P}_L$.

Even More Preliminaries — Divisors and Riemann-Roch

We are getting closer to understanding the characterization this post ultimately is about. But beforehand, we need to grasp the concept of the genus — and to do so, we need to know what divisors are.

For a function field $L/K$, the set of divisors is defined as the weak sum $\sum_{P \in \P_L} \Z P$, that is $\Div(L) := \lbrace \sum_{P\in \P_L} n_P P \mid n_P\in \Z,\ \text{only finitely many entries non-zero}\rbrace$. Alternatively, $\Div(L)$ can be seen as the free abelian group generated by $\P_L$.

The degree of a divisor is defined by $\deg(\sum_{P\in \P_L} n_P P) = \sum_{P \in \P_L} n_P \deg P$, where the degree of a place is the degree of the extension $\mcal{O}_P / P$ over the base field $K$ (the factor ring $\mcal{O}_P / P$ is a field because $P$ is a maximal ideal in the corresponding valuation ring $\mcal{O}_P$). On this set of divisors, we may define a partial order “$\leq$” (component-wise comparison; for $D, D’ \in \Div(L)$ we have $D \leq D’$ if and only if for all places $P \in \P_L$ the coefficient of $P$ with respect to $D$ is less than or equal to the coefficient of $P$ with respect to $D’$. All divisors $D \in \Div(L)$ with $D \geq 0$ are called positive or effective divisors.

Now, for all elements $x \in L^\times$ we may define their corresponding principal divisor, namely $(x) := \sum_{P\in \P_L} v_P(x) P$. For example, the element $u = \frac{(x+1)^2}{x-5}$ from the rational function field $\R(x)$ has principal divisor $(u) = 2\cdot (x+1) + (-1)\cdot (x-5)$. This also illustrates the idea behind divisors: these objects shall, in some sense, represent the location of zeros and poles of regular functions.

In particular, the valuation functions $v_P$ “measure” this property: if $v_P(x) > 0$, then $x$ is said to have a zero at $P \in \P_L$, and if $v_P(x) < 0$, then $x$ is said to have a pole at $P$. With this in mind, it is easy to define the zero divisor ${(x)}_{0} = \sum_{P\in \P_L,\ v_P(x) > 0} v_P(x) P$ as the sum of all weighted places where $x$ has a zero (i.e. $v_P(x) > 0$), and the pole divisor ${(x)}_{\infty}$ analogously.

Finally, we also need the notion of the so-called Riemann-Roch spaces. For $D\in \Div(L)$ set $\mcal{L}(D) := \{ x\in L^\times \mid (x) \geq -D\}$. It is easy to show that these spaces are actually $K$-vector spaces, and thus we may also define the dimension of a divisor $\dim D := \dim_K \mcal{L}(D)$.

The definition of divisors now has some immediate consequences, for example if the Riemann-Roch space $\mcal{L}(D)$ is not equal to $\{0\}$ (i.e. $\dim D > 0$), then $D$ is equivalent to a positive divisor $D’ \geq 0$. This means that there is a principal divisor $(x)$ such that $D’ = D  + (x)$. Furthermore, a quite important result is that the degree of the zero (or the pole) divisor is equal to a field extension degree, concretely: if $x \not\in K$, then $\deg (x)_0 = \deg (x)_\infty = [L : K(x)]$. In particular, this also means that the degree of a principal divisor is always zero.

For divisors $A\in \Div(L)$ with $\deg A \geq 0$ we can show that between the degree and the dimension, there is the relation $\dim A \leq \deg A + 1$. And at this point, the genus of a function field comes into play: we define the genus $g$ of $L/K$ as

$$ g :=  \max\{ \deg A + 1 – \dim A \mid A \in \Div(L)\}. $$

The existence of the genus is guaranteed, and we have $g \in \mathbb{N}_0$. As an example: it can be shown that rational function fields $L/K$ with $L = K(t)$ have genus $0$ (which is already part of the characterization! 🙂 )

By definition of the genus we have $\dim A \geq \deg A + 1 – g$. The remaining difference between the left and the right hand side is called index of speciality, and it turns out that this index is equal to $\dim(W-A)$, where $W$ is a so-called canonical divisor (these are particular divisors of degree $2g-2$ and dimension $g$ that always exist). Overall, we may write

$$ \dim A = \deg A + 1 – g + \dim(W-A), $$

which is the Riemann-Roch theorem. These are (more or less) all the preliminaries necessary for the following proposition.

Characterization of Function Fields of Genus 0

Proposition. Let $L/K$ be an algebraic function field. Then the following conditions are equivalent:

  1. The genus $g$ of $L/K$ is $0$.
  2. There is a divisor $A \in \Div(L)$ with $\deg A = 2$ and $\dim A = 3$.
  3. There is a divisor $A \in \Div(L)$ with $\deg A \geq 1$ and $\dim A > \deg A$.
  4. There is a divisor $A \in \Div(L)$ with $\deg A \geq 1$ and $\dim A = \deg A + 1$.
  5. If additionally $\operatorname{char} K \neq 2$, then this condition is also equivalent to the above: there are elements $x, y\in L$ such that $L = K(x,y)$ and $y^2 = ax^2 + b$ where $a,b\in K^\times$.

So, let us prove this equivalency. We will start with the first four properties (which are relatively easy to prove) and then treat the (e) separately, as this requires more attention.

Proof. First, from (a) to (b). Assume that $g = 0$. Then, every canonical divisor $W\in \Div(L)$ has degree $2g – 2 = -2$ — which, in turn, means that $-W$ has degree $2$. By the Riemann-Roch theorem (if the degree of a divisor is sufficiently high ($\geq 2g-1$), the speciality index is $0$ and we always have $\dim A = \deg A + 1 -g$), the dimension of $-W$ thus is $3$, which proves (b).

Property (b) directly implies (c), so nothing to do here. When assuming (c), we make use of the inequality $\dim A \leq \deg A + 1$ and thus arrive at (d).

Finally, from (d) back to (a): To prove this, we need another result: assume that $g > 0$ and $\dim A > 0$. Then $\dim A = \deg A + 1$ holds if and only if $A$ is a principal divisor. One direction of this equivalence is implied directly by the Riemann-Roch theorem and the fact that canonical divisors have degree $g$ — and the other direction follows from Clifford’s theorem, which states that for a divisor $A$ with $0\leq \deg A\leq 2g-2$ the relation $\dim A \leq 1 + \frac{1}{2} \deg A$ holds. Back to our original setting: assuming (d), we have $\dim A = \deg A + 1$ — but $A$ cannot be a principal divisor, because then we would have $\deg A = 0$, but we have $\deg A \geq 1$. Therefore, $g > 0$ cannot hold — which yields $g = 0$.

The remaining part of this post deals with the equivalence of property (e), so assume that $\operatorname{char} K \neq 2$.

Assume (b), so let $A\in \Div(L)$ be a divisor of dimension 3 and degree 2. Without loss of generality we may assume that $A$ is positive (otherwise we can switch to an equivalent positive divisor). Then the following two cases arise: either $A = P + Q$ for two places $P, Q \in \P_L$, each of degree 1. Or $A$ itself is a place in $\P_L$ with degree 2.

If $A$ is a sum of two (not necessarily distinct) places of degree 1, we know that there is a divisor of degree 1. It can be shown that this is equivalent to $L/K$ being a rational function field, so there is a $t\in L$, transcendental over $K$ such that $L = K(t)$. In this case, define $x = t – \frac{1}{t}$ and $y = t + \frac{1}{t}$, then we have $y^2 = 1\cdot x^2 + 4$, and because $L = K(t)$ also $L = K(x,y)$.

In the other case, $A$ itself is a place of degree 2 and dimension 3. As every non-constant $x \in L\setminus K$ has to have at least one pole at some place, every non-constant $x$ in $\mcal{L}(A)$ has to have a simple pole in $A$. This yields $(x)_\infty = A$, and therefore also $[L : K(x)] = 2$. Because of $\dim A = 3$, we can find a basis $\{1, x, y\}$ of $\mcal{L}(A)$. If we can show that $y \not\in K(x)$, then because of $[L : K(x)] = 2$ we have $L = K(x,y)$.

First, assume that $y \in K(x)$, that is $y = \frac{f(x)}{g(x)}$ for coprime polynomials $f,g\in K[x]$, $g$ not constant. Then

$$ [K(x) : K(y)] = \max (\deg f, \deg g). $$

This can be seen by factoring $f = \prod_i p_i^{\alpha_i}$ and $g = \prod_j q_j^{\beta_j}$, and then making use of the fact that $[K(x) : K(y)] = \deg (y)_\infty$. This yields

$$ [K(x) : K(y)] = \deg (y)_\infty = \underbrace{\sum_j \beta_j\cdot \deg Q_j}_{= \deg g}  + \max(\deg f – \deg g, 0) = \max(\deg f, \deg g). $$

In particular, we used that $K(x)$ is a rational function field and thus all places are the ideals generated by irreducible polynomials (thus $\deg Q_j = \deg (q_j) = \deg q_j$, additionally to the place at $\infty$ (which contributes the summand $\max(\deg f – \deg g, 0)$)).

We can use this as follows: if $K(x) = K(y)$, then we also have $[K(x) : K(y)] = 1$, and thus $y = \frac{ax + b}{cx + d}$ with $ad – bc \neq 0$ (otherwise we have a constant). However, this is equivalent to $cxy = ax + b – dy \in \mcal{L}(A)$, because of the vector space property. But as $xy$ has a pole of order at least 2 in $A$, we have to have $c = 0$. Finally, this means that we have a non-trivial linear combination of $1$, $x$, and $y$ that gives $0$ — which is a contradiction to $\{1,x,y\}$ being a basis of $\mcal{L}(A)$. Therefore we have $y\not\in K(x)$, and thus $L = K(x,y)$.

By investigating the divisor $2A$, we find that by the Riemann-Roch theorem, $\mcal{L}(2A)$ has dimension 5, and it is easy to see that the following set is contained in this space: $\{1, x, y, xy, x^2, y^2\}$. These are 6 elements in a 5-dimensional space, so there have to be coefficients $a,b,c,d,e,f$ (not all of them 0) such that

$$ ax^2 + by^2 + cxy + dx + ey + f = 0 $$

holds. If $a = 0$, then $cy + d \neq 0$ is not possible because it would imply $x \in K(y)$. However, $cy + d = 0$ is also not possible because $y\not\in K$. Therefore, $a\neq 0$ and — by an analogous argument — $b \neq 0$. Without loss of generality we assume $b = 1$.

Linear transformations of $x$ and $y$ (completing the squares) finally yields $\tilde x$ and $\tilde y$ with $L = K(\tilde x, \tilde y)$ where we also have $\tilde y^2 = a \tilde x^2 + b$ for $a, b \in K$. Obviously, $b = 0$ is not possible, otherwise (as $\sqrt{a} \in K$, because $K$ is the full field of constants) we find $x,y\in K$ — and for the same reason, $a = 0$ is not possible. Therefore, $a,b\in K^\times$, and this proves (e).

For the last part we have to show that (e) is equivalent to one of the previous statements. We choose to show that $g = 0$.

So, assume that $L = K(x,y)$ with $y^2 = ax^2 + b$ for $a,b\in K^\times$. If there is a $t\in L$ such that $L = K(t)$, then we immediately know $g = 0$ (the genus of rational function fields is always 0). Otherwise we consider some place $P_\infty$ where $x$ has a pole. For $u \in \mathbb{N}$, we consider the divisor $uP_\infty$. We find $\deg(u P_\infty) = 2u$, because $(x)_\infty = [K(x,y) : K(x)] = 2$, as $Y^2 – ax^2 – b$ acts as a minimal polynomial for $y$ and because there can’t be any places of degree 1 (otherwise we have a rational function field).

The Riemann-Roch space $\mcal{L}(u P_\infty)$ now contains (amongst other elements) all polynomials $f(x) + y g(x)$ where $\deg f \leq u$ and $\deg g\leq u-1$. This results in at least $2u+1$ basis elements, and thus $\dim(u P_\infty) \geq 2u+1$. For sufficiently large $u$, the Riemann-Roch theorem degenerates to

$$ \dim u P_\infty = \deg u P_\infty + 1 – g, $$

which yields $2u + 1 \leq 2u + 1 – g$ or, equivalently, $g \leq 0$. But as $g \in \mathbb{N}_0$ we have shown that $g = 0$, and thus (e) implies (a).

$\square$


I find this statement to be quite interesting, because (apart from $\operatorname{char} K  = 2$), it gives a concise description of algebraic curves of genus 0. This is more complicated in case of $g \geq 1$: the corresponding curves are (hyper-)elliptic curves, and they are practically relevant because of their applications in Elliptic Curve Cryptography. In this context, the structure of algebraic function fields of genus 0 is interesting as well: the proposition proves that all function fields of genus 0 are either rational function fields, or have the structure $L = K(x,y)$ where $x$ and $y$ are connected via a quadratic equation. The condition $\operatorname{char} K \neq 2$ was only needed to simplify the quadratic equation (completing the squares).

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Comments

  • abhishek

    Hi,
    At the end when you are trying to characterize all function fields of genus zero (for char K\neq 2), the divisor associated with y seems inaccurate to me. The valuation associated to place at infinity is f/g goes to deg(g) – deg(f). In this case, since y has a non-negative valuation at each place other than that of A, we must have deg(g)\geq deg(f) and this inequality can be strict.

    • behackl

      Hi!

      As far as I understand your concern, the following argument should help:

      In case of $K(x) = K(y)$, we have $[K(x) : K(y)] = 1$, implying $\max(\deg f, \deg g) = 1$ and thus $y = \frac{ax +b}{cx +d}$.

      However, in case $K(x) \neq K(y)$, as we assume $y\in K(x)$, we find $[K(x) : K(y)] \geq 2$, and thus $[L : K(y)] = [L : K(x)] \cdot [K(x) : K(y)] \geq 4$. Simultaneously, we know that $y$ is not constant and lies in $\mcal{L}(A)$, implying that $[L : K(y)] = 2$, which contradicts the inequality from before. Thus, $K(x) = K(y)$ has to hold.

      Thanks for reading!

      HTH, BH

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