On explicit descent of marked curves and maps

Abstract

We revisit a statement of Birch that the field of moduli for a marked three-point ramified cover is a field of definition. Classical criteria due to Dèbes and Emsalem can be used to prove this statement in the presence of a smooth point, and in fact these results imply more generally that a marked curve descends to its field of moduli. We give a constructive version of their results, based on an algebraic version of the notion of branches of a morphism and allowing us to extend the aforementioned results to the wildly ramified case. Moreover, we give explicit counterexamples for singular curves.

In his Esquisse d’un Programme [10], Grothendieck embarked on a study of ${{\mathrm{Gal}}}(\overline{\mathbb Q}\!\mid \!\mathbb Q)$, the absolute Galois group of $\mathbb Q$, via its action on a set with a geometric description: the set of finite morphisms $f:X \rightarrow \mathbb P^1$ of smooth projective curves over $\overline{\mathbb Q}$ unramified away from $\{0,1,\infty \}$, known as Belyĭ maps. The heart of this program is to understand over what number fields a Belyĭ map is defined and when two Galois-conjugate Belyĭ maps are isomorphic. One of the more subtle aspects of this investigation is the issue of descent: whether or not a Belyĭ map is defined over its field of moduli, which is then necessarily a minimal field of definition. Some concrete descent problems were studied by Couveignes [4] and Couveignes–Granboulan [5]; a more general theoretical approach to the subject of descent of (maps of) curves was developed by Dèbes–Douai [6] and Dèbes–Emsalem [7].

By the classical theory of Weil descent, a Belyĭ map with trivial automorphism group descends to its field of moduli. As such, one typically tries to eliminate descent obstructions by a simple rigidification that eliminates as many non-trivial automorphisms as possible. In his article on dessins, Birch claims [2, Theorem 2] that by equipping a Belyĭ map $f : X \rightarrow \mathbb P^1$ with a point $P \in X(\overline{\mathbb Q})$ satisfying $f(P)=\infty$, the marked tuple (X, f; P) descends to its field of moduli [2, Theorem 2]. Birch provides several references to more general descent results, but upon further reflection we could not see how these implied his particular statement. Obtaining a proof or counterexample was not merely of theoretical importance: indeed, in order to determine explicit equations for Belyĭ maps, one needs a handle on their field of definition [27, §7]. Moreover, if the descent obstruction indeed vanishes, then it is desirable to have general methods to obtain an equation over the field of moduli. The issue under consideration is therefore also very relevant for practical purposes.

Fortunately, it turned out that general results of Dèbes–Emsalem [7, §5] imply that a marked projective curve (X; P) descends to its field of moduli as long as the marked point P is smooth, something that holds by definition for a Belyĭ map $f : X \rightarrow \mathbb P^1$ (as X itself is required to be smooth). Their argument then shows equally well that Birch’s statement holds true: a marked Belyĭ map (X, f; P) descends to its field of moduli.

In this paper, we revisit the results of Dèbes–Emsalem [7] from a slightly different point of view that is more explicit from the point of view of Weil cocycles. Our main theorem (Theorem 1.8) is as follows. We define a marked map $(Y,f; Q_1,\dots ,Q_n ; P_1,\dots ,P_m)$ over F to be a map $f : Y \rightarrow X$ of curves over F equipped with distinct points $Q_1,\dots ,Q_n \in Y(F)$ and $P_1,\dots ,P_m \in X(F)$ with $n,m \ge 0$. (For conventions on curves and maps, see Sect. 1.)

FormalPara Theorem A

Let

$$\begin{aligned} (Y, f : Y \rightarrow X; Q_1,\ldots ,Q_n; P_1,\dots ,P_m) = (Y,f;\mathcal {R}) \end{aligned}$$

be a marked map of curves over a separably closed field $F = {F}^{\mathrm {sep}}$ such that $n \ge 1$ and at least one of the points $Q_1 , \ldots Q_n$ is smooth. Then $(Y,f;\mathcal {R})$ descends to its field of moduli.

Theorem A extends the results of Dèbes–Emsalem [7], where the curve X was assumed to be of genus at least 2 and where moreover the order of ${{\mathrm{Aut}}}(X)$ was assumed to be coprime to the characteristic of the base field.

Our approach to proving Theorem A is essentially self-contained, and we recover the result of Birch on descent of marked Belyĭ maps (Theorem 1.10). We define the branches of a morphism, replacing the choice of a tangential base point with something canonical. This approach has several advantages. First, it provides a more unified treatment, extending results to the wildly ramified (but separable) case. Second, the notion is more conducive to constructive applications: by using branches it becomes straightforward to read off a Weil cocycle from the given rigidification, without having to calculate the canonical model of ${{\mathrm{Aut}}}(X) \backslash X$. From the point of view of a computational Esquisse [27], this gain is quite important. Finally, our approach leads us to construct some explicit counterexamples to descent in the case where the marked point P is not smooth, as follows.

FormalPara Theorem B

There exists a marked curve (X; P) over $\mathbb C$ with X projective whose field of moduli for the extension $\mathbb C\!\mid \!\mathbb R$ is equal $\mathbb R$ but such that (X; P) does not descend to $\mathbb R$.

Similarly, there exists a marked map $(X,f:X \rightarrow \mathbb P^1; P; 0,1,\infty )$ over $\mathbb C$ with X projective and f unramified outside $\{0,1,\infty \}$ whose field of moduli for the extension $\mathbb C\!\mid \!\mathbb R$ is equal $\mathbb R$ but that does not descend to $\mathbb R$.

Our paper is organized as follows. In Sect. 1, we introduce the relevant background and notions under consideration, and we give precise statements of the main results of this article. After a review of the classical Weil descent criterion in Sect. 2, we present our theory of branches in Sect. 3. We give some concrete examples in Sect. 4; these include the explicit descent of marked hyperelliptic curves to their field of moduli as well as the descent a marked Klein quartic (X; P) using branches. In Sect. 5, we conclude the paper by presenting our counterexamples in the singular case. In an appendix, we consider descent of marked Galois Belyĭ maps in genus 0.

1 Background, notation, and statements of main results

In this section, we set up the basic background and notation we will use in the rest of the paper.

1.1 Definitions and notation

Let F be a field with separable closure ${F}^{\mathrm {sep}}$ and absolute Galois group $\Gamma _F = {{\mathrm{Gal}}}({F}^{\mathrm {sep}}\!\mid \!F)$. When ${{\mathrm{char}}}F = 0$, we will also write ${F}^{\mathrm {sep}}=\overline{F}$. A curve X over F is a geometrically integral, separated scheme of finite type over F of dimension 1. A map $f :Y \rightarrow X$ of curves over F is a nonconstant morphism defined over F. If X, Y are projective, this implies that f is finite, hence proper. We will assume throughout and without further mention that all maps are generically separable, so the extension of function fields $F(Y) \!\mid \!F(X)$ is separable.

An isomorphism of maps $f:Y \rightarrow X$ and $f':Y' \rightarrow X'$ of curves over F is a pair $(\psi ,\varphi )$ of isomorphisms of curves $\psi :Y \xrightarrow {\sim } Y'$ and $\varphi :X \xrightarrow {\sim } X'$ over F such that $\varphi f = f' \psi$, i.e., such that the diagram

(1.1)

commutes; if f is isomorphic to $f'$ we write $f \xrightarrow {\sim } f'$.

The absolute Galois group $\Gamma _F$ acts naturally on the set of curves and the set of maps between curves defined over ${F}^{\mathrm {sep}}$—in both cases, applying the Galois action of the automorphism $\sigma$ comes down to applying $\sigma$ to the defining equations of an algebraic model of X (resp. f) over ${F}^{\mathrm {sep}}$. The conjugate of a curve X over ${F}^{\mathrm {sep}}$ by an automorphism $\sigma \in \Gamma _F$ is denoted by $\sigma (X)$, and similarly that of a map f by $\sigma (f)$.

Given a curve X over F and an extension $K \!\mid \!F$ of fields, we will denote by $X_K$ the base extension of X to K. We denote the group of of automorphisms of $X_K$ over K by ${{\mathrm{Aut}}}(X) (K)$. This latter convention reflects the fact that ${{\mathrm{Aut}}}(X)$ is a scheme over F, the K-rational points of which are the K-rational automorphisms of X (by work of Grothendieck on the representability of the Hilbert scheme and related functors [12, 4.c]). As in the introduction, we occasionally write ${{\mathrm{Aut}}}(X)$ for the group of points ${{\mathrm{Aut}}}(X) ({F}^{\mathrm {sep}})$ when no confusion is possible, and for curves Y and X over K, we will often speak of an isomorphism between Y and X over K when more precisely an isomorphism between $Y_K$ and $X_K$ over K is meant.

We call a map of curves $f : Y \rightarrow X$ over F geometrically generically Galois if the group $G = {{\mathrm{Aut}}}_X(Y,f)({F}^{\mathrm {sep}})$ acts transitively on the fibers of f, or equivalently if the extension of function fields ${F}^{\mathrm {sep}}(Y) \!\mid \!{F}^{\mathrm {sep}}(X)$ is Galois. (This terminology reflects the fact that in this version the usual Galois torsor property is only required to hold on the generic points, and that the elements of the group G need only be defined over ${F}^{\mathrm {sep}}$.) In what follows, we abbreviate geometrically generically Galois to simply Galois. If $f : Y \rightarrow X$ is a Galois cover, then it can be identified with a quotient $q_H : Y \rightarrow H \backslash Y$ of Y by a finite F-rational subgroup H of G. (This quotient by G exists by work of Grothendieck [11, §6].)

Given a Galois map $f : Y \rightarrow X$ branching over ${P \in X({F}^{\mathrm {sep}})}$, the ramification indices of the points $Q \in f^{-1}(P)({F}^{\mathrm {sep}})$ are all equal, and we call this common value the branch index of the point x on X.

We will now define certain rigidifications of the maps and curves considered above.

Definition 1.1

Let $n \in \mathbb Z_{\ge 0}$. An n-marked curve (or n-pointed curve) $(X;P_1,\dots ,P_n)$ over F is a curve X equipped with distinct points $P_1,\dots ,P_n \in X(F)$. An isomorphism of n-marked curves $(X;P_1,\dots ,P_n) \xrightarrow {\sim } (X';P_1',\dots ,P_n')$ over F is an isomorphism $\varphi : X \xrightarrow {\sim } X'$ over F such that $\varphi (P_i)=P_i'$ for $i=1,\dots ,n$.

We will often use the simpler terminology marked curve for a 1-marked curve.

Definition 1.2

For $n,m \in \mathbb Z_{\ge 0}$, an n, m-marked map $(Y,f; Q_1,\dots ,Q_n ; P_1,\dots ,P_m)$ over F is a map $f : Y \rightarrow X$ of curves over F equipped with distinct points $Q_1,\dots ,Q_n \in Y(F)$ and $P_1,\dots ,P_m \in X(F)$. An isomorphism between n, m-marked maps over F is an isomorphism of maps $(\psi ,\varphi ) : f \xrightarrow {\sim } f'$ over F such that $\psi (Q_i) = Q_i'$ for $j=1,\dots ,n$ and $\varphi (P_i) = P_i'$ for $i=1,\dots ,m$.

In the definition of an n, m-marked map, no relationship between the points $Q_j$ and $P_i$ under f is assumed; moreover, the points $Q_j$ may or may not be ramification points and the points $P_i$ may or may not be branch points.

We recover the case of marked curves from marked maps by taking $Y = X$ and choosing $f : X \rightarrow X$ to be the identity. To avoid cumbersome notation, we will sometimes abbreviate

$$\begin{aligned} (Y,f;Q_1,\dots ,Q_n;P_1,\dots ,P_m)=(Y,f;\mathcal {R}) \end{aligned}$$

and refer to $\mathcal {R}$ as the rigidification data. We use similar notation $(X;\mathcal {R})$ for marked curves.

1.2 Belyĭ maps

We will be especially interested in the following special class of maps.

Definition 1.3

Suppose ${{\mathrm{char}}}F = 0$. A Belyĭ map over F is a marked map

$$\begin{aligned} (X,f : X \rightarrow \mathbb P^1; - ; 0,1,\infty ) \end{aligned}$$

over F such that X is projective and f is unramified outside $\{0,1,\infty \}$. A marked Belyĭ map $(X,f:X \rightarrow \mathbb P^1;P;0,1,\infty )$ over F is a Belyĭ map that is marked. A Belyĭ map with a marked cusp over F is a marked Belyĭ map $(X,f : X \rightarrow \mathbb P^1; P ; 0,1,\infty )$ over F such that $f(P) = \infty$; the width of the marked point P is the ramification index of P over $\infty$.

Remark 1.4

Replacing f by 1 / f or $1/(1-f)$, one may realize any marked Belyĭ map whose marking lies above 0 or 1 as a Belyĭ map with a marked cusp.

An isomorphism $(\psi ,\varphi )$ of Belyĭ maps fixes the target $\mathbb P^1$: since the automorphism $\varphi$ of $\mathbb P^1$ has to fix each of the points $0,1,\infty$ by definition, $\varphi$ is the identity. When no confusion can result, we will often abbreviate a (marked) Belyĭ map by simply f.

Remark 1.5

Relaxing the notion of isomorphism of Belyĭ maps by removing the marked points $0,1,\infty$ on $\mathbb P^1$ typically leads one to consider covers of conics instead of $\mathbb P^1$, as in work of van Hoeij–Vidunas [28, §§3.3–3.4]; we discuss this setup further in the appendix.

Let $f : X \rightarrow \mathbb P^1$ be a Belyĭ map of degree d over $\overline{\mathbb Q}\subset \mathbb C$. Numbering the sheets of f by $1,\dots ,d$, we define the monodromy group of f to be the group $G \le S_d$ generated by the monodromy elements $s_0,s_1,s_\infty$ of loops around $0,1,\infty$; the monodromy group is well-defined up to conjugation in $S_d$, as are the conjugacy classes $C_0,C_1,C_\infty$ in $S_d$ of the monodromy elements, specified by partitions of d. We organize this data as follows.

Definition 1.6

The passport of f is the tuple $(g;G;C_0,C_1,C_\infty )$, where g is the geometric genus of X and $G \le S_d$ and $C_0,C_1,C_\infty$ are the monodromy group and $S_d$-conjugacy classes associated to f. The size of a passport is the cardinality of the set of $\overline{\mathbb Q}$-isomorphism classes of Belyĭ maps with given genus g and monodromy group G generated by monodromy elements in the $S_d$-conjugacy classes $C_0,C_1,C_\infty$, respectively.

For all $\sigma \in {{\mathrm{Aut}}}(\mathbb C)$, the Belyĭ map $\sigma (f)$ has the same ramification indices and monodromy group (up to conjugation) as f. Therefore the index of the stabilizer of f in ${{\mathrm{Aut}}}(\mathbb C)$ is bounded by the size of its passport. The size of a passport is finite and can be computed by combinatorial or group-theoretic means [26, Theorem 7.2.1]; thus the field of moduli of f over $\overline{\mathbb Q}$ is a number field of degree at most the size of the passport.

We analogously define a marked passport associated to a Belyĭ map with a marked cusp to be the tuple $(g;G;C_0,C_1,C_\infty ;\nu )$ where $\nu$ is the width of the marked cusp, corresponding to a cycle in the partition associated to $C_\infty$. Once more the field of moduli of a Belyĭ map with a marked cusp over $\overline{\mathbb Q}$ is a number field of degree at most the size of its marked passport.

Now let $K\!\mid \!F$ be a Galois field extension, let X be a curve over K, and let $\Gamma ={{\mathrm{Gal}}}(K \!\mid \!F)$. The field of moduli $M^K_F (X)$ of X with respect to $K \!\mid \!F$ is the subfield of K fixed by the subgroup

$$\begin{aligned} \left\{ \sigma \in \Gamma = {{\mathrm{Gal}}}(K \!\mid \!F) : \sigma (X) \simeq X \right\} . \end{aligned}$$

(1.2)

In a similar way, one defines the field of moduli $M^K_F (Y;f;\mathcal {R})$ of a marked map. In this paper we will usually only consider the case where $K = {F}^{\mathrm {sep}}$, and we simply call $M(X) = M^K_F (X)$ the field of moduli of X, and similarly for maps and marked variants.

1.3 Statements

The central question that we consider in this article is the following.

Question 1.7

What conditions ensure that a curve, map, or one of its marked variants is defined over its field of moduli?

For brevity, when an object is defined over its field of moduli with respect to a separable extension $K \!\mid \!F$, we will say that the object descends (to its field of moduli).

We are now in a position to give a precise formulation of the results motivating this paper.

Theorem 1.8

Let

$$\begin{aligned} (Y, f : Y \rightarrow X; Q_1,\ldots ,Q_m; P_1,\dots ,P_n) = (Y,f;\mathcal {R}) \end{aligned}$$

be a marked map of curves over a separably closed field $F = {F}^{\mathrm {sep}}$ with $m \ge 1$ such that at least one of the points $Q_1 , \ldots Q_m$ is smooth. Then $(Y,f;\mathcal {R})$ descends to its field of moduli.

We will prove Theorem 1.8 as a consequence of Theorems 3.6 and 3.13 (the latter dealing with the case of genus zero); it is a special instance of a more general geometric result that shows how to extract a Weil cocycle from the rigidification provided in the theorem. By contrast, marked curves can fail to descend if the marked point is singular: in Sect. 5 we construct two explicit counterexamples, one on a curve X that descends to $\mathbb R$ and another on a curve that does not.

Before proceeding, we deduce several important corollaries.

Corollary 1.9

Let (X; P) be a curve with a smooth marked point over a separably closed field $F = {F}^{\mathrm {sep}}$. Then (X; P) descends.

Another corollary is the following theorem.

Theorem 1.10

A marked Belyĭ map $(X,f:X \rightarrow \mathbb P^1; P; 0,1,\infty )$ over $\overline{\mathbb Q}$ descends.

Theorem 1.10 was claimed by Birch [2, Theorem 2] in the special case of a Belyĭ map with a marked cusp; but his proof is incomplete. In work of Dèbes–Emsalem [7, §5], a proof of Theorem 1.10 is sketched using a suitable embedding in a field of Puiseux series. The following corollary of Theorem 1.10 then follows immediately.

Corollary 1.11

A Belyĭ map $f:X \rightarrow \mathbb P^1$ is defined over a number field of degree at most the minimum of the sizes of the marked passports $(X,f; P; 0,1,\infty )$, where $P \in f^{-1}(\{0,1,\infty \})$.

Corollary 1.11 often enables one to conclude that the Belyĭ map itself descends, as the following result shows.

Corollary 1.12

Let $f:X \rightarrow \mathbb P^1$ be a Belyĭ map over $\overline{\mathbb Q}$ with a marked cusp P whose ramification index is unique in its fiber $f^{-1}(\infty )$. Then the Belyĭ map $(X,f;-;0,1,\infty )$ descends.

Proof

Let F be the field of moduli of the marked map $(X,f;-;0,1,\infty )$ with respect to the extension $\overline{\mathbb Q}\!\mid \!\mathbb Q$. Let $\sigma \in \Gamma _F$ and consider the conjugate Belyĭ map $(\sigma (f) ;-; 0,1,\infty )$ with source $\sigma (X)$. Since F is the field of moduli, there exists an isomorphism

$$\begin{aligned} (\psi _{\sigma },\varphi _{\sigma }) : (\sigma (f) ;-; 0,1,\infty ) \xrightarrow {\sim } (X,f ;-; 0,1,\infty ) \end{aligned}$$

with $\varphi _{\sigma }$ the identity map (as noted after Definition 1.3). We see that

$$\begin{aligned} f (\psi _\sigma (\sigma (P))) = \sigma (f) (\sigma (P)) = \sigma (\infty ) = \infty = f(P) \end{aligned}$$

(1.3)

so $\psi _\sigma (\sigma (P)) \in f^{-1}(\infty )$. The ramification index of $f = \sigma (f) \psi _{\sigma }^{-1}$ at $\psi _{\sigma } (\sigma (P))$ is equal to the ramification of $\sigma (f)$ at $\sigma (P)$, which in turn equals that of f at P. By the uniqueness hypothesis, we must have that $(\psi _{\sigma },\varphi _{\sigma })$ sends $\sigma (P)$ to P.

Since $\sigma$ was arbitrary, this means that the field of moduli of the marked Belyĭ map $(X,f; P; 0,1,\infty )$ coincides with that of $(X,f;-; 0,1,\infty )$. By Theorem 1.10, $(X,f; P; 0,1,\infty )$ descends to this common field of moduli and thus so does $(X,f;-; 0,1,\infty )$. $\square$

Remark 1.13

The hypothesis of Corollary 1.12 is very often satisfied. When it is not, one can still try to obtain a model of a Belyĭ map over a small degree extension of its field of moduli by ensuring that marking a point does not make the size of the passport grow too much; for example, if P is a point of maximal ramification index then the automorphism group of the marked tuple $(X,f;P;0,1,\infty )$ may be of small index in that of $(X,f;-;0,1,\infty )$.

2 Weil cocycles

Our main tool for the construction of examples and counterexamples is the Weil cocycle criterion, which we will give in Theorem 2.1. For more details, we refer to Serre [24, Ch. V, 20, Cor.2] and to Huggins’s thesis [14], which is an excellent exposition on descent of curves.

2.1 Weil cocycle criterion

Throughout this section, we let $K \!\mid \!F$ be a (possibly infinite) Galois extension, let $\Gamma = {{\mathrm{Gal}}}(K \!\mid \!F)$, and let X be a curve over K whose field of moduli with respect to the extension $K \!\mid \!F$ equals F.

Theorem 2.1

(Weil cocycle criterion [29]) The curve X descends if and only if there exist isomorphisms

$$\begin{aligned} \{ \varphi _{\sigma }:\sigma (X) \xrightarrow {\sim } X \}_{\sigma \in \Gamma } \end{aligned}$$

(2.1)

over K such that the cocycle condition

$$\begin{aligned} \varphi _{\sigma \tau } = \varphi _{\sigma } \sigma (\varphi _{\tau })\quad for\,all\,\, \sigma , \tau \in \Gamma \end{aligned}$$

(2.2)

holds.

More precisely, if the isomorphisms (2.1) satisfy (2.2), then there exists a descent $X_0$ of X to F and an isomorphism $\varphi _0 : X \xrightarrow {\sim } X_0$ over K such that $\varphi _{\sigma }$ is given as the coboundary

$$\begin{aligned} \varphi _{\sigma } = \varphi _0^{-1}\sigma (\varphi _0) . \end{aligned}$$

(2.3)

Remark 2.2

The Weil cocycle criterion can also be formulated for arbitrary (possibly transcendental) normal and separable extensions of F; in this case one has to add the condition that the subgroup $\{\sigma \in \Gamma : \sigma (X) = X \; \text {and} \; \varphi _{\sigma } = \mathrm {id}_X \}$ has finite index in $\Gamma$.

Remark 2.3

We warn the reader not to confuse the descent $\varphi _0$ with the isomorphisms $\varphi _\sigma$ for $\sigma \in \Gamma$.

An important corollary of Theorem 2.1, obtained by an immediate uniqueness argument, is the following.

Corollary 2.4

If ${{\mathrm{Aut}}}(X) (K)$ is trivial, then X descends.

The Weil cocycle criterion is especially concrete when the base field F equals $\mathbb R$. In this case we only have to find an isomorphism $\varphi : \overline{X}\xrightarrow {\sim } X$ between X and its complex conjugate $\overline{X}$ and test the single cocycle relation

$$\begin{aligned} \varphi \overline{\varphi } = 1 \end{aligned}$$

(2.4)

that corresponds to the complex conjugation being an involution. Given any $\varphi$ as above, all other are of the form $\alpha \varphi$, where $\alpha \in {{\mathrm{Aut}}}(X)(\mathbb C)$.

Remark 2.5

More generally, if the extension $K \!\mid \!F$ is finite, there is a method to find all possible descents of X with respect to the extension $K \!\mid \!F$; see Method 4.1.

In general, using Theorem 2.1 requires some finesse; when the extension $K \!\mid \!F$ is infinite, it is not immediately clear through which finite subextensions a descent cocycle could factor. However, in Corollary 3.8 below we shall see that for marked curves or marked Belyĭ maps it is possible to reduce considerations to an explicitly computable finite subextension of $K \!\mid \!F$.

2.2 Consequences

Now let $(Y;\mathcal {R})$ be a marked curve or map, with $\mathcal {R}$ the rigidification data, and suppose again that the field of moduli of $(Y;\mathcal {R})$ with respect to the extension $K \!\mid \!F$ equals F. In this case (as mentioned by Dèbes–Emsalem [7, §5]), Theorem 2.1 still applies to the marked curve $(Y;\mathcal {R})$ after replacing ${{\mathrm{Aut}}}(Y)$ by the subgroup ${{\mathrm{Aut}}}(Y;\mathcal {R})$. Moreover, in case a Weil cocycle exists the marked data then descend along with Y to give a model $(Y_0;\mathcal {R}_0)$ of $(Y;\mathcal {R})$ over F. We give a simple example of this general principle.

Example 2.6

Let (X; P) be a marked curve over K. Suppose that X has field of moduli F, with $\varphi _{\sigma }:\sigma (X) \xrightarrow {\sim } X$ in (2.2) having the additional property that $\varphi _{\sigma } (\sigma (P)) = P$. Let $X_0$ be a descent of X to F, with isomorphisms $\varphi _0:X \xrightarrow {\sim } (X_0)_K$ over K. Let $P_0 = \varphi _0(P) \in X_0(K)$. Then by (2.3), for all $\sigma \in \Gamma$ we have

$$\begin{aligned} \sigma (P_0) = \sigma (\varphi _0) (\sigma (P)) = \varphi _0 \varphi _{\sigma } (\sigma (P)) = \varphi _0 (P) = P_0 . \end{aligned}$$

(2.5)

By Galois invariance, we see that $P_0 \in X_0(F)$. Therefore the marked curve (X; P) descends.

The following special case of Theorem 1.8, an analogue of Corollary 2.4, is then clear from Example 2.6.

Corollary 2.7

If the group ${{\mathrm{Aut}}}(Y,f;\mathcal {R})(K)$ is trivial, then $(Y,f;\mathcal {R})$ descends.

Corollary 2.8

Suppose that $(Y,f;\mathcal {R})$ is given by

$$\begin{aligned} (Y,f;Q_1,\dots ,Q_n;P_1,\dots ,P_m)=(Y,f;\mathcal {R}) \end{aligned}$$

with $n \ge 1$, and moreover suppose that $Q_1$ is not a ramification point of f. Then $(Y,f;\mathcal {R})$ descends.

Proof

We have ${{\mathrm{Aut}}}(Y,f;\mathcal {R})(K) \le {{\mathrm{Aut}}}(Y,f)(K)$ and the latter group acts freely on those orbits that contains a non-ramifying point, so indeed ${{\mathrm{Aut}}}(Y,f;\mathcal {R})(K)$ is trivial. $\square$

The moral is that when rigidification trivializes the automorphism group, then the obstruction to descent vanishes. At the other extreme, when the original curve descends and the marked curve has the same automorphism group, then the marked curve descends as well.

Proposition 2.9

Suppose that ${{\mathrm{Aut}}}(Y,f;\mathcal {R})(K) = {{\mathrm{Aut}}}(Y)(K)$. Let F be the field of moduli of $(Y,f;\mathcal {R})$. Then $(Y,f;\mathcal {R})$ descends if Y descends, and any descent datum for Y gives rise to one for $(Y,f;\mathcal {R})$.

Proof

Let $Y_0$ be a descent of Y to F with isomorphism $\psi _0 : Y \xrightarrow {\sim } Y_0$ over K, and let $\sigma \in {{\mathrm{Gal}}}(K \!\mid \!F)$. We now interpret the map f as a further rigidification of the curve Y, and correspondingly write $(Y;\mathcal {R}')$ for $(Y,f;\mathcal {R})$. By hypothesis, there exists an isomorphism $\psi _{\sigma } : \sigma (Y) \xrightarrow {\sim } Y$ over K that sends the conjugate rigidification $\sigma (\mathcal {R}')$ to $\mathcal {R}'$. Composing, we get an isomorphism

$$\begin{aligned} \psi '_{\sigma } : \sigma (Y_0) \xrightarrow {\sigma (\psi _0^{-1})} \sigma (Y) \mathop {\longrightarrow }\limits ^{\psi _{\sigma }} Y \mathop {\longrightarrow }\limits ^{\psi _0} Y_0 . \end{aligned}$$

(2.6)

Moreover, if we let $\mathcal {R}'_0 = \psi _0 (\mathcal {R}')$, then the cocycle relation $\psi _{\sigma }=\psi _0^{-1} \sigma (\psi _0)$ shows that $\psi '_{\sigma }$ sends $\sigma (\mathcal {R}'_0)$ to $\mathcal {R}'_0$: we have

$$\begin{aligned} \psi '_{\sigma } (\mathcal {R}'_0) = \psi _0 (\psi _{\sigma } (\sigma (\psi _0^{-1}) (\psi _0 (\mathcal {R}')))) = \psi _0 (\psi _{\sigma } (\psi _{\sigma }^{-1} (\mathcal {R}'))) = \psi _0 (\mathcal {R}') = \mathcal {R}'_0 . \end{aligned}$$

(2.7)

Now we have assumed that $Y_0$ is defined over F. Therefore $\sigma (Y_0) = Y_0$, which shows that $\psi '_{\sigma }$ actually belongs to ${{\mathrm{Aut}}}(Y_0)(K)$. Since ${{\mathrm{Aut}}}(Y;\mathcal {R}')(K) = {{\mathrm{Aut}}}(Y)(K)$ we equally well have ${{\mathrm{Aut}}}(Y_0;\mathcal {R}'_0)(K) = {{\mathrm{Aut}}}(Y_0)(K)$. Hence in fact $\sigma (\mathcal {R}'_0) = (\psi '_{\sigma })^{-1} (\mathcal {R}'_0) = \mathcal {R}'_0$. Since $\sigma$ was arbitrary, we see that the rigidification $\mathcal {R}'_0$ is rational on $Y_0$, as desired. $\square$

Remark 2.10

In case of proper inclusions $\{1\} \subsetneq {{\mathrm{Aut}}}(X;\mathcal {R})(K) \subsetneq {{\mathrm{Aut}}}(X)(K)$, it is possible that X may descend to F while $(X;\mathcal {R})$ does not: we will see an example of this in Sect. 5 which is minimal in the sense that the chain of inclusions

$$\begin{aligned} \{1\} \subsetneq {{\mathrm{Aut}}}(X;\mathcal {R})(\mathbb C) \simeq \mathbb Z/ 2 \mathbb Z\subsetneq {{\mathrm{Aut}}}(X) (\mathbb C) \simeq \mathbb Z/ 4 \mathbb Z\end{aligned}$$

is as small as possible.

3 Branches

The geometric equivalent of a fundamental tool of Dèbes–Emsalem [7] is the consideration of what we will define as the branches of a morphism of curves over a point P. For a map tamely ramified at P, branches can be interpreted by embeddings into certain rings of Puiseux series. We revisit this definition in a more general geometric context, which will extend to the wildly ramified case.

3.1 Definitions

Let X be a curve over F, and let $P \in X ({F}^{\mathrm {sep}})$ be a geometric point of X. Let $\mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ be the local ring of X at P for the étale topology. Using idempotents, we have a canonical decomposition

$$\begin{aligned} \mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P} = \prod _i \mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P,i} \end{aligned}$$

(3.1)

into a finite product of domains. Let $\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P,i}$ be the integral closure of the domain $\mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P,i}$ in ${{\mathrm{Frac}}}(\mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P,i})^{\mathrm {sep}}$, the separable closure of its quotient field. We get a ring

$$\begin{aligned} \mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P} = \prod _i \mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P,i} \end{aligned}$$

(3.2)

along with a canonical morphism ${{\mathrm{Spec}}}\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P} \rightarrow X$.

Definition 3.1

Let $f : Y \rightarrow X$ be a map of curves over F. A branch of f over P is a morphism $b:{{\mathrm{Spec}}}\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P} \rightarrow Y$ such that the following diagram commutes:

(3.3)

The set of branches of f over P is denoted B (f, P).

Remark 3.2

Geometrically, a branch of f can be seen as an equivalence class of sections $V \rightarrow Y \mathop {\rightarrow }\limits ^{f} X$, where the composition $V \rightarrow X$ is a “separable neighborhood” of P in the sense that it factors as $V \rightarrow U \rightarrow X$, where $V \rightarrow U$ is separable and where $U \rightarrow X$ is an étale neighborhood of P. Note that we could also have defined branches by using the total quotient ring of $\mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},x}$ instead of explicitly employing the factorization in the current definition. Another alternative at a smooth point is to use the completion of the usual local ring $\mathscr {O}_{X, x}$.

Definition 3.1 is most concrete when P is a smooth point of X. In this case $\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ is the closure of $\mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ in ${{\mathrm{Frac}}}(\mathscr {O}_{X,P})^{\mathrm {sep}}$. The ring $\mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ is the subring of ${F}^{\mathrm {sep}}[[ t ]]$ consisting of power series that are algebraic over the field of rational functions ${F}^{\mathrm {sep}}(t)$ in t [21, Prop. 4.10]. Therefore, if ${{\mathrm{char}}}F = 0$, then the elements of $\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ can be seen as elements of the ring of Puiseux series ${F}^{\mathrm {sep}}[[t^{1/\infty }]]$, or in other words as elements of the field of Puiseux series ${F}^{\mathrm {sep}}((t^{1/\infty }))$ whose monomials all have non-negative exponent. (This isomorphism explains our notation for the ring $\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$.)

If ${{\mathrm{char}}}F = p$, then wild ramification can occur, as for example when considering Artin–Schreier extensions; in this case the ring $\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ is no longer simply a subring of a field of Puiseux series, and we have to use the field of generalized power series $F ((t^{\mathbb Q}))$ instead. These are the formal linear combinations $\sum _{i \in \mathbb Q} c_i t^i$ whose support is a well-ordered subset of $\mathbb Q$. Kedlaya [19, Theorem 10.4] has characterized the integral closure of F (t) inside $F ((t^{\mathbb Q}))$ as those generalized power series that are p -automatic; in particular, branches will give rise to such p-automatic power series. A practical illustration of how these generalized power series are obtained is given in Example 4.14.

Remark 3.3

By taking the fiber over the closed point of ${{\mathrm{Spec}}}\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ (in the case of a valuation ring in a field of Puiseux series, we are setting $t=0$), we recover a lift of the point $P \in X({F}^{\mathrm {sep}})$. We can see b as an infinitesimal thickening of this lift; the fact that we cannot use the local ring for the étale topology reflects that we need slightly thinner thickenings than that in this topology to obtain enough sections.

In general, the decomposition in (3.1) has more than one factor. Using the categorical properties of products we see that giving a branch amounts to specifying, for every i, a morphism $b_i$ making the following diagram commute:

(3.4)

In this way, we have effectively passed to the normalization of X. For the identity map $X \rightarrow X$, the $b_i$ give what is classically called the set of branches of X at P.

Remark 3.4

Alternatively, one can first define branches in the case of a smooth point and then for a general morphism $f : Y \rightarrow X$ at a point P to be a branch of the induced map $\widetilde{f} : \widetilde{Y} \rightarrow \widetilde{X}$ of normalizations at some point $\widetilde{P} \in \widetilde{X} ({F}^{\mathrm {sep}})$ over P.

From here on, we will restrict to the case where $P \in X ({F}^{\mathrm {sep}})$ is a smooth point.

3.2 Galois action and descent

Given an element $\sigma$ of the absolute Galois group ${{\mathrm{Gal}}}({F}^{\mathrm {sep}}\!\mid \!F)$, there is a canonical map of sets

$$\begin{aligned} \begin{array}{ll} B(f,P) \rightarrow B(\sigma (f),\sigma (P)) \\ \qquad b \mapsto \sigma (b) \end{array} \end{aligned}$$

(3.5)

Here $\sigma (f)$ is the morphism $\sigma (Y) \rightarrow \sigma (X)$ induced by the action of $\sigma$ on the coefficients of f. The map on branches is defined as follows. Given a branch $b : {{\mathrm{Spec}}}\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P} \rightarrow Y$ of f over P, we can conjugate it to obtain a morphism $\sigma (b) : {{\mathrm{Spec}}}\sigma (\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}) \rightarrow \sigma (Y)$. We compose with the canonical isomorphism

$$\begin{aligned} \sigma (\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}) \simeq \mathscr {P}_{\sigma (X)_{{{\mathrm{\mathrm {\acute{e}t}}}}},\sigma (P)} \end{aligned}$$

(3.6)

where the ${F}^{\mathrm {sep}}$-algebra structure on this ring has been changed by conjugation with $\sigma$.

In the upcoming proposition, we show that the set B (f, P) has two very pleasant properties, which should be thought of as generalizing the properties of usual fibers over non-branch points (cf. Corollary 2.8).

Proposition 3.5

Let $f : Y \rightarrow X$ be a proper separable map of curves of degree d, and let $P \in X({F}^{\mathrm {sep}})$ be a smooth geometric point of X. Then the set B (f, P) has cardinality d, and the action of ${{\mathrm{Aut}}}(Y,f)$ on B (f, P) is faithful.

Proof

Let $R = \mathscr {O}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ and $S=\mathscr {P}_{X_{{{\mathrm{\mathrm {\acute{e}t}}}}},P}$ be as above. Since P is smooth, the rings R and S are integral domains; we denote their quotient fields by K and L. By definition, L is a separable closure of K, and the point P yields canonically a point $P_K \in X(K)$. Pulling back the map f by $P_K$, we obtain another separable map of degree d, corresponding to a field extension of K. By Galois theory, there are d points $Q_1 , \dots , Q_d$ in Y (L) that lie over P.

We claim that the points $Q_1,\dots ,Q_d$ are the generic points of unique points in Y (S). To see this, note that every point $Q_i$ is in fact defined over a finite subextension $K_i$ of K contained in L. Let $R_i$ be the integral closure of the strictly henselian discrete valuation ring R in $K_i$. Then $R_i$ is again a discrete valuation ring. The morphism f is proper because Y and X both are. Therefore, by the valuative criterion of properness applied to the morphism f, the points $Q_i \in Y (K_i)$ extend uniquely to points in $Y (R_i)$, proving our claim.

To conclude, we show that the action of ${{\mathrm{Aut}}}(Y,f)$ is faithful: this follows since it is a specialization of the action of the automorphism group on the generic fiber. $\square$

The following theorem can then be applied to give a constructive proof of Theorem A (Theorem 1.8) in the case where the automorphism group of the marked map is finite. We finish the proof below.

Theorem 3.6

Let $(Y,f;\mathcal {R})$ be a marked curve over ${F}^{\mathrm {sep}}$ with finite automorphism group $G = {{\mathrm{Aut}}}(Y,f;\mathcal {R})({F}^{\mathrm {sep}})$. Let $\pi : Y \rightarrow G \backslash Y = W$ be the quotient map.

• (a) Let $Q \in Y ({F}^{\mathrm {sep}})$ be smooth. Then $S = \pi (Q)$ is a smooth point of W, and the set of branches $B(\pi ,S)$ is a torsor under G.

• (b) Suppose that $\mathcal {R}$ contains at least one smooth point Q on Y. Then $(Y,f;\mathcal {R})$ descends. More precisely, let $S = \pi (Q)$, and suppose that F is the field of moduli of $(Y,f;\mathcal {R})$. Then given $b \in B(\pi ,S)$, for all $\sigma \in \Gamma ={{\mathrm{Gal}}}({F}^{\mathrm {sep}}\!\mid \!F)$ there exists a unique morphism $\varphi _{\sigma }:\sigma (Y) \xrightarrow {\sim } Y$ such that $\sigma (b)$ is sent to b, and the collection $\left\{ \varphi _{\sigma } \right\} _{\sigma \in \Gamma }$ defines a Weil cocycle.

Proof

First, part (a). The smoothness of the point S follows from the fact that a quotient of the power series ring ${F}^{\mathrm {sep}}[[x]]$ by a finite group action is isomorphic to this same ring by the structure theorem for complete discrete valuation rings [25, II, Th. 2]. That the set $B(\pi ,S)$ is a G-torsor then follows from Proposition 3.5, since the map $Y \rightarrow W$ is (generically) Galois. Note that the quotient morphism $\pi$ is separable and finite, hence proper.

Now we prove (b). The existence of $\varphi _{\sigma }$ follows because choosing just any $\varphi '_{\sigma } : \sigma (Y) \rightarrow Y$ maps $\sigma (b)$ to some branch over S, which we can then modify to be b by invoking the transitivity part of being a torsor. The $\varphi _{\sigma }$ give a Weil cocycle because of the uniqueness part of being a torsor; both $\varphi _{\sigma \tau }$ and $\varphi _{\sigma } \sigma (\varphi _{\tau })$ send $\sigma \tau (b)$ to b via pullback. $\square$

Remark 3.7

Theorem 3.6 in fact shows that there exists a descent $(Y_0;f_0;\mathcal {R}_0)$ with a branch at the image $S_0$ of S that is defined over F. Indeed, by a similar argument as that in (2.5), the coboundary corresponding to the Weil cocycle $\varphi _{\sigma }$ maps b to a branch defined over F (in the sense that b is sent to itself under the action (3.5)).

The proof of Theorem 3.6 also shows how to obtain a finite extension of F over which a Weil cocycle can be constructed.

Corollary 3.8

Let $(Y,f;\mathcal {R})$ be a marked map with a smooth point P as part of the rigidification data. Let G be the automorphism group of $(Y,f;\mathcal {R})$. Then there exists a finite Galois extension $K \!\mid \!F$ over which a Weil cocycle for $(Y,f;\mathcal {R})$ can be constructed.

Proof

Let $K \!\mid \!F$ be a finite Galois extension over which all of $(Y,f;\mathcal {R})$, the elements of the group G, and the branches of the canonical morphism $\pi : Y \rightarrow G \backslash Y$ at $\pi (P)$ are defined. Such an extension exists: defining expressions for $Y,f,\mathcal {R}$ and the elements of the finite group G have finitely many coefficients, and the finitely many branches are defined over a finite extension of F. Working over K instead of ${F}^{\mathrm {sep}}$ in Theorem 3.6 gives a descent relative to the extension $K \!\mid \!F$. $\square$

Remark 3.9

As we will see explicitly later, computing the splitting field of the branches at a given point requires no more than determining the leading term of certain (generalized) power series whose order is known explicitly.

Remark 3.10

The counterexamples in Sect. 5 all share the property that the number of branches over the points of the base curve is not constant, because of the merging of these branches over the singular points, which are (crucially) the only rational points of the base curve. This makes it impossible to read off a uniquely determined cocycle as in Theorem 3.6.

Remark 3.11

As in Birch’s original article [2] we can consider the case where X is a modular curve associated with a subgroup ${{\mathrm{SL}}}_2 (\mathbb Z)$, which we suppose to be defined over some number field F. Using a cusp of X then allows one to use q-expansions with respect to a uniformizer q of the appropriate width with respect to the cusp.

Echelonizing a basis of modular forms gives a defining equation for X over F; this does indeed turn out to lead to F-rational q-expansions, with the slight subtlety that one may need to twist X to have a rational branch over F.

3.3 Descent under infinite automorphism group

To finish the proof of Theorem 1.8, it remains to deal with the case where the automorphism group of the marked curve $(Y,f;\mathcal {R})$ is infinite. In this case the map f has to be the identity map, so we are reduced to considering marked curves $(X;\mathcal {R}) = (X;P_1, \ldots P_n)$ with $n \ge 1$ for which one of the $P_i$, say $P_1$, is smooth.

Lemma 3.12

Let $(X;\mathcal {R}) = (X;P_1, \ldots P_n)$ be a marked projective curve with $P_1$ smooth and $\#{{\mathrm{Aut}}}(X;\mathcal {R})=\infty$. Let $\Sigma =X_{sing }({F}^{\mathrm {sep}})$ be the set of singular points of X. Then the following statements hold.

• (a) The geometric genus g of X is equal to 0.

• (b) We have $n + \#\Sigma \le 2$, and the inverse image of $\mathcal {R}\cup \Sigma$ with respect to the normalization $\widetilde{X} \rightarrow X$ has cardinality at most 2.

Proof

An automorphism of X extends to a unique automorphism of the normalization $\widetilde{X}$, so ${{\mathrm{Aut}}}(X;\mathcal {R}) \le {{\mathrm{Aut}}}(X) \le {{\mathrm{Aut}}}(\widetilde{X})$. If $g \ge 2$ then ${{\mathrm{Aut}}}(\widetilde{X})$ is finite, so immediately we have $g=0,1$.

Since $P_1$ is smooth, it lifts to a unique point $\widetilde{P}_1$ on $\widetilde{X}$, and any automorphism of X fixing $P_1$ lifts to a unique automorphism of $\widetilde{X}$ fixing $\widetilde{P}_1$. If $g=1$, then $(\widetilde{X},\widetilde{P}_1)$ has the structure of an elliptic curve so ${{\mathrm{Aut}}}(\widetilde{X},\widetilde{P}_1)$ is finite, and so ${{\mathrm{Aut}}}(X;\mathcal {R}) \le {{\mathrm{Aut}}}(X;P_1) \le {{\mathrm{Aut}}}(\widetilde{X},\widetilde{P}_1)$. So $g=0$, proving (a).

Statement (b) then follows, since an automorphism of a curve of genus 0 is determined by the image of three distinct points on this curve. $\square$

Theorem 3.13

Let $(X;\mathcal {R})$ be a marked curve over ${F}^{\mathrm {sep}}$ such that $\mathcal {R}$ contains at least one smooth point on X. Then $(X;\mathcal {R})$ descends.

Proof

Let $G={{\mathrm{Aut}}}(X;\mathcal {R})$. If $\#G<\infty$, we can apply Theorem 3.6. If $\#G=\infty$, we apply Lemma 3.12 and have two cases.

First suppose X is projective and that $(X;\mathcal {R}) = (X;P_1)$ for a smooth point $P_1$. By Lemma 3.12(b), X has at most one singular point. If X is smooth, we can then take $(\mathbb P^1; \infty )$ as a descent (cf. Lemma 4.5). This leaves the case where X has a unique singular point $P_2$, and again by Lemma 3.12(b), $P_2$ has a unique inverse image $\widetilde{P}_2$ on $\widetilde{X}$. The automorphism group ${{\mathrm{Aut}}}(X;\mathcal {R})$ is contained in the group of automorphisms ${{\mathrm{Aut}}}(\widetilde{X}; \widetilde{P}_1, \widetilde{P}_2) \simeq \mathbb G_m$ since $P_2$ is the unique singular point on X. Moreover, this inclusion is functorial. Therefore ${{\mathrm{Aut}}}(X;\mathcal {R})$ is isomorphic to a subvariety of $\mathbb G_m$, and since we assumed that it was infinite, it coincides with ${{\mathrm{Aut}}}(\widetilde{X}; \widetilde{P}_1, \widetilde{P}_2)$. Now $(\widetilde{X}; \widetilde{P}_1, \widetilde{P}_2)$ descends because $\widetilde{X}$ is smooth. Let $\left\{ \varphi _{\sigma } \right\} _{\sigma \in \Gamma }$ be a set of isomorphisms of $(X ; P_1, P_2)$ with its conjugates. By the property of the normalization, these lift to yield a set of isomorphisms $\left\{ \widetilde{\varphi }_{\sigma } \right\} _{\sigma \in \Gamma }$ of $(\widetilde{X} ; \widetilde{P}_1, \widetilde{P}_2)$. Because $(\widetilde{X}; \widetilde{P}_1, \widetilde{P}_2)$ descends we can modify this latter set by automorphisms $\left\{ \widetilde{\alpha }_{\sigma } \right\} _{\sigma \in \Gamma }$, to obtain a Weil cocycle $\left\{ \widetilde{\varphi }_{\sigma } \widetilde{\alpha }_{\sigma } \right\} _{\sigma \in \Gamma }$ for $(\widetilde{X}; \widetilde{P}_1, \widetilde{P}_2)$. Our statement on automorphism groups above means that the set of automorphisms $\left\{ \widetilde{\alpha }_{\sigma } \right\} _{\sigma \in \Gamma }$ yields a set of automorphisms $\left\{ \alpha _{\sigma } \right\} _{\sigma \in \Gamma }$, and by construction $\left\{ \varphi _{\sigma } \alpha _{\sigma } \right\} _{\sigma \in \Gamma }$ is then a Weil cocycle for $(X;P_1,P_2)$.

Now suppose that X is projective with two marked points $(X;\mathcal {R}) = (X;P_1,P_2)$. Again we are done if X is smooth, since then a descent is given by $(\mathbb P^1;\infty ,0)$. In the case where we admit singular points, we are reduced to the same case as that in the previous paragraph: $P_2$ is singular and has a unique inverse image $\widetilde{P}_2$ on $\widetilde{X}$.

The case where X is not projective follows by taking its smooth completion and applying the results above. $\square$

Remark 3.14

It is indeed possible for the automorphism group ${{\mathrm{Aut}}}(X;P_1,P_2)$ to be finite and non-trivial when X is singular and of (geometric) genus 0. Take $\mathbb P^1$ with affine coordinate t, pinch together the points $t = -1,1$ to create a node, and let X be the resulting singular curve. Then ${{\mathrm{Aut}}}(X;0,\infty )$ is the subgroup of ${{\mathrm{Aut}}}(\mathbb P^1;0,\infty )$ that preserves $\{-1,1\}$, and is therefore the subgroup $t \mapsto \pm t$. Examples where $P_1$ and $P_2$ are singular on X can be constructed by constructing morphisms from suitable sublinear systems on $\mathbb P^1$.

3.4 Descent and the canonical model

To conclude this section, we phrase our results in terms of a notion from Dèbes–Emsalem [7]. Let $(Y,f;\mathcal {R})$ be a marked map over ${F}^{\mathrm {sep}}$ with field of moduli F and finite automorphism group $G = {{\mathrm{Aut}}}(Y,f;\mathcal {R})$, and let $W = G \backslash Y$. Then by construction any choice of isomorphisms (2.1) gives rise to a Weil cocycle on W.

Definition 3.15

The canonical model $W_0$ is the model of W defined over F determined by the cocycle on W induced by any choice of isomorphisms (2.1).

The canonical model depends only on the isomorphism class of $(Y,f;\mathcal {R})$ over ${F}^{\mathrm {sep}}$. In the canonical model, we denote the quotient morphism $\pi :Y \rightarrow G \backslash Y$ and let $\psi _0 : W \xrightarrow {\sim } W_0$ be a coboundary corresponding to the uniquely determined collection of isomorphisms $\{ \psi _{\sigma } \}_{\sigma \in \Gamma }$ induced by (2.1). Set $\pi _0 = \psi _0 \pi$.

Proposition 3.16

Let $Q \in Y({F}^{\mathrm {sep}})$ be smooth. Then the field of moduli of $(Y,f;\mathcal {R};Q)$ is the field of definition of the point $\pi _0 (Q)$ on $W_0$.

Proof

Let $F'=F(\pi _0(Q))$ be the field of definition of $\pi _0(Q)$ on $W_0$.

First, we claim that $F'$ is contained in the field of moduli M of $(Y,f;\mathcal {R};Q)$. By Theorem 1.8, we may assume without loss of generality that $(Y,f;\mathcal {R};Q)=(Y_0;f_0;\mathcal {R}_0;Q_0)$ is defined over M. Thus $Q \in Y(M)$. This implies $\pi _0 (Q) \in W_0(M)$: we have $W_0 = G \backslash Y$ where $G={{\mathrm{Aut}}}(Y,f;\mathcal {R})$ so that $\pi _0$ coincides with the natural projection $Y \rightarrow W = G \backslash Y$ defined over M. Thus $M \supseteq F'=F(\pi _0(Q))$.

To conclude, we show $M \subseteq F'$. We then have to show that for all $\sigma \in \Gamma ' = {{\mathrm{Gal}}}({F}^{\mathrm {sep}}\!\mid \!F')$, there exist isomorphisms

$$\begin{aligned} \varphi _{\sigma } : (\sigma (Y);\sigma (f);\sigma (\mathcal {R});\sigma (Q)) \xrightarrow {\sim } (Y,f;\mathcal {R};Q) \end{aligned}$$

(3.7)

so that M is contained in $F'$. We are given that $(Y,f;\mathcal {R})$ has field of moduli F, so we have isomorphisms $\varphi _{\sigma } : (\sigma (Y) ; \sigma (f) ; \sigma (\mathcal {R})) \xrightarrow {\sim } (Y,f;\mathcal {R})$ for all $\sigma \in {{\mathrm{Gal}}}({F}^{\mathrm {sep}}\!\mid \!F)$.

Let $\sigma \in \Gamma '$. We will show that $\varphi _{\sigma }$ can be chosen in such a way that additionally $\varphi _{\sigma } (\sigma (Q)) = Q$. Let $\psi _{\sigma } = \psi _0^{-1} \sigma (\psi _0)$. Then by construction of the canonical Weil cocycle for W we know that $\pi \varphi _{\sigma } = \psi _{\sigma } \sigma (\pi )$. We obtain the following commutative diagram:

(3.8)

The rationality of $\pi _0(Q)$ over $F'$ implies that

$$\begin{aligned} \pi _0 (Q) = \sigma (\pi _0(Q)) = \sigma (\pi _0) (\sigma (Q)). \end{aligned}$$

(3.9)

We claim that $Q,\varphi _{\sigma }(\sigma (Q)) \in Y({F}^{\mathrm {sep}})$ are in the same fiber of the map $\pi _0$. Indeed, from (3.9) and tracing through the diagram (3.8), we obtain

$$\begin{aligned} \pi _0(Q) = \sigma (\pi _0) (\sigma (Q)) = \pi _0(\varphi _{\sigma } (\sigma (Q))) . \end{aligned}$$

(3.10)

Because ${{\mathrm{Aut}}}(Y,f;\mathcal {R})$ acts transitively on the fibers of $\pi$ and thus on those of $\pi _0$, we see that we can compose the chosen $\varphi _{\sigma }$ with an element of this group to obtain an isomorphism as in (3.7). $\square$

Theorem 3.17

Let $Q \in Y(K)$ be smooth. Then Q is F-rational on some descent of $(Y,f;\mathcal {R})$ to F if and only if $\pi _0 (Q)$ is F-rational.

As a corollary, we obtain the following theorem, alluded to by Dèbes–Emsalem [7, §5].

Theorem 3.18

Suppose that $W_0$ admits a smooth F-rational point. Then $(Y,f;\mathcal {R})$ descends.

Proof

Let $Q \in Y({F}^{\mathrm {sep}})$ be such that $\pi _0(Q) \in V_0 (F)$. Then the field of moduli of $(Y,f;\mathcal {R};Q)$ equals F by Proposition 3.16. We get a descent $(Y_0;f_0;\mathcal {R}_0;Q_0)$ of $(Y,f;\mathcal {R};Q)$ to F by Theorem 1.8, which also gives a descent of $(Y,f;\mathcal {R})$ to F. $\square$

Remark 3.19

Theorem 3.18 applies in particular when Y is not equipped with a rigidification. Consider for example the marked curve case $Y = X$ where X is hyperelliptic, with hyperelliptic involution $\iota$; one can then show [18] that a descent to F induced by a point P as in Theorem 3.17 always exists, except possibly if g(X) and $\# {{\mathrm{Aut}}}(X) / \langle \iota \rangle$ are both odd. That is, except in these special cases any descent can be obtained by marking a point on X. Note that the descents obtained will be of a rather special kind; indeed, Remark 3.7 shows that these all have a rational branch over Q, and hence in particular a rational point.

Remark 3.20

The results in this section can also be obtained by using a tangential base point (a germ of a regular function at a given point) as in Deligne [8, §15] by splitting an exact sequence of fundamental groups, as explained by Dèbes–Emsalem [7]. We prefer our new approach for two reasons. First, a choice of tangential base point is noncanonical, so it is cleaner (and more intuitive) to instead refine points into branches, which does not require such a choice. Second, the formalism of tangential base points has not been extended to the wildly ramified case to our knowledge, whereas our definition immediately covers this case. Phrased differently, we do not need to split the exact sequence of fundamental groups [7], which is obtained only after restricting attention to prime-to-p parts.

4 Descent of marked curves

This section focuses on the explicit descent on marked curves, notably hyperelliptic curves and plane quartics. Before treating these examples, we indicate a general method for descending marked curves.

4.1 Determining a Weil descent

Let $K \!\mid \!F$ be a Galois extension, and let $(Y,f;\mathcal {R})$ be a marked map over K. Suppose that the extension $K \!\mid \!F$ and the automorphism group ${{\mathrm{Aut}}}(Y,f;\mathcal {R})$ are both finite. Then the following method finds all possible descents of $(Y,f;\mathcal {R})$ with respect to the extension $K \!\mid \!F$.

Method 4.1

This method takes as input a marked map $(Y,f;\mathcal {R})$ over K and produces as output all descents $(Y_0,f_0;\mathcal {R}_0)$ of $(Y,f;\mathcal {R})$ to F up to isomorphism over F.

• (1) Compute a presentation $\Gamma = \langle \Sigma \mid \Pi \rangle$ of the Galois group $\Gamma = {{\mathrm{Gal}}}(K \!\mid \!F)$ in finitely many generators $\Sigma$ and relations $\Pi$.

• (2) Compute $G = {{\mathrm{Aut}}}(Y,f;\mathcal {R})(K)$.

• (3) For all $\sigma \in \Sigma$, compute an isomorphism $\psi _{\sigma } : (\sigma (Y),\sigma (f);\sigma (\mathcal {R})) \xrightarrow {\sim } (Y,f;\mathcal {R})$ over K. If for one of these generators no such isomorphism exists, output the empty set and terminate.

• (4) For all tuples $(\varphi _{\sigma })_{\sigma \in \Sigma }$ in the product $\prod _{\sigma \in \Sigma } G \psi _{\sigma }$, use the relation (2.2) to determine the value of the corresponding cocycle on the relations in $\Pi$; retain a list $\mathcal {L}$ of those tuples for which the corresponding cocycle is trivial on all elements of $\Pi$.

• (1) For the tuples $(\varphi _{\sigma })_{\sigma \in \Sigma }$ in $\mathcal {L}$, construct a coboundary morphism $\varphi _0 : (Y,f;\mathcal {R}) \xrightarrow {\sim } (Y_0,f_0;\mathcal {R}_0)$ to obtain a descent of $(Y,f;\mathcal {R})$. Output this set of descents.

Remark 4.2

Method 4.1 applies to smoothly marked maps $(Y,f;\mathcal {R})$ over the separable closure ${F}^{\mathrm {sep}}$ to detect the existence of a descent: by Corollary 3.8, we can always reduce the existence of a descent from ${F}^{\mathrm {sep}}$ to a finite extension $K \!\mid \!F$. (There may be infinitely many descents of $(Y,f;\mathcal {R})$ over ${F}^{\mathrm {sep}}$, but only finitely many coming from a given finite extension $K \!\mid \!F$.)

Remark 4.3

In step 1 of the above method, one can work instead with the full group $\Gamma$ itself rather than generators and relations; then one loops over the elements of the group and checks compatibility with products.

The correctness of this method follows by the general approach in the previous section. For now, we present only a ‘method’ rather than a true algorithm, as each of these steps may involve a rather difficult computer algebra problem in general; however, efficient algorithms for finding isomorphisms exist for smooth hyperelliptic or plane curves, making all but the final step of the method algorithmic. The construction of a coboundary is somewhat more involved but can be obtained in the aforementioned cases as well, essentially by finding points on a certain variety over the ground field. We do not go into these general problems here, as they form a substantial challenge of their own, but we do indicate how to proceed in special cases below.

Remark 4.4

If $(Y,f;\mathcal {R})$ is already defined over F, the method above will find the twists of $(Y,f;\mathcal {R})$ over K, that is, the set of F-isomorphism classes of marked maps curves over F that become isomorphic to $(Y,f;\mathcal {R})$ over K. In this case the isomorphisms $\varphi _{\sigma }$ are themselves already in G, so this computation is slightly easier.

4.2 Descent of marked smooth curves of genus at most one

We begin by disposing of two easy cases in low genus.

Lemma 4.5

Let (X; P) be marked curve over ${F}^{\mathrm {sep}}$ with X smooth of genus $\le 1$. Then (X; P) has field of moduli F. If $g=0$, a descent to F is given by $(X_0;P_0) = (\mathbb P^1; \infty )$; if $g=1$, then the field of moduli is equal to $F(j({{\mathrm{Jac}}}(X)))$.

Proof

The case $g=0$ is immediate. So suppose $g=1$. By translation on the genus 1 curve X, we can take $(X_0;P_0) = (J ; \infty )$, where $J={{\mathrm{Jac}}}(X)$ is the Jacobian of X with origin $\infty$, which is defined over F(j) where j is the j-invariant of J. $\square$

4.3 Hyperelliptic curves

We now pass to a class of examples that can be treated in some generality, namely that of smooth hyperelliptic curves of genus at least 2. Throughout, we suppose that the base field F does not have characteristic 2.

Let X be a smooth hyperelliptic curve over ${F}^{\mathrm {sep}}$ and let $\iota :X \rightarrow X$ be the hyperelliptic involution. Let $W=\langle \iota \rangle \backslash X$ and $V = {{\mathrm{Aut}}}(X) \backslash X$. We get a sequence of natural projection maps

$$\begin{aligned} X \mathop {\longrightarrow }\limits ^{q} W \mathop {\longrightarrow }\limits ^{p} V; \end{aligned}$$

(4.1)

the map p is the quotient of W by the reduced automorphism group $G = {{\mathrm{Aut}}}(X)({F}^{\mathrm {sep}}) / \langle \iota \rangle$ of W (or of X, by abuse).

We can identify G with a subgroup of ${{\mathrm{PGL}}}_2 ({F}^{\mathrm {sep}})={{\mathrm{Aut}}}(W)({F}^{\mathrm {sep}})$. For simplicity, we also suppose that G does not contain any non-trivial unipotent elements. This hypothesis implies that G is isomorphic to one of the finite groups $C_n, D_n, A_4, S_4$ and $A_5$.

The reduced automorphism group G is also described as $G = {{\mathrm{Aut}}}(W;D)({F}^{\mathrm {sep}})$, where D is the branch divisor of q, or equivalently, the image on W of the divisor of Weierstrass points on X; thus X can be recovered from (W; D) over ${F}^{\mathrm {sep}}$ by taking a degree 2 cover of W ramified over D.

Remark 4.6

Even if $(W;D) = (W_0;D_0)$ is defined over F, it is not always possible to construct a cover $X_0$ of $(W_0;D_0)$ over F whose branch divisor equals $D_0$. Indeed, let X be a hyperelliptic curve with field of moduli F and ${{\mathrm{Aut}}}(X)({F}^{\mathrm {sep}})=\langle \iota \rangle$. Then $W = V$, and there exists a canonical descent $W_0$ of W. Moreover, by the same argument that we used to obtain (2.5), D transforms to a F-rational divisor $D_0$. If the aforementioned cover could be constructed over F, then $X_0$ would be a descent of X. But as is known classically (see also Remark 5.3), there exist hyperelliptic curves of genus 2 with generic automorphism group that do not descend.

Let $R \in X({F}^{\mathrm {sep}})$ and consider the marked curve (X; R) over ${F}^{\mathrm {sep}}$. Let $Q = q(R)$ be the image of R on W, and similarly let $P = p(q(R))$ be its image on V. Then we can also reconstruct (X; R) from the datum (W; Q; D) obtained by rigidifying (W; Q) with the branch divisor D of q.

Proposition 4.7

With the above notation, let $X'$ be a curve over ${F}^{\mathrm {sep}}$ obtained as a degree 2 cover of W ramifying over D, and let $R'$ be any preimage of Q under the corresponding covering map. Then $(X';R')$ is isomorphic to (X; R) over ${F}^{\mathrm {sep}}$.

Proof

This follows from the fact that a hyperelliptic curve over ${F}^{\mathrm {sep}}$ is uniquely determined by the branch locus of its hyperelliptic involution $\iota$, combined with the fact that $\iota$ acts transitively on the fibers of the quotient $X \rightarrow W$. $\square$

We now suppose that (X; R) has field of moduli F, which we can do without loss of generality. By Theorem 1.8, we know that (X; R) descends to a marked curve $(X_0;R_0)$ over F, which upon taking quotients by normal subgroups gives rise to a sequence of marked curves and morphisms

$$\begin{aligned} (X_0; R_0) \mathop {\longrightarrow }\limits ^{p_0} (W_0; Q_0) \mathop {\longrightarrow }\limits ^{q_0} (V_0; P_0) . \end{aligned}$$

(4.2)

over F. In what follows, we will show how to calculate the descent $(X_0; R_0)$ explicitly. Our methods do not yet need branches; for pointed hyperelliptic curves, it suffices to use more elementary techniques involving morphisms between curves of genus 0.

We will first construct the descent $(V_0; P_0)$ of (V, P). This is done as in Dèbes–Emsalem [7]; we start with any collection of isomorphisms $\{ \varphi _{\sigma } : \sigma (X) \rightarrow X \}_{\sigma \in \Gamma _F}$ and consider the induced maps $\chi _{\sigma } : \sigma (V) \xrightarrow {\sim } V$. By construction, the collection $\{ \chi _{\sigma } \}_{\sigma \in \Gamma _F}$ satisfies the Weil cocycle relation. Therefore there exists a map $\chi _0 : V \xrightarrow {\sim } V_0$ to a descent $V_0$ of V over F such that $\chi _{\sigma } = \chi _0^{-1}\sigma (\chi _0)$. By the same argument that we used to obtain (2.5), we see that $P_0 = \chi _0 (P)$ becomes an F-rational point on $V_0$.

The isomorphisms $\varphi _{\sigma } : \sigma (X) \xrightarrow {\sim } X$ can be explicitly calculated [17]. After determining the induced maps $\chi _{\sigma } : \sigma (V) \xrightarrow {\sim } V$ between genus 0 curves, we can explicitly calculate the map $\chi _0$, and with it the point $P_0$ [13, 23]. We therefore assume the descent $\chi _0 : (V; P) \xrightarrow {\sim } (V_0; P_0)$ to be given, and will now discuss how to reconstruct a descent $(X_0; R_0)$ of (X; R) from it.

In one case, determining $(X_0; R_0)$ turns out to be particularly easy. Suppose that the reduced automorphism group of the pair (X, R) is trivial, so that $(W_0; Q_0) = (V_0; P_0)$. Then the image $D_0$ of the branch divisor D on $V_0$ is defined over F. Since $V_0$ admits the point $P_0$ over F, it is isomorphic to $\mathbb P^1$ over F. Let $V_0 \xrightarrow {\sim } \mathbb P^1$ be any isomorphism over F that sends $P_0$ to $\infty$, and let $p_0$ be a monic polynomial cutting out the image of ${{\mathrm{Supp}}}(D_0) \backslash \infty$. Let

$$\begin{aligned} X_0 : y^2 = p_0 (x) \end{aligned}$$

(4.3)

If R is not a Weierstrass point of X, then $\infty \notin {{\mathrm{Supp}}}(D_0)$, so that $p_0$ is of even degree. In this case we let $R_0$ be the point at infinity corresponding to the image of (0, 1) under the change of coordinates $(x,y) \leftarrow (1/x,y/x^{g+1})$. On the other hand, if R is a Weierstrass point of X, then $\infty \in {{\mathrm{Supp}}}(D_0)$ and $p_0$ is of odd degree, in which case we let $R_0$ be the point corresponding to (0, 0) under the aforementioned change of coordinates.

The following proposition is then immediate.

Proposition 4.8

Let (X; R) be a marked hyperelliptic curve over ${F}^{\mathrm {sep}}$ whose reduced automorphism group is trivial. Then the pair $(X_0; R_0)$ defined in (4.3) is a descent of (X; R).

Remark 4.9

In Proposition 4.8, the indicated point $R_0$ on $X_0$ admits a rational branch. In the case when R is a not a Weierstrass point of X, this is because $R_0$ is unramified, whereas in the latter case we get a rational branch for the map $(x,y) \mapsto x$. This follows from a local power series expansion at infinity; since $y^2 = x + O(x^2)$, we get the rational branch corresponding to the root $y = \sqrt{x} + O(x)$.

The quadratic twist defined by $c_0 y^2 = p_0 (x)$ also gives a descent of the marked curve (X; R); but if $c_0 \in F^{\times } \setminus F^{\times 2}$, the map $(x,y) \rightarrow x$ does not admit a rational branch at the point at infinity, rather the morphism $(x,y) \mapsto c_0 x$ gives a rational branch. In general, the passage from points to branches gives rise to further cohomological considerations, and any branch on a descended curve $(X_0; R_0)$ for the quotient map ${{\mathrm{Aut}}}(X_0; R_0) \backslash (X_0; R_0)$ becomes rational on some uniquely determined twist of $(X_0; R_0)$.

Now suppose that the pair (X; R) has nontrivial reduced automorphism group $G = {{\mathrm{Aut}}}(X) / \langle \iota \rangle$. This automorphism group is then cyclic, of order $n=\# G$. We will now show how to obtain a descent of (X; R) in this case.

Let $\alpha$ be a generator of G, and let $Q = q (R)$ be the image of R on the quotient $W = \langle \iota \rangle \backslash X$. Since we assumed that $\alpha$ was not unipotent, we can identify W with a projective line $\mathbb P^1$ with affine coordinate x in such a way that Q corresponds to the point $\infty \in \mathbb P^1 ({F}^{\mathrm {sep}})$ and such that $\alpha$ is defined by

$$\begin{aligned} \alpha (x) = \zeta _n x \end{aligned}$$

(4.4)

with $\zeta _n$ a primitive nth root of unity in ${F}^{\mathrm {sep}}$. Let p be the polynomial defining X as a cover of W. Then because of our normalizations, we either have

$$\begin{aligned} p (x) = \pi (x^n) \end{aligned}$$

(4.5)

or

$$\begin{aligned} p (x) = x \pi (x^n) . \end{aligned}$$

(4.6)

for some polynomial $\pi$. In the latter case, there is a single point at infinity that is a Weierstrass point of X, whereas in the former there are two ordinary points at infinity.

We normalize $\pi$ to be monic by applying a scaling in x (which does not affect the above assumptions on $\alpha$ and Q) and denote its degree by d. By our assumption that the reduced automorphism group is of order n, if we write

$$\begin{aligned} \pi (x) = \sum _{i=0}^{d} \pi _{i} x^{d-i} = x^d + \pi _1 x^{d-1} + \pi _2 x^{d-2} + \dots + \pi _{d-1} x + \pi _d \end{aligned}$$

(4.7)

then the subgroup of $\mathbb Z$ generated by $\{ i : \pi _i \ne 0 \}$ equals $\mathbb Z$.

Given a monic polynomial $\pi$ as above, we define the hyperelliptic curves $X_{\pi }$ and $X'_{\pi }$ by

$$\begin{aligned} X_{\pi } : y^2 = \pi (x) \end{aligned}$$

(4.8)

and

$$\begin{aligned} X'_{\pi } : y^2 = x \pi (x) . \end{aligned}$$

(4.9)

The former curve admits a distinguished point $R_{\pi } = (0:1)$ at infinity after the coordinate change $(x,y) \leftarrow (1/x,y/x^{g+1})$, whereas the latter admits a Weierstrass point at infinity (which corresponds to $R'_{\pi } = (0,0)$ under the same coordinate transformation).

Our only freedom left to modify the coefficients of $\pi$ is scaling in the x-coordinate: this induces an action of ${F}^{\mathrm {sep}}$ on the coefficients $\pi _i$, given by

$$\begin{aligned} \alpha (\pi _1 , \pi _2, \ldots , \pi _d) = (\alpha ^1 \pi _1 , \alpha ^2 \pi _2 , \ldots , \alpha ^d \pi _d). \end{aligned}$$

(4.10)

We can interpret the coefficients of $\pi$ as a representative of a point in the corresponding weighted projective space. Now there is a notion of a normalized representative of such a point [16]. We call a polynomial $\pi$ normalized if its tuple of coefficients is normalized.

The uniqueness of the normalized representative then shows the following theorem.

Theorem 4.10

Let (X; R) be a marked hyperelliptic curve over ${F}^{\mathrm {sep}}$ whose reduced automorphism group is tamely cyclic of order n. Then (X; R) admits a unique model of the form $(X_{\pi }, R_{\pi })$ or $(X'_{\pi }, R'_{\pi })$, with $\pi$ normalized, depending on whether R is a Weierstrass point of X or not.

Proof

It is clear from the above that (X; R) admits a model of the indicated form. On the other hand, let $\sigma \in G_F$ and consider the polynomial $\sigma (\pi )$ obtained by conjugating the coefficients of $\pi$ by $\sigma$. If the coefficients of $\pi$ were not stable under Galois, then we would find two distinct normalized representatives of the same weighed point, a contradiction. Therefore $\sigma (\pi ) = \pi$, and since $\sigma$ was arbitrary, our theorem is proved. $\square$

Remark 4.11

If we do not happen to know the field of moduli of the pair (X; R), then the approach above calculates it as well. Indeed, $M^{{F}^{\mathrm {sep}}}_F (X;R)$ is generated by the coordinates of the normalized representative $c_{I,0}$. A similar remark applies to the upcoming Proposition A.2.

4.4 The Klein quartic

We now apply the method of branches to a plane quartic, namely the Klein quartic, defined by the equation

$$\begin{aligned} x^3 y + y^3 z + z^3 x = 0. \end{aligned}$$

(4.11)

By classical results (see the complete exposition by Elkies [9]), this curve admits Belyĭ map of degree 168 that realizes the quotient map by its full automorphism group. In what follows, we will determine a model of the Klein quartic that admits a rational ramification point for this morphism that is of index 2. The geometry of the Klein quartic is exceedingly well-understood due to its moduli interpretation. As such, our result is not new; it can be found in a different form in work of Poonen–Schaefer–Stoll [22]. However, what makes our method new is the purely algebraic approach to the problem, which is available in rather greater generality.

Before starting, we modify our model slightly and instead consider the rational $S_3$ model from Elkies [9], which is given by

$$\begin{aligned} X : x^4+ & {} y^4 + z^4 + 6 (x y^3 + y z^3 + z x^3) - 3 (x^2 y^2 + y^2 z^2 + z^2 x^2) \nonumber \\+ & {} \ 3xyz (x + y + z) = 0 . \end{aligned}$$

(4.12)

Consider the field $K = \mathbb Q(i, \theta )$ where $\theta ^4=-7$. Over K, the curve X admits the involution defined by the matrix

$$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c} 2 &{} -3 &{} -6 \\ -3 &{} -6 &{} 2 \\ -6 &{} 2 &{} -3 \end{array}\right) \in {{\mathrm{PGL}}}_3(\mathbb Q) = {{\mathrm{Aut}}}(\mathbb P^2)(\mathbb Q) \end{aligned}$$

(4.13)

This involution has the fixed point

$$\begin{aligned} P= & {} ( (i + 1) \theta ^3 + 12 i \theta ^2 -21 (i-1) \theta + 66 : 3(i + 1) \theta ^3 + 36i\theta ^2\nonumber \\&-63 (i-1) \theta - 50 : 124 ) . \end{aligned}$$

(4.14)

Using branches, we will construct an isomorphism between the curve (X; P) and its conjugates that satisfies the cocycle relation.

Let $\sigma$ be the automorphism of K such that $\sigma (\theta ) = i\theta$ and $\sigma (i)=i$; and similarly let $\tau (\theta ) = \theta$ and $\tau (i) = -i$. Then $\sigma$ and $\tau$ satisfy the standard relations for the dihedral group of order 8, the Galois group of the extension K.

One surprise here is that it is impossible to construct a Weil cocycle for the full extension $K \!\mid \!\mathbb Q$. In fact there exist isomorphisms $\varphi _{\sigma } : (\sigma (X);\sigma (P))$ and $\varphi _{\tau } : (\tau (X); \tau (P))$ but no matter which we take, the automorphism

$$\begin{aligned} \varphi _{\sigma } \sigma (\varphi _{\sigma }) \sigma ^2(\varphi _{\sigma }) \sigma ^3(\varphi _{\sigma }) \end{aligned}$$

(4.15)

of X that corresponds to the cocycle relation for $\sigma$ is never trivial, but always of order 2. This strongly suggests that we should find an extension L of $\mathbb Q$ with Galois group dihedral of order 16 into which K embeds. Such extensions do indeed exist; one can be found in the ray class field of $\mathbb Q(i)$ of conductor 42, obtained as the splitting field of the polynomial $63 t^8 - 70 t^4 - 9$ over $\mathbb Q$.

The above extension is exactly the one that shows up when considering the branches of the Belyĭ map $q : X \rightarrow {{\mathrm{Aut}}}(X) \backslash X$, which can be calculated by using the methods in Poonen–Schaefer–Stoll [22]. Note that it would have sufficed for our purposes to quotient out by the involution fixing P; however, we use the Belyĭ map to show that it, too, can be used for purposes of descent.

We use $x = s$ as a local parameter at the point P, and modify the coordinate on ${{\mathrm{Aut}}}(X) \backslash X$ such that P is sent to 0. After dehomogenizing by putting $z = 1$ and determining $y = y(s)$ as a power series in s, we obtain a power series expression in s for q by composition, that is, by evaluating the Belyĭ map q, considered as an explicit function q (x, y) of x and y, in $( x(s), y(s) ) = (s, y(s))$. This power series looks like

$$\begin{aligned} s \rightarrow c_2 s^2 + O(s^3) \end{aligned}$$

(4.16)

The leading coefficient $c_2$ of this power series has minimal polynomial

$$\begin{aligned} 27889 t^4 - 1869588 t^3 - 18805122 t^2 + 1869795900 t - 25658183943 \end{aligned}$$

(4.17)

over the rationals. Since the branch is an inverse of the power series above under composition, it will look like

$$\begin{aligned} s \rightarrow (1/\sqrt{c_2}) s^{1/2} + O(s) . \end{aligned}$$

(4.18)

The field generated by its leading coefficient can therefore be obtained as the splitting field of the polynomial (4.17) evaluated at $t^2$. This is indeed the extension L constructed above: we have

$$\begin{aligned} \frac{1}{\sqrt{c_2}} = \frac{1}{16032}(211239\eta ^7 - 66339\eta ^5 - 163835\eta ^3 + 98343\eta ) \end{aligned}$$

(4.19)

where $\eta$ is a root of $63 t^8 - 70 t^4 - 9$.

We now evaluate all Puiseux series for x, y, z involved at the branch $(1/\sqrt{c_2}) s^{1/2} + O(s)$ instead of s. This allows us to study the Galois action; by dehomogenizing and comparing the conjugated Puiseux series $(\sigma (X)(s), \sigma (y)(s))$ with the fractional linear transformation of (x, y) under the automorphisms of X, we can read off which $\alpha _{\sigma }$ maps the former branch to the latter.

There exist elements $\widetilde{\sigma },\widetilde{\tau }$ satisfying the relations for the dihedral group of order 16, generating the Galois group of L, and restricting to $\sigma ,\tau$ on K. For these elements, the conjugate branch $(\widetilde{\sigma }(x)(s), \widetilde{\sigma }(y)(s))$, respectively $(\widetilde{\tau }(x)(s), \widetilde{\tau }(y)(s))$, is sent to (x(s), y(s)) by the ambient transformation

$$\begin{aligned} \alpha _{\widetilde{\sigma }} = \left( \begin{array}{c@{\quad }c@{\quad }c} -6 &{} -2 \theta ^2 + 2 &{} \theta ^2 + 11 \\ 2 \theta ^2 + 2 &{} -10 &{} 3 \theta ^2 + 1 \\ -\theta ^2 + 11 &{} -3 \theta ^2 + 1 &{} 2 \end{array}\right) , \end{aligned}$$

(4.20)

respectively

$$\begin{aligned} \alpha _{\widetilde{\tau }} = \left( \begin{array}{c@{\quad }c@{\quad }c} -6 &{} 2 \theta ^2 + 2 &{} -\theta ^2 + 11 \\ -2 \theta ^2 + 2 &{} -10 &{} -3 \theta ^2 + 1 \\ \theta ^2 + 11 &{} 3 \theta ^2 + 1 &{} 2 \end{array}\right) . \end{aligned}$$

(4.21)

In particular $\sigma (P)$, respectively $\tau (P)$, is sent to P by $\alpha _{\widetilde{\sigma }}$, respectively $\alpha _{\widetilde{\tau }}$. By uniqueness of $\alpha _{\widetilde{\sigma }}$ and $\alpha _{\widetilde{\tau }}$ when considering its action on the branches, we do get a Weil cocycle this time. Moreover, the corresponding descent leads to a rational branch of the Belyĭ map q. After calculating a coboundary and polishing the result, we get the model

$$\begin{aligned} X_0 : x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + 18 x y z^2 + 9 y^2 z^2 - 9 z^4 = 0 \end{aligned}$$

(4.22)

as in Poonen–Schaefer–Stoll [22]. The curve $X_0$ admits the point (0 : 1 : 0), which is a rational branch for the quotient map $(x:y:z) \mapsto (x:y:z^2)$ to the curve

$$\begin{aligned} x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + 18 x y z + 9 y^2 z - 9 z^2 = 0 \end{aligned}$$

(4.23)

in $\mathbb P(1,1,2)$.

We do not give explicit isomorphisms with the models (4.11) and (4.12) here, as they are slightly unwieldy to write down; a matrix $\alpha _0$ inducing an isomorphism $X \xrightarrow {\sim } X_0$ can be found by solving the equations

$$\begin{aligned} \begin{array}{ll} \alpha _0 \alpha _{\sigma } &{} = \sigma (\alpha _0) \\ \alpha _0 \alpha _{\tau } &{} = \sigma (\alpha _0) \end{array} \end{aligned}$$

(4.24)

up to a scalar, where $\alpha _\sigma ,\alpha _\tau \in {{\mathrm{PGL}}}_3(L)={{\mathrm{Aut}}}(\mathbb P^2)(L)$ induce the isomorphisms $\sigma ,\tau$, respectively. The equations (4.24) are none other than (2.3), and are the explicit version of Step 5 in Method 4.1. Once we lift the ${{\mathrm{PGL}}}_3 (L)$-cocycle $\sigma \mapsto \alpha _{\sigma }$ to ${{\mathrm{GL}}}_3 (L)$, then we can forget about the scalar factor, so that (4.24) can be solved by expressing the entries of $\alpha _0$ as combinations of a $\mathbb Q$-basis of L. This in turn reduces solving (4.24) to solving a system of linear equations; it then remains to find a solution of relatively small height to find a reasonable descent morphism.

Remark 4.12

Lifting the cocycle $\sigma \mapsto \alpha _{\sigma }$ mentioned above to ${{\mathrm{GL}}}_3 (L)$ is not trivial; for general plane curves it requires modifying an arbitrary set-theoretic lift by suitable scalars, which in turn comes down to determining the rational points on a certain conic. For plane quartics, however, things are slightly easier, since we can consider the morphism on the space of differentials that our cocycle induces.

The case of a ramification point of index 3 is considerably easier and yields the model

$$\begin{aligned} 7 x^3 z + 3 x^2 y^2 - 3 x y z^2 + y^3 z - z^4 = 0 , \end{aligned}$$

(4.25)

on which the points (1 : 0 : 0) and (0 : 1 : 0) are fixed for the automorphism $(x:y:z) \mapsto (\omega ^2 x: \omega y: z)$ where $\omega$ is a primitive cube root of unity. In this case, we do not insist on the branch being rational in order to get a nicer model for the marked curve.

Finally, the original model (4.11) already admits the rational point (0 : 0 : 1), which is stable under the automorphism $(x:y:z) \rightarrow (\zeta _7^4 x : \zeta _7^2 y : \zeta _7 z)$ of order 7.

Remark 4.13

Yet another way to find the descent above is to diagonalize the automorphism 4.13 to the standard diagonal matrix with diagonal $(1,1,-1)$, which can be done by a $\mathbb Q$-rational transformation. The 4 fixed points of the involution are then on the line $z = 0$. By transforming the coordinates x and y over $\overline{\mathbb Q}$ we can ensure that the involution remains given by the diagonal matrix $(1,1,-1)$ and the set 4 points remains still defined over $\mathbb Q$, while also putting one of these points at (0 : 1 : 0). This normalization reduces the automorphism group sufficiently to make the rest of the descent straightforward.

4.5 Wild branches and wild descent

Determining branches in the wild case is less intuitive than in the tame case. It turns out that branches can still be represented by certain power series, but this time the denominators of the exponents involved are unbounded—so although the branch belongs to the separable closure of the power series ring, it cannot be represented by a Puiseux series. We instead have to take recourse to generalized power series [19], in which the denominators of the exponents form an unbounded set. We illustrate what this involves by considering two examples.

Example 4.14

We start off with the standard Artin–Schreier extension

$$\begin{aligned} y^p - y = x . \end{aligned}$$

(4.26)

The projection $f : (y, x) \rightarrow x$ is then unramified outside $\infty$, and in fact this is a Galois cover, with automorphism group generated by $(x, y) \mapsto (x, y + 1)$. Transforming coordinates to interchange 0 and $\infty$, the extension (4.26) becomes

$$\begin{aligned} y^p - y z^{p - 1} = z^{p - 1} . \end{aligned}$$

(4.27)

Now the map f has $(y, z) \mapsto z$ is a Galois cover, unramified outside 0. Its automorphism group is isomorphic to $\mathbb Z/ p \mathbb Z$, and a generator is given by $(y,z) \mapsto (y + z, z)$.

We want to find a solution to (4.27) in a generalized power series (with positive exponents). For this, we first set $y = c z^e$. This gives the equation

$$\begin{aligned} c^p z^{p e} - c z^{p - 1 + e} = z^{p - 1} . \end{aligned}$$

(4.28)

To get cancellation with $z^{p - 1}$, the smallest among the exponents pe and $p - 1 + e$ has to equal $p - 1$. Now since $e > 0$ we can never have that the latter exponent does the job. We get that $e = 1 - p^{-1}$ and $c^p = 1$, so we have our leading monomial $z^{1-p^{-1}}$ of y. Writing $y = z^{1-p^{-1}} + c z^e$ and continuing iteratively, we get rational exponents of z with arbitrarily large powers of p in the denominator. More precisely, we obtain the generalized power series

$$\begin{aligned} y = \sum _{n = 1}^{\infty } z^{1 - p^{-n}}; \end{aligned}$$

(4.29)

to be precise, we think of (4.29) as encoding a sequence given by its partial sums, each of which belongs to a ring $k(t^{1/p^n})$ for $n \in \mathbb Z_{\ge 0}$. These partial sums “converge” to a root of (4.27), but only in a formal sense.

The exponents pe and $p - 1 + e$ cancel mutually precisely when $e = 1$, confirming Proposition 3.5 and recovering exactly the images of (4.29) under the automorphism group, to wit

$$\begin{aligned} y = c z + \sum _{n = 1}^{\infty } z^{1 - p^{-n}}, \quad \; c \in \mathbb F_p . \end{aligned}$$

(4.30)

We see how the wildness of the ramification leads to the denominators of the exponents $1 - p^{-n}$ being unbounded.

Example 4.15

As a second illustration, we use a family of hyperelliptic curves in characteristic 2 considered by Igusa [15], namely

$$\begin{aligned} y^2 - y = x^3 + a x + b x^{-1} . \end{aligned}$$

(4.31)

This curve has a distinguished point at infinity. Igusa [15] proved that the family (4.31) contains duplicates; more precisely, modifying $(a , b) \mapsto (\zeta _3 a, \zeta _3^{-1} b)$ does not change the isomorphism class. The corresponding invariant subfield of $\mathbb F_2 (a,b)$ is $\mathbb F_2 (a^3,a b)$, and we will use the marked point at infinity to descend to this subfield.

Transforming projectively as before, we get the equation

$$\begin{aligned} X : x - x z = (b + a x^2 + x^4) z^2 . \end{aligned}$$

(4.32)

The hyperelliptic quotient map is given by $(x,z) \mapsto x$, and the automorphism group (of either the curve or the marked curve) is generated by $(x,z) \mapsto (x,z/(z + 1))$. Determining a branch, we obtain one generalized power series that starts as

$$\begin{aligned} z = b^{-1/2} x^{1/2} + b^{-3/4} x^{3/4} + \cdots \end{aligned}$$

(4.33)

Now note that from (4.32) it follows that the second branch can be obtained by replacing z by $z + c$, where

$$\begin{aligned} c = \frac{x}{b + a x^2 + x^4} = b^{-1} x + a b^{-2} x^3 + \cdots \end{aligned}$$

(4.34)

Regardless, note that if we let $\sigma$ be the automorphism sending (a, b) to $(\zeta _3 a, \zeta _3^{-1} b)$, there is an obvious isomorphism $\varphi _{\sigma } : X^{\sigma } \rightarrow X$ given by $(x,z) \mapsto (\zeta _3 x, z)$. There is a second isomorphism $(x,z) \rightarrow (\zeta _3 x, z / (z + 1))$. However, the first isomorphism respects the branches, as can indeed be seen by looking at the exponents in the corresponding expansions alone; since only the second branch above contains a linear term in x, only $\varphi _{\sigma }$ can send it to its conjugate.

Taking a corresponding coboundary corresponding to the cocycle generated by $\sigma$ is easy and comes down to substituting bx for x in (4.32). We get a descent

$$\begin{aligned} x - x z = (1 + a b x^2 + b^3 x^4) z^2 \end{aligned}$$

(4.35)

in which the rational branch is clearly visible. Transforming back to Igusa’s form, we get the representative family

$$\begin{aligned} y^2 - y = b^3 x^3 + a b x + x^{-1} . \end{aligned}$$

(4.36)

This family can also be obtained directly; our purpose in this example was to illustrate how branches can be used explicitly as a tool for descent. Descending other families of hyperelliptic curves in this way is a topic for future work.

5 Counterexamples

In this final section, we prove Theorem B and consider two (counter)examples that show that for general (non-singular) marked curves the conclusion of Theorem 1.8 is false. That is, we exhibit pointed curves (and associated pointed Belyĭ maps) that are defined over $\mathbb C$ and have field of moduli $\mathbb R$ with respect to the extension $\mathbb C\!\mid \!\mathbb R$, yet do not descend to $\mathbb R$. In the first example, neither the singular curve nor its normalization descends to $\mathbb R$. In the second example, the curve itself is defined over $\mathbb R$ but the marked curve is not, so rigidifying by marking a point creates a descent problem where previously none existed.

5.1 First example: the curve does not descend

We will employ the criterion (2.4) of Weil descent with respect to the extension $\mathbb C\!\mid \!\mathbb R$.

Lemma 5.1

Let W be a variety over $\mathbb R$ and let $\varrho \in {{\mathrm{Aut}}}(W)(\mathbb R)$ have order 4. Suppose that $X \subseteq W$ is a curve over $\mathbb C$ such that $\varrho (\overline{X}) = X$ and ${{\mathrm{Aut}}}(X)(\mathbb C) = \langle \varrho ^2|_X \rangle \simeq \mathbb Z/2\mathbb Z$. Suppose $P \in W(\mathbb R) \cap X(\mathbb C)$ has $\varrho (P) = P$. Then both X and (X; P) have field of moduli $\mathbb R$ with respect to the extension $\mathbb C\mid \mathbb R$ but neither descends to $\mathbb R$.

Proof

By hypothesis, the map $\varphi = \varrho |_{\overline{X}}$ gives an isomorphism $\varphi : \overline{X} \xrightarrow {\sim } X$, and $\varphi (P) = \varrho (\overline{P}) = \overline{P} = P$ since $P \in W(\mathbb R)$ and $\varrho$ is defined over $\mathbb R$. We have $\overline{\varphi } \varphi = (\varrho ^2)|_{\overline{X}} \ne 1$ on both W and X. Therefore $\varphi$ does not give rise to a descent of X to $\mathbb R$. Neither does $\varphi ' = \varrho ^3|_{\overline{X}}$, since again it similarly satisfies $\overline{\varphi }'\varphi ' = (\varrho ^2)|_{\overline{X}} \ne 1$. The maps $\varphi ,\varphi '$ are the only isomorphisms $\overline{X} \rightarrow X$ since any two would differ by an element of ${{\mathrm{Aut}}}(X)(\mathbb C)=\{1,\varrho ^2|_X\}$, so the result follows by the necessity part of the Weil coycle criterion (Theorem 2.1). $\square$

To give an explicit example of Lemma 5.1, we take $W=\mathbb P^2$ and the automorphism

$$\begin{aligned} \begin{array}{ll} \varrho :\mathbb P^2 \rightarrow \mathbb P^2 \\ (x:y:z)\rightarrow (y:-x:z). \end{array} \end{aligned}$$

(5.1)

Lemma 5.2

We have

$$\begin{aligned} \{P \in \mathbb P^2(\mathbb C) : \varrho (P)=P \} = \{ (0:0:1) , (\pm i:1:0) \}. \end{aligned}$$

(5.2)

Proof

In the affine open where $z = 1$ we get the equations $y = x = -x$, whose unique solution is given by $x = y = 0$. On the other hand, when $z = 0$ we find $(x : y) = (y : -x) \in \mathbb P^1(\mathbb C)$. This implies $x = iy$, and the result follows. $\square$

By Lemma 5.2, the only point $P \in \mathbb P^2(\mathbb R)$ with $\varrho (P)=P$ is $P=(0:0:1)$. Our curve $X \subseteq \mathbb P^2$ is a projective plane curve, defined by a homogeneous polynomial $h(x,y,z) \in \mathbb R[x,y,z]$ with $h(0:0:1)=0$. The condition that $\varrho (\overline{X}) = X$ is equivalent to the condition that $h(y:-x:z)$ is a scalar multiple of $\overline{h}(x:y:z)$; for simplicity, we assume that

$$\begin{aligned} h(y:-x:z)=\overline{h}(x:y:z). \end{aligned}$$

The condition that $\varrho ^2|_X \in {{\mathrm{Aut}}}(X)(\mathbb C)$ implies that

$$\begin{aligned} h(x:y:-z)=h(x:y:z). \end{aligned}$$

We take $\deg h=4$ so that ${{\mathrm{Aut}}}(X)(\mathbb C)$ is finite. The above conditions then imply that

$$\begin{aligned} h(x,y,z) = ax^4 + \overline{a}y^4 + (bx^2 - \overline{b}y^2) x y + (cx^2 + \overline{c}y^2)z^2 + rx^2y^2 + sixyz^2 \end{aligned}$$

(5.3)

with $a,b,c \in \mathbb C$ and $r,s \in \mathbb R$. We observe that $P = (0:0:1)$ is a singular point of X and P is a nodal double point if and only if $s^2 \ne -4|c|^2$.

To apply Lemma 5.1, we need to select coefficients of h such that X has no automorphisms beyond $\varrho ^2|_X$; we will show this is true for a particular choice, which will then in fact imply that the resulting computations are true for a general choice, i.e., for the curve defined by equation (5.3) over $\mathbb C[a,b,c,r,s,\Delta ^{-1}]$ where $\Delta$ is the discriminant of h. We consider the curve with

$$\begin{aligned} (a,b,c,r,s)=(0,i,1+i,1,2) \end{aligned}$$

so that X is described by the homogeneous equation

$$\begin{aligned} X : i (x^2 + y^2) x y + ((1 + i) x^2 + (1 - i) y^2) z^2 + x^2 y^2 + 2 i x y z^2 = 0 . \end{aligned}$$

(5.4)

Standard techniques in computational algebraic geometry (we performed our computation in Magma [3]) show that P is the only singular point of X, so that X has geometric genus 2. We verify that ${{\mathrm{Aut}}}(X)(\mathbb C) \simeq \mathbb Z/2\mathbb Z$ by computing that the normalization of X over $\mathbb R$ is given by

$$\begin{aligned} Y : y^2 = (i+1)x^5 + (-i - 1)x^4 + 4x^3 + (-i + 1)x^2 + (-i + 1)x \end{aligned}$$

(5.5)

and by a computation using invariant theory (again a computation in Magma [17]) we see that indeed Y has automorphism group $\mathbb Z/ 2 \mathbb Z$ generated by the hyperelliptic involution. Since every automorphism group of a singular curve lifts to its normalization, we see that indeed ${{\mathrm{Aut}}}(X) (\mathbb C) \simeq \mathbb Z/ 2 \mathbb Z$. Lemma 5.1 therefore applies to show that 5.4 is an explicit counterexample.

Alternatively, one can compute that the Igusa–Clebsch invariants of Y are defined over $\mathbb R$, so that the field of moduli is $\mathbb R$, but that there is an obstruction to the curve Y being defined over $\mathbb R$, as described by Mestre [20]. On the other hand, one shows that the existence of a descent of Y is equivalent with that for X since P is the unique singular point of X.

Remark 5.3

With the equation (5.5) in hand, one sees how to recover this class of examples in a different way, as follows. Let $W_0$ be the conic defined by $x^2+y^2+z^2=0$ in $\mathbb P^2$ over $\mathbb R$. Then $W_0(\mathbb R) = \emptyset$. Let $S \subseteq W_0(\mathbb C)$ be a subset of 6 distinct points over $\mathbb C$ forming 3 complex conjugate pairs. Then there exists a smooth hyperelliptic curve $X'$ over $\mathbb C$ branched over the set S that does not descend to $\mathbb R$: this follows from work of Mestre [20], but it is also a rephrasing of the results by Shimura and Earle reviewed by Huggins [14, Chapter 5].

Now let $Q_1,Q_2 \in S$ be such that $Q_2 = \overline{Q_1}$ and let $P_1,P_2 \in X'(\mathbb C)$ be their ramified preimages. By the cocycle condition, the points $\overline{P_1}, \overline{P_2} \in \overline{X'} (\mathbb C)$ are mapped to $P_1,P_2 \in X' (\mathbb C)$ under the isomorphism between $\overline{X'}$ and $X'$. Now pinch together the points $P_1,P_2$ to obtain a curve X over $\mathbb C$ with a double point P. Then $\overline{P} \in \overline{X} (\mathbb C)$ is mapped to its conjugate P on X (“itself” on $W_0$) by construction, and so (X; P) also has field of moduli $\mathbb R$.

To obtain a map $f:X \rightarrow \mathbb P^1$ unramified away from $\{0,1,\infty \}$, thereby completing the first proof of Theorem B, we first choose a function $\pi$ on W defined over $\mathbb R$ that is invariant under $\varrho$ and that is nonconstant on X. This yields a morphism $\pi |_X:X \rightarrow \mathbb P^1$, and because $\varrho$ maps X to $\overline{X}$ and $\pi$ is invariant under $\varrho$, the branch locus of $\pi |_X$ is invariant under complex conjugation.

In the specific example (5.4) above, we take $\pi = x^2 + y^2$ and compute that the branch locus is described by the vanishing of the homogeneous polynomial

$$\begin{aligned} 3072u^7v + 4352u^6v^2 + 5840u^5v^3 + 3424u^4v^4 + 920u^3v^5 + 104u^2v^6 + 5uv^7, \end{aligned}$$

(5.6)

which shows that it is in fact even stable under ${{\mathrm{Gal}}}(\overline{\mathbb Q}\!\mid \!\mathbb Q)$.

We can now apply the following slight strengthening of the result of Belyĭ.

Theorem 5.4

(Belyĭ [1]) Let $g : X \rightarrow \mathbb P^1$ be a map of curves over $\overline{\mathbb Q}$ whose branch locus in $\mathbb P^1(\overline{\mathbb Q})$ is defined over a number field $F \subset \overline{\mathbb Q}$. Then there exists a morphism $\alpha : \mathbb P^1 \rightarrow \mathbb P^1$ defined over F such that $f = \alpha \circ g$ is a Belyĭ map.

Theorem 5.4 applies equally well to the extension $\mathbb C\!\mid \!\mathbb R$, essentially since every Belyĭ map is already defined over $\overline{\mathbb Q}$. We apply the above for $g = \pi$, whose branch locus is $\mathbb R$-rational, and let $f = \alpha \pi : X \rightarrow \mathbb P^1$ be the map thus obtained, unramified away from $\{0,1,\infty \}$. Now f is $\varrho$-invariant since $\pi$ is, and because we have $\overline{\pi } = \pi \varrho = \pi$ on W we also have

$$\begin{aligned} \overline{f} \varrho = \overline{\alpha \pi } \varrho = \alpha \overline{\pi } \varrho = \alpha \pi \varrho ^2 = \alpha \pi = f. \end{aligned}$$

(5.7)

So indeed the field of moduli of the pointed map $(X;f:X \rightarrow \mathbb P^1; P; 0,1,\infty )$ with respect to the extension $\mathbb C\!\mid \!\mathbb R$ equals $\mathbb R$. However, because X does not descend to $\mathbb R$, neither does $(X;f;P;0,1,\infty )$.

5.2 Second example: the curve descends

In this second example, the curve itself descends but the marked curve does not. The setup is described by the following lemma.

Lemma 5.5

Let X be a curve over $\mathbb R$ such that

$$\begin{aligned} {{\mathrm{Aut}}}(X) (\mathbb C) = {{\mathrm{Aut}}}(X) (\mathbb R) = \langle \varrho \rangle \simeq \mathbb Z/4\mathbb Z. \end{aligned}$$

(5.8)

Suppose that $P \in X(\mathbb C) \setminus X(\mathbb R)$ satisfies $\overline{P} = \varrho (P)$. Then $(X_{\mathbb C} ; P)$ has field of moduli $\mathbb R$ with respect to the extension $\mathbb C\!\mid \! \mathbb R$ but does not descend to $\mathbb R$.

Proof

The field of moduli claim follows by taking $\varphi = \varrho$ to be the identity map, since $\overline{X} = X$ and $\varrho (\overline{P}) = P$ as $\varrho$ is defined over $\mathbb R$. Moreover, we have

$$\begin{aligned} P = \overline{\overline{P}} = \overline{\varrho (P)} = \varrho (\overline{P}) = \varrho ^2(P) . \end{aligned}$$

(5.9)

As in the proof of Lemma 5.1, we show that no element $\omega \in {{\mathrm{Aut}}}(X)(\mathbb C) = \langle \varrho \rangle$ gives a descent. The cocycle condition implies $\omega \overline{\omega } = \omega ^2 = 1$, so $\omega$ is an involution. And yet by (5.9) and the assumption that $\overline{P} \ne P$ implies that neither the identity nor $\varrho ^2$ gives a descent of (X; P), since they do not give an isomorphism to $(X;\overline{P})$. Therefore no descent of (X; P) exists. $\square$

Remark 5.6

Admitting for a moment that curves as in Lemma 5.5 do exist, we see that rigidifying by marking a point may very well lead to a descent problem where previously none existed: while both $(X_{\mathbb C},P)$ and a fortiori $X_{\mathbb C}$ have field of moduli $\mathbb R$ with respect to the extension $\mathbb C\!\mid \! \mathbb R$, the obstruction vanishes for the non-marked curve $X_{\mathbb C}$, whereas it is non-trivial for the pair $(X_{\mathbb C},P)$. So while non-marked curves $X_{\mathbb C}$ may not descend, as was noted by Birch [2], neither may a given rigidification of $X_{\mathbb C}$, as we saw in the last subsection, and in fact it may even complicate descent matters further when the automorphism group remains nontrivial.

To exhibit an explicit curve meeting the requirements of Lemma 5.5, we first construct a smooth curve Y defined over $\mathbb R$ with

$$\begin{aligned} {{\mathrm{Aut}}}(Y) (\mathbb R) = {{\mathrm{Aut}}}(Y) (\mathbb C) \simeq \mathbb Z/4\mathbb Z. \end{aligned}$$

To do this, we choose distinct $x_1, x_2 \in \mathbb C$ and let

$$\begin{aligned} p (x) = (x^2 + 1) \prod _{i = 1}^2 (x - x_i) (x - \overline{x}_i) (x + 1/x_i) (x + 1/\overline{x}_i) \in \mathbb R[x]. \end{aligned}$$

(5.10)

We see that $p(-1/x)=p(x)/x^{10}$. One then expects that generically the hyperelliptic curve

$$\begin{aligned} Y : y^2 = p(x) \end{aligned}$$

will have automorphism group generated by $\varrho (x,y)=(-1/x, y/x^5)$. It is possible to verify [17] that this indeed happens for the choice $x_1 = 1 + i , x_2 = 2 + i$, which (after scaling) yields the genus 4 hyperelliptic curve

$$\begin{aligned} Y : y^2 = 10x^{10} - 42x^9 + 67x^8 - 36x^7 + 23x^6 + 23x^4 + 36x^3 + 67x^2 + 42x + 10.\nonumber \\ \end{aligned}$$

(5.11)

Consider the point $Q = (1 + i,0) \in Y(\mathbb C)$. To get Lemma 5.5 to apply, we need $\overline{Q}=\varrho (Q)$. Of course this cannot be the case on Y in light of Theorem 1.8, but we can make it true by construction if we pinch together the points $\overline{Q}$ and $\varrho (Q)$. Moreover, if we take care to pinch Q with $\varrho (\overline{Q})$ similarly, then the resulting contraction morphism $c : Y \rightarrow X$ will be defined over $\mathbb R$. If we let $P = c(Q)$, then by construction, the pair (X; P) will satisfy the conditions of Lemma 5.5. Indeed, any automorphism of a singular curve lifts to its normalization, and conversely the automorphisms of Y all transfer to X because of our construction of the contraction, so that indeed

$$\begin{aligned} {{\mathrm{Aut}}}(X)(\mathbb R)={{\mathrm{Aut}}}(X)(\mathbb C) \simeq {{\mathrm{Aut}}}(Y)(\mathbb C) \simeq \mathbb Z/ 4 \mathbb Z; \end{aligned}$$

the stabilizer of P again corresponds to the subgroup of $\mathbb Z/ 4 \mathbb Z$ of order 2.

A contraction morphism c can be constructed in many different ways, none of which is particularly pleasing from the point of view of finding an equation. The first approach is to find a sufficiently ample linear system and extract the sublinear system of sections whose values at $\overline{Q}$ and $\varrho (Q)$ (resp. Q and $\varrho (\overline{Q})$) coincide. This leads to ambient spaces that are too large to give rise to attractive equations.

An alternative approach is as follows. Define

$$\begin{aligned} \begin{array}{ll} q(x) &{} = (x - x_1)(x - \overline{x}_1) (x + 1/x_1)(x + 1/\overline{x}_1) = x^4 - x^3 + \dfrac{1}{2} x^2 + x + 1, \\ r(x) &{} = \dfrac{2}{3} x^3 - 2 x^2 + x. \end{array} \end{aligned}$$

(5.12)

and let $c : \mathbb A^2 \rightarrow \mathbb A^4$ be given by

$$\begin{aligned} c(x,y) = (q(x), x q(x), r(x), y) = (t,u,v,w). \end{aligned}$$

(5.13)

Given a point in the image of Y under c, we can always read off the original y-coordinate from w. We can also recover the x-coordinate u / t as long as $t \ne 0$. The latter only occurs if x is a root of q; these roots all have the property that $t = u = w = 0$. On the other hand, one verifies that r assumes exactly two distinct values on this set of roots; moreover, the preimage of one of these values is given by $\{ Q, \varrho (\overline{Q}) \}$, and the other by $\{ \overline{Q}, \varrho (Q) \}$. We see that the morphism c constracts the pairs $\{ Q , \varrho (\overline{Q}) \}$ and $\{ \overline{Q}, \varrho (Q) \}$ exactly in the way that we wanted.

This method gives only an affine open of X, but this open contains all the singular points that we are interested in, so that completing the corresponding model smoothly at infinity will give the desired model of X. On the patch $t \ne 0$ we can describe the image of c by the following equations:

$$\begin{aligned} \begin{array}{l} t^5 - t^4 - t^3 u - 1/2 t^2 u^2 + t u^3 - u^4 = 0, \\ t^3 v - t^2 u + 2 t u^2 - 2/3 u^3 = 0, \\ 10t^{10} w^2 - 10t^{10} - 42 t^9 u - 67 t^8 u^2 - 36 t^7 u^3 - 23 t^6 u^4 \\ \quad - 23 t^4 u^6 + 36 t^3 u^7 - 67 t^2 u^8 + 42 t u^9 - 10u^{10} = 0. \end{array} \end{aligned}$$

(5.14)

Adding a few more coordinates, we can also describe the automorphism $\varrho$: using the contraction

$$\begin{aligned} c (x,y)= & {} (q(x), x q(x), r(x), y, q(-1/x), -1/x q(-1/x), r(-1/x), y/x^5)\nonumber \\= & {} (t,u,v,w,t',u',v',w'), \end{aligned}$$

(5.15)

it can be described by

$$\begin{aligned} \varrho (t,u,v,w,t',u',v',w') = (t',u',v',w',t,u,v,-w). \end{aligned}$$

(5.16)

In both cases, the full ideal of the image of c can be recovered by using Gröbner bases.

To conclude our argument; by Theorem 5.4, we can find a map $f:X \rightarrow \mathbb P^1$ unramified outside $\{0,1,\infty \}$ such that (X, f; P) has the same automorphism group: one takes any function $\varphi$ defined over $\mathbb R$ that is invariant under ${{\mathrm{Aut}}}(X,P)$ and postcompose with a function h following Belyĭ to get a map $\varphi = h \varphi _0$ with the same field of moduli and the same obstruction.

Remark 5.7

The curve in the first counterexample had exactly one singular point—in a sense, it was “marked” anyway. By contrast, the curve in this subsection has two singular points, which makes it important to keep track of the chosen marking.

Remark 5.8

There should be many more ways to construct singular curves over $\mathbb R$ for whom the field of moduli is no longer a field of definition after rigidification by a singular point.

For example, if we take the automorphism $\varrho :\mathbb P^3 \rightarrow \mathbb P^3$ given by the cyclic permutation of the coordinates $\varrho (x:y:z:w)=(y:z:w:x)$ and considers the generic complete intersection X of two polynomials invariant under $\varrho$ containing the point $P=(1:i:1:i)$ with $\varrho (P)=\overline{P}$, then we expect that the curve X will satisfy the hypotheses above. For example, if we define

$$\begin{aligned} \varrho (m)=m+\varrho ^*(m)+(\varrho ^{*2})(m)+(\varrho ^{*3})(m) \end{aligned}$$

(5.17)

for $m \in k[x,y,z,w]$ a monomial, then we believe that the curve X of genus 7 defined by

$$\begin{aligned} \begin{array}{ll} \varrho (x^2)=x^2 + y^2 + z^2 + w^2 = 0 \\ 5 \varrho (x^4) \!+\! \varrho (x^3y) + \varrho (x^3z) -4\varrho (x^2yw) -2 \varrho (x^2zw) \!+\! 3\varrho (x^2y^2) \!+\! 7\varrho (xyzw) = 0 \end{array} \end{aligned}$$

(5.18)

has this property, as we can verify that ${{\mathrm{Aut}}}(X)(\mathbb F_p) \simeq \mathbb Z/4\mathbb Z$ for a number of large primes p; however, it is much more involved to provably compute the automorphism group for such a curve, which is why we have preferred to stick with this admittedly more special case.

Appendix A: Descent of marked Galois Belyĭ maps in genus zero

In this appendix, we show how branches can be used to provide explicit descent of marked Galois Belyĭ maps in genus zero. The results are classical, but the method to derive them is quite pleasing so we provide it here.

In the absence of unipotent automorphisms, a (geometrically generically) Galois map ${W \rightarrow V}$ between curves of genus 0 over ${F}^{\mathrm {sep}}$ is always a quotient by one of the groups $C_n, D_n, A_4, S_4, A_5$ for $n \ge 2$. Because we want to use the language of Belyĭ maps, throughout this appendix we suppose ${{\mathrm{char}}}F = 0$; the results hold equally well when $\#G$ is coprime to ${{\mathrm{char}}}F$, but we are reluctant to use the term Belyĭ maps in this case. The quotients mentioned previously are then all Belyĭ maps over ${F}^{\mathrm {sep}}$, and in what follows we wish to consider some special descents of these maps. This will expand on the results in Couveignes–Granboulan [5], in which forms of these Belyĭ maps were already given.

1.1 A.1 Descent of lax Belyĭ maps

We will slightly broaden our notion of Belyĭ maps over F in order to allow their branch divisor to be an arbitrary F-rational divisor instead of merely consisting of multiples of points individually defined over F, which is to say want to allow the situation where the points on $\mathbb P^1$ are not marked.

Definition A.1

A lax Belyĭ map over F is a map $f : X \rightarrow \mathbb P^1$ of curves over F that is ramified above at most three points. A marked lax Belyĭ map over F is a pair (X; f; P), where $f:X \rightarrow \mathbb P^1$ is a lax Belyĭ map over F and $P \in X(F)$ is a point of X over F.

By Birch’s theorem, any marked lax Belyĭ map descends to its field of moduli, and the method of branches gives a descent of marked lax Belyĭ maps.

However, it turns out that if we consider the case where X has genus 0, then things are much easier, essentially because we have an obvious descent $(\mathbb P^1;\infty )$ of (X; P) that we can exploit. This leads to a useful and easily memorizable trick that resembles our approach to hyperelliptic curves in the previous section.

In what follows, we suppose that $f:X \rightarrow \mathbb P^1$ is a lax Belyĭ map with X of genus 0 such that f is (geometrically generically) Galois with field of moduli F with respect to the extension ${F}^{\mathrm {sep}}\!\mid \!F$. We suppose P is a ramification point of f of non-trivial order $n \in \mathbb Z_{\ge 2}$, which we suppose maps to $0 \in \mathbb P^1$ for the sake of simplicity. Our reasoning will now be very similar to that in Sect. 4.3.

After applying an automorphism if necessary, there exists a model $\mathbb P^1 \simeq X$ such that:

• $P=\infty =(1:0) \in \mathbb P^1$, and

• The automorphism group of (X, f; P) is generated by $(x:z) \mapsto (\zeta _n x:z)$ with $\zeta _n$ a primitive nth root of unity.

This means that if we let t be the affine coordinate z / x the function f can be identified with a power series expansion

$$\begin{aligned} f (t) = \sum _{i = 1}^{\infty } c_i t^{i n} . \end{aligned}$$

(A.1)

Since ${{\mathrm{Aut}}}(X,f;P)({F}^{\mathrm {sep}}) \simeq \mathbb Z/n\mathbb Z$, the set $\{ i \ge 1 : c_i \ne 0 \}$ generates $\mathbb Z$ as a group. So there exists a finite subset I of this set with this property as well. Let $c_I = ( c_i )_{i \in I}$ be the corresponding point in the projective space weighted by its indices, and choose a normalized representative $c_{I,0}$ of $c_I$ [16]. There exists a scaling $t \mapsto \alpha t$ of t that transforms $c_I$ into $c_{I,0}$, which is uniquely determined up to a power of the automorphism $t \mapsto \zeta _n t$ since the elements in I generate $\mathbb Z$.

Proposition A.2

With notations as above, let

$$\begin{aligned} f_0 (t) = f (\alpha t) = \sum _{i = 1}^{\infty } c_{i,0} t^{i n} . \end{aligned}$$

(A.2)

Then $f_0$ has coefficients in F and $(X,f_0; \infty )$ is a descent of (X, f; P).

Proof

Any isomorphism between $f_0$ and its conjugate $\sigma (f_0)$ normalizes the group of automorphisms generated by $(x:z) \mapsto (\zeta _n x:z)$ and fixes $(1:0) \in \mathbb P^1$. Therefore it is given by a scaling $t \mapsto \alpha _{\sigma } t$. The subset $c_I$ is defined over F, and in particular it is fixed by Galois conjugation. By uniqueness of the normalized representative, we therefore see that $\alpha _{\sigma }$ is a power of $\zeta _n$, and in particular that $f_0$ is stable under conjugation by $\sigma$. Since $\sigma$ was arbitrary, we see that indeed $f_0$ is defined over F. $\square$

Remark A.2

Note that contrary to the descent obtained in the proof of Theorem 3.6 the descent constructed in Proposition A.2 may not give rise to an F-rational branch. If desired, such a branch can be found by further scaling the t-coordinate.

1.2 A.2 Explicit descents

In what follows, we exhibit descents of all the Galois lax Belyĭ maps of genus 0, ordered by their automorphism group G. As mentioned above, G is then one of the groups $C_n, D_n, A_4, S_4, A_5$. Given such a group G, let $f : X \rightarrow \mathbb P^1$ be a Belyĭ map with Galois group G. Then f branches over 2 or 3 points. The former possibility occurs only for $G = C_n$, and in this case we slightly abusively add one more point to the set of branch points, over which only trivial ramification (of index 1) occurs. By transivity of ${{\mathrm{Aut}}}(\mathbb P^1)$ on triples of points of $\mathbb P^1$, we may assume the branch points in $\mathbb P^1$ to be $0, 1, \infty$. We denote the uniform ramification indices of f over these points by $e_0, e_1, e_{\infty }$. Any two Belyĭ maps with the same Galois group and the same values of $e_0, e_1, e_{\infty }$ are in fact isomorphic; we choose an ordering of these values, which therefore amounts to specifying the map f up to isomorphism.

We now want to determine the corresponding descents $f_0$ of marked lax Belyi maps. Since G acts transitively on the fibers, it suffices to determine the descents for any choice of a point P over one of $0, 1, \infty$. The proof of Proposition A.2 shows that given such a point it is always possible to find a descent $f_0$ with the same ramification over $0, 1, \infty$. We give equations for such descents by indicating triples $p_0, p_1, p_{\infty }$ with the property that

$$\begin{aligned} p_0 - p_{\infty } = p_1 \end{aligned}$$

with $p_i (t) \in F [t]$ is an $e_i$-th power. These polynomials describe a rational function $f_0$ via

$$\begin{aligned} f_0 = \frac{p_0}{p_{\infty }} = 1 + \frac{p_1}{p_{\infty }} . \end{aligned}$$

The function $f_0$ is a marked Belyĭ map defined over F, hence in particular a marked lax Belyĭ map. We give three such triples $p_0, p_1, p_{\infty }$ in general, corresponding to the choice of $0, 1, \infty$ over which we mark a point P. Occasionally a descent for one of these choices also works for others.

In general, more descents $f_0$ exist. In fact the branch point on $\mathbb P^1$ that is the image of the marked point P will always be rational over the field of moduli; this was also the reason that we did not consider conics as the target space in our Definition A.1. However, the definition of a lax Belyĭ map allows for the possibility that the other two branch points are conjugate over F. This can occur if these points have the same branch indices.

We can determine corresponding twists as follows. Suppose for simplicity that the coinciding ramification indices are $e_0$ and $e_{\infty }$, as indeed we may up to reordering. Then we can consider the rational function $f_0$ over F with branch points $0, 1, \infty$ constructed above, for a choice of P over 1. We can then construct the function $\sqrt{d} (f_0 + 1)/(f_0 - 1)$, which moves the branch point 1 to $\infty$ and the pair $0,\infty$ to $\pm \sqrt{d}$. Normalizing as in Proposition A.2, we get twists over F for all non-zero values of d; the individual points of their branch divisor are defined over the quadratic extension of F determined by d. Having additionally determined such equations, all twists of marked lax Belyĭ maps can then be recovered, if so desired, by scaling in the t-coordinate. Note that such a scaling changes the field of definition of the branch at the marked point.

Case $G \simeq C_n$ Here we can take $e_0 = e_{\infty } = n$ and $e_1 = 1$ to get the triple

$$\begin{aligned} p_0 = t^n, \, p_{\infty } = 1, \, p_1 = t^n - 1 . \end{aligned}$$

(A.3)

The corresponding rational function

$$\begin{aligned} f_0 = t^n = 1 + (t^n - 1) \end{aligned}$$

(A.4)

in fact has rational preimages over all three points $0, 1, \infty$, so that the given triple works for all markings.

Twisting this map to get it to ramify over $\pm \sqrt{d}$ is actually not so straightforward as the procedure above, essentially because there is no branching over 1, so that we cannot apply our normalization argument. However, a corresponding lax Belyĭ map was constructed by Lercier–Ritzenthaler–Sijsling [18, Proposition 3.16].

Case $G \simeq D_n$ Taking $e_0 = e_{\infty } = 2$ and $e_1 = n$ we can use the triples

$$\begin{aligned} p_0 = (t^n + 1)^2 , \, p_{\infty } = (t^n - 1)^2 , \, p_1 = -4 t^n \end{aligned}$$

with corresponding rational function

$$\begin{aligned} f_0 = \frac{(t^n + 1)^2}{(t^n - 1)^2} = 1 + \frac{4 t^n}{(t^n - 1)^2} . \end{aligned}$$

(A.5)

This gives marked descents over $\infty$ and 1 since over these points f has the rational preimages $t = 1$ respectively $t = 0, \infty$. Exchanging the roles of 0 and $\infty$ gives a marked descent over 0. Twisting gives the functions

$$\begin{aligned} f_0 = \frac{d t^{2n} + 1}{2 t^{n}} \end{aligned}$$

(A.6)

which indeed branch of index n over $\infty$ and moreover of index 2 over $\pm \sqrt{d}$ since

$$\begin{aligned} (d t^{2n} + 1) \pm 2 \sqrt{d} t^n = (\pm \sqrt{d} t^n + 1)^2 . \end{aligned}$$

(A.7)

Case $G \simeq A_4$ In this case we take $e_0 = e_{\infty } = 3$ and $e_1 = 2$. We get a triple

$$\begin{aligned} p_0 = \bigl ( t(t^3 + 8) \bigr )^3 , \, p_{\infty } = 2^6 (t^3 - 1)^3 , \, p_1 = (t^6 - 20 t^3 - 8)^2, \end{aligned}$$

(A.8)

which gives rise to the rational points $t = 0$ over 0 and $t = \infty$ over $\infty$. A triple that has the rational point $t = \infty$ over 1 is given by

$$\begin{aligned} p_0 = (3 t^4 + 6 t^2 - 1)^3 , \, p_{\infty } = (3 t^4 - 6 t^2 - 1)^3 , \, p_1 = 2^2 3^2 \bigl ( t (3 t^4 + 1) \bigr )^2 . \end{aligned}$$

(A.9)

A corresponding family of twists branching over $\infty$ and $\pm \sqrt{d}$ is given by

$$\begin{aligned} f_0 = \frac{d^3 t^{12} + 99 d^2 t^8 - 297 d t^4 - 27}{18 (t (d t^4 + 3))^2} = \pm \sqrt{d} + \frac{(d t^4 \mp 6 \sqrt{d} t^2 - 3)^3}{18 (t (d t^4 + 3))^2} . \end{aligned}$$

(A.10)

Case $G \simeq S_4$ For this case, we take $e_{\infty } = 4$, $e_0 = 3$, $e_1 = 2$. Now the branch indices are distinct, so we do not get extra twists, but only the three following triples for rational points over $\infty , 0, 1$ respectively:

$$\begin{aligned} \begin{array}{ll} p_0 &{} = (t^8 + 14 t^4 + 1)^3 \\ p_{\infty } &{} = 2^2 3^3 \bigl ( t(t^4 - 1) \bigr )^4, \\ p_1 &{} = (t^{12} - 33 t^8 - 33 t^4 + 1)^2 , \\ \end{array} \end{aligned}$$

(A.11)

$$\begin{aligned} \begin{array}{ll} p_0 &{} = 2^8 (t^7 - 7 t^4 - 8 t)^3 \\ p_{\infty } &{} = (t^6 + 20 t^3 - 8)^4 \\ p_1 &{} = - (t^{12} - 88 t^9 - 704 t^3 - 64)^2 , \end{array} \end{aligned}$$

(A.12)

and

$$\begin{aligned} \begin{array}{ll} p_0 &{} = (3 t^8 + 28 t^6 - 14 t^4 + 28 t^2 + 3)^3 \\ p_{\infty } &{} = 3^3 (t^6 - 5 t^4 - 5 t^2 + 1)^4 \\ p_1 &{} = 2^4 (9 t^{11} + 11 t^9 + 66 t^7 - 66 t^5 - 11 t^3 - 9 t)^2 . \end{array} \end{aligned}$$

(A.13)

Case $G \simeq A_5$ We take $e_{\infty } = 5$, $e_0 = 3$, $e_1 = 2$. Once more the branch triples are distinct, and the triples with rational points over $\infty , 0, 1$ are given by

$$\begin{aligned} \begin{array}{ll} p_0 &{} = (t^{20} + 228 t^{15} + 494 t^{10} - 228 t^5 + 1)^3 \\ p_{\infty } &{} = 2^6 3^3 \bigl ( t(t^{10} - 11 t^5 - 1) \bigr )^5 \\ p_1 &{} = (t^{30} - 522 t^{25} - 10005 t^{20} - 10005 t^{10} + 522 t^5 + 1)^2 , \end{array} \end{aligned}$$

(A.14)

$$\begin{aligned} \begin{array}{ll} p_0 &{} = 5^3 \bigl ( t (5 t^6 + 5 t^3 + 8) (5 t^6 - 40 t^3 - 1) (40 t^6 - 5 t^3 + 1) \bigr )^3 \\ p_{\infty } &{} = 2^6 (25 t^{12} + 275 t^9 - 165 t^6 - 55 t^3 + 1)^5 \\ p_1 &{} = - \bigl ( (5 t^6 + 1) (25 t^{12} - 1750 t^9 - 2190 t^6 + 350 t^3 + 1) \\ &{} \quad \quad \quad \cdot (200 t^{12} - 500 t^9 + 2055 t^6 + 100 t^3 + 8) \bigr )^2 , \end{array} \end{aligned}$$

(A.15)

and finally

$$\begin{aligned} \begin{array}{ll} p_0 &{} = \bigl ((3 t^4 + 10 t^2 + 15)(t^8 + 70 t^4 + 25)(t^8 + 60 t^6 - 370 t^4 + 300 t^2 + 25) \bigr )^3 \\ p_{\infty } &{} = 3^3 \bigl ((t^4 - 2 t^2 + 5)(t^8 - 20 t^6 - 210 t^4 - 100 t^2 + 25)\bigr )^5 \\ p_1 &{} = 2^2 \bigl (t(t^4 - 5)(t^4 - 10 t^2 + 45)(9 t^4 - 10 t^2 + 5) \\ &{} \quad \quad \quad \cdot (t^8 + 20 t^6 + 470 t^4 + 500 t^2 + 625) (5 t^8 + 20 t^6 + 94 t^4 + 20 t^2 + 5)\bigr )^2 . \end{array} \end{aligned}$$

(A.16)

The above method extends to curves and Belyĭ maps of genus 0 with unipotent automorphism group, and to higher genus (hyperelliptic or not). We expect this will be useful in building databases of Belyĭ maps with small defining equations, a topic for future work.