Stable models of plane quartics with hyperelliptic reduction

TL;DR

This paper presents an algorithm to compute stable models of smooth plane quartics with hyperelliptic reduction over discrete valuation fields, utilizing valuations of theta constants for classification.

Contribution

It introduces a new criterion based on theta constants for determining reduction type and provides an algorithm for approximating stable models in hyperelliptic cases.

Findings

Algorithm successfully computes stable models in hyperelliptic reduction cases.

New valuation criterion for reduction type based on theta constants.

Examples demonstrate practical computation of models.

Abstract

Let C/K: F = 0 be a smooth plane quartic over a complete discrete valuation field K. In a previous paper the authors togetehr with Q. Liu give various characterizations of the reduction (i.e. non-hyperelliptic genus 3 curve, hyperelliptic genus 3 curve or bad) of the stable model of C: in terms of the existence of a special plane quartic model and in terms of the valuations of the Dixmier-Ohno invariants of C. The last one gives in particular an easy computable criterion for the reduction type. However, it does not produce a stable model, even in the case of good reduction. In this paper we give an algorithm to obtain (an approximation of) the stable model when the reduction of the latter is hyperelliptic and the characteristic of the residue field is not 2. This is based on a new criterion giving the reduction type in terms of the valuations of the theta constants of C. Some examples of the computation of these models are given.

Equations

$$(x_1^2+x_2^2+x_3^2{)}^2+\sqrt{-7}{\alpha }^2\cdot (x_1^2{x}_2^2+x_2^2{x}_3^2+x_3^2{x}_1^2)=0$$

$$\mathcal{C}:\{\begin{array}{l}{y}^2=-(x_1^2{x}_2^2+x_2^2{x}_3^2+x_3^2{x}_1^2),\\ \sqrt[4]{-7}\alpha \cdot y=x_1^2+x_2^2+x_3^2\end{array}$$

$$\{\begin{array}{l}{y}^2=-(x_1^2{x}_2^2+x_2^2{x}_3^2+x_3^2{x}_1^2),\\ 0=x_1^2+x_2^2+x_3^2\end{array}$$

$$\{\begin{array}{ll}{y}^2+G=0,& \\ {\pi }^{s}y-Q=0.& \end{array}$$

$${\vartheta }_s(x)={\vartheta }^{\alpha }\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right](\tau )=e^{-i\pi x_1.x_2}\cdot \vartheta \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right](\tau )$$

$$F=(x_1{u}_1+x_2{u}_2-x_3{u}_3{)}^2-4x_1{u}_1{x}_2{u}_2$$

$$\left[\begin{array}{ccc}1& 1& 1\\ \frac{1}{a_{11}}& \frac{1}{a_{12}}& \frac{1}{a_{13}}\\ \frac{1}{a_{21}}& \frac{1}{a_{22}}& \frac{1}{a_{23}}\end{array}\right]\cdot \left[\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]\cdot \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right].$$

$$M=\left[\begin{array}{ccc}1& 1& 1\\ \frac{1}{a_{11}}& \frac{1}{a_{12}}& \frac{1}{a_{13}}\\ \frac{1}{a_{21}}& \frac{1}{a_{22}}& \frac{1}{a_{23}}\end{array}\right]$$

$$\frac{-a_{11}{a}_{12}{a}_{21}{a}_{23}+a_{11}{a}_{12}{a}_{22}{a}_{23}+a_{11}{a}_{13}{a}_{21}{a}_{22}-a_{11}{a}_{13}{a}_{22}{a}_{23}-a_{12}{a}_{13}{a}_{21}{a}_{22}+a_{12}{a}_{13}{a}_{21}{a}_{23}}{a_{11}{a}_{12}{a}_{13}{a}_{21}{a}_{22}{a}_{23}}=$$

$$=\frac{a_{11}(-a_{12}{a}_{21}{a}_{23}+a_{12}{a}_{22}{a}_{23}+a_{13}{a}_{21}{a}_{22}-a_{13}{a}_{22}{a}_{23})-a_{12}{a}_{13}{a}_{21}(a_{22}-a_{23})}{a_{11}{a}_{12}{a}_{13}{a}_{21}{a}_{22}{a}_{23}}$$

$$i\frac{\vartheta \left[\begin{array}{c}000\\ 100\end{array}\right]\vartheta \left[\begin{array}{c}110\\ 111\end{array}\right]\vartheta \left[\begin{array}{c}001\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}111\\ 011\end{array}\right]}{\vartheta \left[\begin{array}{c}101\\ 111\end{array}\right]\vartheta \left[\begin{array}{c}011\\ 100\end{array}\right]\vartheta \left[\begin{array}{c}111\\ 101\end{array}\right]\vartheta \left[\begin{array}{c}001\\ 110\end{array}\right]}$$

$$\left[\begin{array}{ccc}{v}_0& v_0& v_0\\ 0& v_0& v_0\\ v_0& v_0& v_0\end{array}\right],\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]\text{for the LHS, and}\left[\begin{array}{ccc}{v}_0& v_0& v_0\\ 2v_0& v_0& v_0\\ v_0& v_0& v_0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\text{for the RHS,}$$

$$F=\underset{Q}{\underbrace{({\pi }^{v_1}{x}_1{u}_1^{\prime }+x_2{u}_2-x_3{u}_3{)}^2}}-4{\pi }^{v_1}\underset{G}{\underbrace{x_1{u}_1^{\prime }{x}_2{u}_2}}=0,$$

$${\bar{u}}_2+{\bar{u}}_3$$

$$\epsilon \frac{{\bar{u}}_1^{\prime }}{a_{11}^{\prime }}+\frac{{\bar{u}}_2}{a_{12}}+\frac{{\bar{u}}_3}{a_{13}}$$

$$\frac{{\bar{u}}_2}{a_{22}}+\frac{{\bar{u}}_3}{a_{23}}$$

$$\frac{u_1}{a_{31}}+\frac{u_2}{a_{32}}+\frac{u_3}{a_{33}}=a_{31}{x}_1+a_{32}{x}_2+a_{33}{x}_3,$$

$$\epsilon \frac{{\bar{u}}_1}{a_{31}^{\prime }}+\frac{{\bar{u}}_2}{a_{32}}+\frac{{\bar{u}}_3}{a_{33}}=a_{32}{x}_2+a_{33}{x}_3.$$

$$\{\bar{Q}=0\}\cap \{x_2=0\}=\{x_3{\bar{u}}_3=0\}\cap \{x_2=0\}=\{P_1=(1:0:0)\}\cup (\{{\bar{u}}_3=0\}\cap \{x_2=0\}).$$

$$\left(\frac{1}{a_{23}}-\frac{1}{a_{22}}\right)u_3=\left(a_{21}-\frac{1}{a_{22}}\right)x_1+\left(a_{22}-\frac{1}{a_{22}}\right)x_2+\left(a_{23}-\frac{1}{a_{22}}\right)x_3.$$

$$(a_{21}{a}_{22}-1)(a_{22}{)}^{-1}=-\frac{\vartheta \left[\begin{array}{c}110\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 011\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 100\end{array}\right]\vartheta \left[\begin{array}{c}110\\ 111\end{array}\right]}{\vartheta \left[\begin{array}{c}011\\ 011\end{array}\right]\vartheta \left[\begin{array}{c}101\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}101\\ 111\end{array}\right]\vartheta \left[\begin{array}{c}011\\ 100\end{array}\right]}(-i)\frac{\vartheta \left[\begin{array}{c}101\\ 111\end{array}\right]\vartheta \left[\begin{array}{c}011\\ 100\end{array}\right]}{\vartheta \left[\begin{array}{c}110\\ 001\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 010\end{array}\right]}=i\frac{\vartheta \left[\begin{array}{c}110\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 011\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 100\end{array}\right]\vartheta \left[\begin{array}{c}110\\ 111\end{array}\right]}{\vartheta \left[\begin{array}{c}011\\ 011\end{array}\right]\vartheta \left[\begin{array}{c}101\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}110\\ 001\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 010\end{array}\right]}$$

$$\left(\frac{1}{a_{22}}-\frac{1}{a_{23}}\right)u_2=\left(a_{21}-\frac{1}{a_{23}}\right)x_1+\left(a_{22}-\frac{1}{a_{23}}\right)x_2+\left(a_{23}-\frac{1}{a_{23}}\right)x_3,$$

$$a_{22}{x}_2^2+(a_{22}+a_{23})x_2{x}_3+a_{23}{x}_3^2=0.$$

$$\{\bar{Q}=0\}\cap \{x_1=0\}=\{P_3=(0:-1:1),P_4=(0:-\frac{a_{23}}{a_{22}}:1)\}.$$

$$P_5=\left(1:-\frac{a_{23}{a}_{22}-1}{a_{23}{a}_{21}-1}:0\right).$$

$$P_6=(a_{22}-a_{23}:a_{23}-a_{21}:a_{21}-a_{22}).$$

$$(a_{12}{a}_{13}-a_{32}{a}_{33})\left(\frac{1}{a_{13}{a}_{33}}{x}_2^2+\left(\frac{1}{a_{12}{a}_{33}}+\frac{1}{a_{13}{a}_{32}}\right)x_2+\frac{1}{a_{12}{a}_{32}}\right)=0.$$

$$a_{12}{a}_{13}-a_{32}{a}_{33}=\frac{\vartheta \left[\begin{array}{c}000\\ 110\end{array}\right]\vartheta \left[\begin{array}{c}010\\ 100\end{array}\right]}{\vartheta \left[\begin{array}{c}111\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}101\\ 010\end{array}\right]}\left(\frac{\vartheta \left[\begin{array}{c}010\\ 001\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 011\end{array}\right]}{\vartheta \left[\begin{array}{c}101\\ 111\end{array}\right]\vartheta \left[\begin{array}{c}111\\ 101\end{array}\right]}-\frac{\vartheta \left[\begin{array}{c}100\\ 010\end{array}\right]\vartheta \left[\begin{array}{c}110\\ 000\end{array}\right]}{\vartheta \left[\begin{array}{c}011\\ 100\end{array}\right]\vartheta \left[\begin{array}{c}001\\ 110\end{array}\right]}\right).$$

$$D=\vartheta \left[\begin{array}{c}010\\ 001\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 011\end{array}\right]\vartheta \left[\begin{array}{c}011\\ 100\end{array}\right]\vartheta \left[\begin{array}{c}001\\ 110\end{array}\right]-\vartheta \left[\begin{array}{c}100\\ 010\end{array}\right]\vartheta \left[\begin{array}{c}110\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}101\\ 111\end{array}\right]\vartheta \left[\begin{array}{c}111\\ 101\end{array}\right]$$

$$D=\pm \vartheta \left[\begin{array}{c}011\\ 000\end{array}\right]\vartheta \left[\begin{array}{c}001\\ 010\end{array}\right]\vartheta \left[\begin{array}{c}010\\ 101\end{array}\right]\vartheta \left[\begin{array}{c}000\\ 111\end{array}\right]$$

$$\left[\begin{array}{c}\epsilon (1)\\ {\epsilon }^{\prime }(1)\end{array}\right]=\left[\begin{array}{c}010\\ 001\end{array}\right],\left[\begin{array}{c}\epsilon (2)\\ {\epsilon }^{\prime }(2)\end{array}\right]=\left[\begin{array}{c}100\\ 010\end{array}\right],\left[\begin{array}{c}\epsilon (3)\\ {\epsilon }^{\prime }(3)\end{array}\right]=\left[\begin{array}{c}011\\ 000\end{array}\right],\left[\begin{array}{c}\lambda \\ {\lambda }^{\prime }\end{array}\right]=\left[\begin{array}{c}010\\ 010\end{array}\right]\text{and}\left[\begin{array}{c}\mu \\ {\mu }^{\prime }\end{array}\right]=\left[\begin{array}{c}001\\ 101\end{array}\right]$$

Full text

1. Introduction and main result

Let $K$ be a complete discrete valuation field with valuation $v$ and valuation ring $\mathcal{O}$ containing a maximal ideal generated by $\pi$. Let $k=\mathcal{O}/\langle\pi\rangle$ be the residual field. In the following, the expression “after a possible extension of $K$” means that we are allowed to take a finite extension of $K$ and still call $K,\mathcal{O},v,\pi$ the corresponding notions. When $F$ is an integral polynomial, i.e. with coefficients in $\mathcal{O}$, we denote $\bar{F}$ its reduction modulo $\pi$.

Given a genus 3 non-hyperelliptic curve $C/K$, in [LLLGR18], the authors answer the following question: after a possible extension of $K$, what is the reduction type of the stable model $\mathcal{C}/\mathcal{O}$ of $C$? By this, we mean to distinguish between

• •

$\mathcal{C}_{k}$ is still a non-hyperelliptic curve of genus 3. We say that $C$ has potentially good quartic reduction;

• •

$\mathcal{C}_{k}$ is a hyperelliptic curve of genus 3. We say that $C$ has potentially good hyperelliptic reduction;

• •

$\mathcal{C}_{k}$ is not a curve of genus 3. We say that $C$ has geometrically bad reduction.

In the first case, the special fiber is again a smooth plane quartic over $k$, whereas in the second case it is isomorphic over $\bar{k}$ to $y^{2}=f(x,z)$ where $f$ is a binary octic with no multiple roots.

Example 1.1*.*

The discriminant of the Klein quartic given by $C:\,x_{1}^{3}x_{2}+x_{2}^{3}x_{3}+x_{3}^{3}x_{1}=0$ is equal to $7^{7}$, hence the model $x_{1}^{3}x_{2}+x_{2}^{3}x_{3}+x_{3}^{3}x_{1}=0$ over $\mathbb{Z}$ has good reduction everywhere except at $\pi=7$. To study the reduction type of the stable model at $7$, notice that $C$ is $\bar{\mathbb{Q}}$-isomorphic to the curve [Elk99, pp.56]

[TABLE]

with $\alpha=\frac{-1+\sqrt{-7}}{2}$. Consider now the scheme

[TABLE]

in the weighted projective space $\mathbb{P}\,^{(1,1,1,2)}$ over the ring of integers $\mathcal{O}$ of $K=\mathbb{Q}_{7}(\sqrt[4]{-7})$. Its generic fiber is isomorphic over $K$ to $C$ whereas $\mathcal{C}_{k}$ is isomorphic over $\bar{\mathbb{F}}_{7}$ to

[TABLE]

which turns out to be the hyperelliptic curve $y^{2}=x^{8}+14x^{4}z^{4}+z^{8}$.

The shape of the stable model in the previous example is symptomatic of the situation and motivates the following definition.

Definition 1.2**.**

(Def. 1.3, [LLLGR18]) Let $C/K$ be a smooth plane quartic. We say that $C$ admits a toggle model if there exist an integer $s>0$, a primitive (i.e. the gcd of its coefficients is 1) quartic form $G\in\mathcal{O}[x_{1},x_{2},x_{3}]$ and a primitive quadric $Q\in\mathcal{O}[x_{1},x_{2},x_{3}]$ with $\bar{Q}$ irreducible such that $Q^{2}+\pi^{s}G=0$ is $K$-isomorphic to $C$. If moreover $\bar{Q}=0$ intersects $\bar{G}=0$ transversely in $8$ distinct $\overline{k}$-points, we say that $C$ admits a good toggle model.

Proposition 1.3**.**

(Prop. 1.2, [LLLGR18]) Suppose $p\neq 2$. Let $C/K$ be a smooth plane quartic having a good toggle model $Q^{2}+\pi^{2s}G=0$. Let us denote $\mathcal{C}/\mathcal{O}$ the subscheme of the weighted projective space $\mathbb{P}\,^{(1,1,1,2)}$ defined by

[TABLE]

Then the generic fiber is isomorphic to the plane smooth quartic $C/K$ which has good hyperelliptic reduction. The special fiber of $\mathcal{C}$ is isomorphic to the double cover of $\bar{Q}=0$ ramified over the 8 distinct intersection $\overline{k}$-points of $\bar{Q}=0$ with $\bar{G}=0$.

Remark 1.4*.*

In [LLLGR18, Thm. 2.10] there is an equivalent result for the characteristic $2$ case.

Theorem 1.5**.**

([LLLGR18, Thm. 1.4]) Let $C/K$ be a plane smooth quartic. Then $C$ has good hyperelliptic reduction if and only if $C$ has a good toggle model over $K$.

Over $\mathbb{C}$, it is well-known that one can associate to the Jacobian of a genus $g$ curve $N=2^{g-1}(2^{g}+1)$ values, which are called theta constants. We will need a fancy version of these values over $K$ (see Section 2) but the intuition remains the same. Now, over a DVR, we can multiply the theta constants by a common factor such that they become integral and that the minimum of their valuations is [math]. These new values $(\vartheta_{i})_{i=0,\ldots,N-1}$ are uniquely defined up to a unit in $\mathcal{O}$ and we call them the integral theta constants of the curve. In this paper we prove the following result:

Theorem 1.6**.**

Let $C$ be a smooth plane quartic over $K$ with residue field of characteristic different from $2$. The curve $C$ has potentially good hyperelliptic reduction if and only if there is a unique integral theta constant of $C$ with positive valuation.

The direct implication of Theorem 1.6 is relatively straightforward, for the converse we construct an explicit toggle model given by a Riemann model of the quartic and we show that it is good using the relations between the theta constants given in [Web76]. The construction of the toggle model is completely explicit in the proof and allows us to compute a stable model in this case. We implemented this construction in Magma  [BCP97] and we performed numerical experiments of some example curves.

Remark 1.7*.*

A similar procedure could actually be realized over a number field (and would be easier to implement) when the $2$-structure of the curve is not defined over a too large extension. Unfortunately, in general, this structure is defined over an extension of degree $\#\operatorname{Sp}_{6}(\mathbb{F}_{2})=1451520$ ([Har79]).

Remark 1.8*.*

In [LLLGR18], the distinction between potentially good hyperelliptic reduction and geometrically bad reduction in terms of the Dixmier-Ohno is possible only when $p>7$. Our present algorithm present the advantage to allow the characteristics $p=3,5$ and $7$.

Acknowledgement

We thank Tristan Vaccon for helpful discussions.

2. Link with theta constants

Let $R$ be a ring and $S=\operatorname{Spec}R$. Let $X/R$ be an abelian scheme and $\mathcal{L}$ be a relatively ample line bundle on $X$ such that $(-1_{X})^{*}\mathcal{L}\simeq\mathcal{L}$. Fix an isomorphism $\varepsilon:0^{*}\mathcal{L}\simeq\mathcal{O}_{S}$ where $0:S\to X$ is the zero section. To any $s\in\Gamma(X,\mathcal{L})$, Mumford associates (see [Mum91, Appendix I]) a morphism $\vartheta_{s}:X[2]\to\mathbb{A}^{1}_{S}$. Following [Mum91, Prop. 5.11] (see also loc. cit. Definition. 5.8), in the special case where $S=\operatorname{Spec}\mathbb{C}$ and $\mathcal{L}$ is the basic line bundle on $X_{\tau}=\mathbb{C}^{g}/(\mathbb{Z}^{g}+\tau\mathbb{Z}^{g})$ (see loc. cit. p. 36), then $s$ is uniquely defined up to a multiplicative constant and there is a unique choice of $\varepsilon$ such that

[TABLE]

for any $x\in X[n]\simeq(\frac{1}{n}\mathbb{Z}/\mathbb{Z})^{2g}\ni(x_{1},x_{2})$ (after a specific isomorphism of the $n$-torsion) where $\vartheta\left[{\begin{subarray}{c}x_{1}\\ x_{2}\end{subarray}}\right](\tau)$ is the value at [math] of the classical theta function with characteristic $\left[{\begin{subarray}{c}x_{1}\\ x_{2}\end{subarray}}\right]$ [Mum83, p.192].

Let $\mathcal{C}/R$ be a proper and flat scheme whose fibers are smooth curves of genus $g>0$. Then $Pic^{0}(\mathcal{C})/R$ is an abelian scheme and the previous theory can be applied. Let $D_{0}$ be a theta characteristic divisor on $\mathcal{C}$. Recall that a theta characteristic divisor $D$ is a divisor such that $2D\sim\kappa$, where $\kappa$ is the canonical divisor (relative to $S$). The divisor $\operatorname{Sym}^{g-1}\mathcal{C}-D_{0}\subset\operatorname{Pic}^{0}(\mathcal{C})$ is symmetric and defines a symmetric line bundle $\mathcal{L}$ on $\operatorname{Pic}^{0}(\mathcal{C})$ which induces the canonical principal polarization on $\operatorname{Pic}^{0}(\mathcal{C})$. In particular there exists a unique (up to a multiplicative constant) $s\in\Gamma(\operatorname{Pic}^{0}(\mathcal{C}),\mathcal{L})$.

Let $R=\mathbbm{k}$ where $\mathbbm{k}$ is a field of characteristic different from $2$ and for a divisor $D$ on $\mathcal{C}/\mathbbm{k}$, let us denote $L(D)$ its Riemann-Roch space. Mumford showed in [Mum71] that the function $e_{*}:D\mapsto\dim L(D)\pmod{2}$ is a quadratic function on the set of theta characteristic divisors $D$. We say that $D$ is even (resp. odd) if $e_{*}(D)=0$ (resp. $1$). Since the map $D\mapsto D-D_{0}$ is a bijection between the set of theta characteristic divisors and the set of $2$-torsion points, one can say that the corresponding $\vartheta_{D}=\vartheta_{s}(D-D_{0})$ is even (resp. odd). The study of Arf invariants of quadratic forms shows that there are $2^{g-1}(2+1)$ even (resp. $2^{g-1}(2-1)$ odd) theta characteristic divisors and the corresponding $\vartheta_{D}$ are called theta constants – or Thetanullwerte–. As stated in loc. cit. p.182 and refined in [Kem73], the classical Riemann singularity theorem over $\mathbb{C}$ extends to the present setting and in particular we get that $\vartheta_{D}=0$ if and only if $\dim L(D)>0$ (the odd $\vartheta_{D}$ are therefore always equal to zero).

In the particular case where $\mathcal{C}/\mathbbm{k}$ is a curve of genus $3$, Clifford’s theorem [Har77, IV.Th.5.4] shows that $\dim L(D)\leq 2$, the equality being possible only when $C$ is hyperelliptic and for a unique even theta characteristic divisor. Hence one recovers the classical result that $C$ has $36$ (resp. 35) non-zero theta constants when $C$ is non-hyperelliptic (resp. hyperelliptic).

Coming back to the case where $R=\mathcal{O}$ is a DVR and $C/K$ a non-hyperelliptic genus $3$ we denote (after a possible extension of $K$) $(\vartheta_{0},\ldots,\vartheta_{35})\in\mathcal{O}^{36}$ its integral theta constants.

Proof.

(of Theorem 1.6) Let us assume that $C$ has potentially good hyperelliptic reduction and let $\mathcal{C}/\mathcal{O}$ be a a smooth model of $C$ such that $\mathcal{C}_{k}$ is hyperelliptic. Since $\operatorname{Pic}^{0}(\mathcal{C})/R$ is an abelian scheme, we can use the algebraic thetas $\vartheta_{s}(x)\in\mathcal{O}$ defined above. Mumford [Mum91, Appendix I] has proved that these values are equal up to a constant $\lambda\in K^{*}$ to the $\vartheta_{D}$ on $C$, therefore there are exactly $36$ of them which are non-zero. Moreover they shall also reduce to the theta constants on $\mathcal{C}_{k}$ which is hyperelliptic of genus $3$ and therefore exactly one of the $36$ non-zero ones has positive valuation. Therefore after ordering them, the non-zero $\vartheta_{s}(x)$ coincides up to a unit in $\mathcal{O}$ with $(\vartheta_{0},\ldots,\vartheta_{35})$. We therefore get that exactly one of the latter has positive valuation.

In the opposite direction, we assume now that among the $(\vartheta_{i})_{i=0,\ldots,35}$, there is a unique one with positive valuation. We are going to use the beautiful work of Weber [Web76] to obtain a good toggle model. Although this work is of course written over $\mathbb{C}$ in the language of classical theta functions, the link (2.1) between classical theta constants and algebraic theta constants shows that all the algebraic homogeneous relations can be used with the latter ones.

Weber ([Web76, p.108], see also [Fio16]) introduces $9$ values $(a_{ij})_{i,j=1,2,3}$ 111In Weber’s notation $a_{1i}=a_{i}$, $a_{2i}=a^{\prime}_{i}$ and $a_{3i}=a^{\prime\prime}_{i}$. given as homogeneous quotient of products of theta constants. In Section 4, we will see the relations of these constants with the geometry of the curve (through its bitangents) but we will not need them right now. We will assume that the theta constant denoted $\vartheta\left[{\begin{subarray}{c}100\\ 001\end{subarray}}\right]$ in Weber’s book is the one which corresponds to the one with positive valuation (this choice can be made without loss of generality after a choice of a right symplectic basis of the $2$-torsion for the Weil pairing). The expressions of the $a_{ij}$ in terms of the theta constants imply that only $a_{11}$ and $a_{31}$ have positive valuation. Weber gives then an explicit construction of a quartic form $F\in K[x_{1},x_{2},x_{3}]$ such that $F=0$ is isomorphic to $C$ from the $a_{ij}$. This form is

[TABLE]

where the $u_{i}$ are linear forms, solutions of the linear system

[TABLE]

Let us denote $Q=x_{1}u_{1}+x_{2}u_{2}-x_{3}u_{3}$ and $G=-4x_{1}u_{1}x_{2}u_{2}$. A series of computations (see Section 3) using relations between the theta constants shows that

1. (i)

the $u_{i}$ are defined over $\mathcal{O}$ (Lemma 3.1); 2. (ii)

$G=\pi^{s}G_{0}$ for $s>0$ an integer, $G_{0}$ integral and primitive and $\bar{G_{0}}=0$ intersects $\bar{Q}=0$ in 8 distinct points (Lemma 3.2); 3. (iii)

$\bar{Q}$ is a non-degenerate quadric (end of Lemma 3.2).

We therefore get that $F$ is a good toggle model of $C$ and so from Proposition 1.3 we get that $C$ has potentially good hyperelliptic reduction. ∎

3. A Riemann model providing a good toggle model

We resume with the notation of previous section. In the sequel, we denote $a_{11}=\pi^{v_{0}}a_{11}^{\prime}$ and $a_{31}=\pi^{v_{0}}a_{31}^{\prime}$ with $v_{0}>0$ an integer and $a_{11}^{\prime}$ and $a_{31}^{\prime}$ units. If $\ell=ax_{1}+bx_{2}+cx_{3}$ is a linear form over $K$, we denote $v(\ell)=\min(v(a),v(b),v(c))$. In particular for the linear form $u_{i}$ defined in (2.3), we let $v_{i}=v(u_{i})$ for $i=1,2,3$.

Lemma 3.1**.**

The valuations $v_{i}$ are non-negative.

Proof.

We already know that $v(a_{ij})\geq 0$ and $v(x_{i})=0$, so it is enough to check that the valuations of all the entries of the inverse of the matrix

[TABLE]

are also non-negative. We compute the inverse of $M$ by computing the adjoint matrix and dividing by the determinant of $M$.

The determinant of $M$ is equal to

[TABLE]

[TABLE]

where the factor $a_{22}-a_{23}$ equals

[TABLE]

by [Web76, 16.14, pp.110]. Hence, it has zero valuation. Therefore, $v(\operatorname{det}M)=-v_{0}$. On the other hand, the valuation of the entries of the adjoint matrix are greater or equal to $-v_{0}$. So, the result follows. ∎

If we multiply both sides of eq. (2.3) by $a_{11}$ and we look at the valuation of the entries in the equation, we get

[TABLE]

from which we easily read that $v_{1}\geq v_{0}$ since $v_{2},\,v_{3}$ are non-negative by Lemma 3.1. We can write $u_{1}=\pi^{v_{1}}u^{\prime}_{1}$. This proves that the Riemann model

[TABLE]

is defined over $\mathcal{O}$. We are going to show that it is a good toggle model.

Lemma 3.2**.**

The intersection of $\bar{Q}=0$ with $\bar{G}=0$ is transverse and the quadratic form $\bar{Q}=\,x_{2}\bar{u}_{2}-x_{3}\bar{u}_{3}$ is non-degenerate.

Proof.

Let us write down the reduction of eq. (2.3) before studying the intersection points (here $\varepsilon=0$ if $v_{1}>v_{0}$ and 1 otherwise),

[TABLE]

One can also add the following equation (see [Fio16, Prop. 2]):

[TABLE]

which reduces to

[TABLE]

We compute the intersection of $\bar{Q}=0$ successively with $x_{2}=0,x_{1}=0,\bar{u}_{2}=0$ and then $\bar{u}_{1}^{\prime}=0$.

$\bullet$ Intersection with $x_{2}=0$.

[TABLE]

To compute the second intersection point above, denoted $P_{2}$, we see from equations (3.1) and (3.3) that

[TABLE]

Letting $x_{2}=0$ and $\bar{u}_{3}=0$ we need to look at the valuation of $a_{21}-{1}/{a_{22}}$. As in Lemma 3.1, Weber [Web76, 16.11, 16.14] gives an expression for this difference in terms of the theta constants

[TABLE]

and one can check that the valuation is [math]. Hence we get $P_{2}=\displaystyle\left(\frac{a_{23}-{1}/{a_{22}}}{{1}/{a_{22}}-a_{21}}:0:1\right)$.

$\bullet$ Intersection with $x_{1}=0$. We deal with $\{\bar{Q}=0\}\cap\{x_{1}=0\}$ in the same way. We use eq. (3.5) in conjunction with

[TABLE]

and $x_{1}=0$ to get that the coordinates of the intersection points must satisfy

[TABLE]

The discriminant of this quadratic form has valuation [math] since $v(a_{22}-a_{23})=0$, still using [Web76, 16.14]. We have that $-1$ and $-\frac{a_{23}}{a_{22}}$ are the two different roots of this polynomial with $x_{3}=1$, and we get that

[TABLE]

$\bullet$ Intersection with $\bar{u}_{2}=0$. The case $\{\bar{Q}=0\}\cap\{\bar{u}_{2}=0\}$ is also similar. Thanks to [Web76, 16.14], $v(a_{23}a_{22}-1)=v(a_{23}a_{21}-1)=0$ and so if $x_{3}=0$, eq. (3.6) gives the intersection point

[TABLE]

If $\bar{u}_{3}=0$, we use equations (3.1) and (3.3) to compute a second intersection point

[TABLE]

$\bullet$ Intersection with $\bar{u}^{\prime}_{1}=0$. The case $\{\bar{Q}=0\}\cap\{\bar{u}^{\prime}_{1}=0\}$ is a bit more delicate as we miss some relations from Weber which we have to work out. Letting $x_{3}=1$ we can write $\bar{u}_{3}=x_{2}\bar{u}_{2}$. We use this to write $\bar{u}_{2}$ in terms of $x_{2}$ in eq. (3.4) and in eq. (3.2). Equating both expressions for $\bar{u}_{2}$, we get a degree two equation for $x_{2}$:

[TABLE]

We first need to prove that it indeed defines a degree $2$ equation, that is, that the leading coefficient $a_{12}a_{13}-a_{32}a_{33}$ is not zero. Secondly, in order to prove that we obtain two different solutions for $x_{2}$, we need to prove that the discriminant $\left(\frac{a_{13}a_{32}-a_{12}a_{33}}{a_{12}a_{13}a_{32}a_{33}}\right)^{2}$ is also different from zero.

We first prove that the valuation of the leading coefficient is [math]. By [Web76, 16.14],

[TABLE]

All the theta constants involved in the formula are units, so we only need to check that

[TABLE]

is also a unit. We claim that

[TABLE]

and so we get the result. The proof of the claim is a well-known game with (classical) theta constants that we will now play. By [RF74, Chap. II, Theorem 18] with the notation from there

[TABLE]

one gets

[TABLE]

We need to prove that the first sign is negative to get our expression of $D$. We consider this relation for Riemann matrices222using a diagonal matrix $\tau$ we would have only got that the second sign is negative, which is not what we want. $\tau=\begin{pmatrix}*&0&*\\ 0&*&0\\ *&0&*\end{pmatrix}$. By [RF74, Chap. I, Theorem 11] the equation (3.7) simplifies to

[TABLE]

Now we use [RF74, Chap. I, Theorem 5] with the notation there

[TABLE]

to rewrite the expression inside the parentheses above as

[TABLE]

The third term is zero since the characteristic $\left[{\begin{subarray}{c}11\\ 01\end{subarray}}\right]$ is odd. Finally, since by [RF74, Chap. I, Theorem 3] $\vartheta\left[{\begin{subarray}{c}20\\ 01\end{subarray}}\right]=-\vartheta\left[{\begin{subarray}{c}00\\ 01\end{subarray}}\right]$, the previous equation can be realized only if the first sign is a minus sign. This proves our claim on $D$ and we get a degree $2$ equation for $x_{2}$.

The same arguments work for proving that the discriminant has valuation [math]. Hence, the intersection $\{\bar{Q}=0\}\cap\{\bar{u}^{\prime}_{1}=0\}$ consists of two distinct points $\{P_{7},\,P_{8}\}$.

$\bullet$ The intersection is transverse. We need to prove that $P_{i}\neq P_{j}$ for all distinct $i,j\in\{1,...,8\}$. Recall that:

[TABLE]

[TABLE]

The six first point are clearly distinct, and $P_{7}$ and $P_{8}$ are distinct. We will check now that they are also distinct from the first six points. For that we study the following intersections:

$\{\bar{Q}=0\}\cap\{\bar{u}^{\prime}_{1}=0\}\cap\{x_{2}=0\}$: we have here $\bar{u}^{\prime}_{1}=x_{2}=0$ and $x_{3}=1$, hence $\bar{u}_{3}=0$ from $\bar{Q}=0$ and from equations (3.2) and (3.4) we get $a_{12}a_{13}-a_{32}a_{33}=0$, which we have just seen is not. Hence, the intersection in empty and the points $P_{7}$ and $P_{8}$ are distinct from the points $P_{1}$ and $P_{2}$.

$\{\bar{Q}=0\}\cap\{\bar{u}^{\prime}_{1}=0\}\cap\{x_{2}=0\}$: we get $x_{1}=\bar{u}^{\prime}_{1}=0$, $x_{3}=1$ and $u_{3}=x_{2}u_{2}$. From eq. (3.3) we get $\bar{u}_{3}=x_{2}$ and hence equations (3.1) and (3.3) implies $a_{13}a_{32}-a_{12}a_{33}=0$. We have also already seen that this is not possible. Therefore, the points $P_{7},\,P_{8}$ are distinct from the points $P_{3},\,P_{4}$.

$\{\bar{Q}=0\}\cap\{\bar{u}^{\prime}_{1}=0\}\cap\{\bar{u}_{2}=0\}$: in this situation we get $\bar{u}^{\prime}_{1}=\bar{u}_{2}=\bar{u}_{3}=0$, which is not possible since those forms are linearly independent. We then conclude that $P_{7},\,P_{8}$ are also distinct from $P_{5}$ and $P_{6}$.

$\bullet$ The conic $\bar{Q}=0$ is non-singular. If it were, $\bar{Q}$ would be the product of two linear forms $L_{1}$ and $L_{2}$. None of these lines $L_{i}=0$ is equal to $x_{1}=0,\,x_{2}=0,\,\bar{u}^{\prime}_{1}=0$ or $\bar{u}_{2}=0$, since the intersection of the conic $\bar{Q}=0$ with each of them is two distinct points. Moreover, $L_{1}$ and $L_{2}$ define distinct lines. Hence, we have that $4$ of the points $P_{1},...,P_{8}$ are on $L_{1}=0$ and the other $4$ on $L_{2}=0$, where points in the pairs $\{P_{1},P_{2}\}$, $\{P_{3},P_{4}\}$, $\{P_{5},P_{6}\}$, $\{P_{7},P_{8}\}$ are on different lines. Assume that $P_{1}$ is on $L_{1}$ and $P_{2}$ on $L_{2}$. Then $P_{5}$ is on $L_{2}$ since otherwise $L_{1}=0$ would be $\{x_{3}=0\}$ but $\{x_{3}=0\}$ is not contained in $\bar{Q}=0$ by eq. (3.6). Hence, we get

[TABLE]

Now, it is easy to check that neither $P_{3}$ or $P_{4}$ are on $L_{1}=0$: for $P_{3}$ it is enough to check that $a_{22}-a_{23}\neq 0$ and for $P_{4}$ that $a_{21}(a_{22}-a_{23})\neq 0$, which is true by [Web76, 16.14].

This gives a contradiction with the assumption $\bar{Q}=L_{1}\cdot L_{2}$. Hence, $\bar{Q}=0$ defines a non-singular conic.

∎

4. Bitangents of a smooth plane quartic

Let $C/K$ be a smooth plane quartic curve given by $F=0$. As $C$ is canonically embedded in the projective plane, the canonical divisors on $C$ are the intersection of the lines with $C$. In particular we can describe the $28$ odd theta characteristic divisors using bitangents.

Definition 4.1**.**

A line $\beta$ is called a bitangent of $C$ if the intersection divisor $(\beta\cdot C)$ is of the form $2P+2Q$ for some not necessarily distinct points $P,Q$ of $C$. The divisor $D_{\beta}:=P+Q$ is the odd theta characteristic divisor associated to $\beta$.

Remark 4.2*.*

The bitangents of a curve can be computed by looking at the singular points of the dual curve, but this is rather expensive in terms of computations since the singularities also contains the tangents at inflexion points. A better approach is to work out the two algebraic conditions in $a$ and $b$ under which the form $F(x_{1},ax_{1}+b,1)$ is a perfect square and to look for the solutions of the corresponding system. Of course, one has to take care of the bitangents which are not of the form $x_{2}=ax_{1}+b$.

Definition 4.3**.**

Let $S=\{\beta_{i}\}$ be a set of bitangents of $C$. The set $S$ is called azygetic if there is no conic $Q$ such that the intersection divisor $Q\cdot C\geq D_{\beta_{1}}+D_{\beta_{2}}+D_{\beta_{3}}$ for $\beta_{1},\beta_{2},\beta_{3}\in S$. An azygetic set of $7$ elements is called an Aronhold set.

Among the $\binom{28}{7}$ subsets of $7$ bitangents of $C$, 288 of them form an Aronhold set, see [GH04, after Cor. 2.5]. Note that if one knows equations for the $28$ bitangents and the curve, a tedious but straighforward computation using the definition of azygeticness allow to exhibit an Aronhold set.

We resume with the notation from the previous sections. Let us recall the following result [Rie76].

Theorem 4.4**.**

Let $C$ be a smooth plane quartic curve and $S=\{\beta_{i}\}_{i=1,\ldots,7}$ be an Aronhold set. After a linear change of variables, we may assume that the $\beta_{i}$ are given by the equations:

[TABLE]

where $i\in\{1,2,3\}$. The coefficients $a^{\prime}_{ij}$ are multiples of the $a_{ij}$ (defined in Section 2), i.e. $a^{\prime}_{ij}=\eta_{i}a_{ij}$ for some $\eta_{i}\neq 0$, that are determined, up to sign, by the linear system:

[TABLE]

where $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$ are given by

[TABLE]

After the previous change of variables, the plane quartic $C$ is given by a Riemann model

[TABLE]

where $u_{1},u_{2},u_{3}$ are given as in Section 2 by

[TABLE]

Moreover, we can express all the bitangents for this model as:

[TABLE]

[TABLE]

[TABLE]

where $i,l\in\{1,2,3\}$ and $\{i,j,k\}=\{1,2,3\}$.

Proposition 4.5**.**

Let assume that $C$ has potentially good hyperelliptic reduction. Given an Aronhold set $S$ as in Theorem 4.4, there exists a constant $v_{0}>0$ such that either

• (1)

two of the $a_{ij}$ (with the same value of $j$) have positive valuation $v_{0}$ and the rest have valuation equal to zero;

• (2)

or two of the $a_{ij}$ (with the same value of $j$) have negative valuation $-v_{0}$ and the rest have valuation equal to zero;

• (3)

or all the $a_{ij}$ have valuation equal to zero.

Proof.

Look at the expressions for the coefficients $a_{ij}$ in terms of the theta constants in page $108$ formula $(11^{\prime})$ in [Web76] and consider the different cases. For example, if the theta constant that has positive valuation $v_{0}$ is $\vartheta\left[{\begin{subarray}{c}100\\ 001\end{subarray}}\right]$ as in the proof of Theorem 1.6, we can see that all $a_{ij}$ have zero valuation except

[TABLE]

that have also positive valuation $v_{0}$, so we are in case $(1)$ of the Proposition. ∎

Theorem 4.6**.**

Let assume that $C$ has potentially good hyperelliptic reduction. Given an Aronhold set $S=\{\beta_{i}\}_{i=1,\ldots,7}$ we can construct another Aronhold set $\{\gamma_{i}\}_{i=1,\ldots,7}$ for which $v(a^{\prime}_{11})=v(a^{\prime}_{31})>0$.

Proof.

Notice that if we are already in the first case, then we may assume that $v(a_{1j})=v(a_{3j})=v_{0}$ for $v_{0}>0$ and some $j$ after permuting the set $\{\beta_{1},\beta_{2},\beta_{3}\}$ of bitangents, and indeed, that $j=1$ after permuting the set $\{\beta_{4},\beta_{5},\beta_{6}\}$.

If we are in the second case, and after a permutation of the sets $\{\beta_{1},\beta_{2},\beta_{3}\}$ and $\{\beta_{4},\beta_{5},\beta_{6}\}$, we may assume that $a_{11}$ and $a_{31}$ are the ones with negative valuations, that is, $\vartheta\left[{\begin{subarray}{c}001\\ 100\end{subarray}}\right]$ is the theta constant with positive valuation equal to $v_{0}$. Then we can take the Aronhold system:

[TABLE]

[TABLE]

So, $a^{\prime}_{i1}=1/a_{i1}$ and $a^{\prime}_{ij}=a_{ij}$ for $i\in\{1,2,3\}$ and $j\in\{2,3\}$, and we fall in the first case with $v(a^{\prime}_{11})=v(a^{\prime}_{31})=v_{0}$.

In the third case we have to distinguish several cases. If any of the differences $1-a_{ij}a_{i(j+1)}$ has positive valuation, then after permutations of the sets $\{\beta_{1},\beta_{2},\beta_{3}\}$ and $\{\beta_{4},\beta_{5},\beta_{6}\}$, we can assume that $\vartheta\left[{\begin{subarray}{c}100\\ 000\end{subarray}}\right]$ is the theta constant with positive valuation, then again by using Weber’s formulas [Web76, pp. 109], we have that $\{\gamma_{1},\ldots,\gamma_{7}\}=\{\beta_{23},\,\beta_{13}\,\,\beta_{12},\,\beta_{4},\,\beta_{45},\,\beta_{46},\,\beta_{47}\}$ is an Aronhold system with $v(a^{\prime}_{11})=v(a^{\prime}_{31})=v_{0}$. If none of the differences $1-a_{ij}a_{i(j+1)}$ has positive valuation, then the differences $a_{ij}a_{i(j+1)}-a_{(i+1)j}a_{(i+1)(j+1)}$ have positive valuations. Hence, the Aronhold system $\{\gamma_{1},\ldots,\gamma_{7}\}=\{\beta_{5},\,\beta_{6}\,\,\beta_{7},\,\beta_{4},\,\beta_{1},\,\beta_{2},\,\beta_{3}\}$ falls in the first case. ∎

5. The algorithm and an example

In this section we start with a smooth plane quartic $C/K:\,F=0$, where $K$ is a finite extension of $\mathbb{Q}_{p}$ with $p\neq 2$. The previous sections lead to the following algorithm to compute a stable model when the curve has potentially good hyperelliptic reduction.

[TABLE]

Remark 5.1*.*

The azygetic test on line $(5)$ is perform by checking if the $6$ points of intersection of $3$ of the bitangents in $S\cup\{b\}$ with the curve lie on a conic. Notice that as we are over a non-exact field, this step is harder than over a number field.

Example 5.2*.*

Let us consider the plane smooth quartic $C/\mathbb{Q}:F=0$ where

[TABLE]

This curve is an equation of $X_{ns}(13)\simeq X_{s}(13)\simeq X_{0}^{+}(169)$ which is studied in [BDS*+*17, Cor.6.8]. Using a general result from [Edi90], they prove that this curve has good reduction everywhere and potentially good reduction at 13 after a ramified extension of degree 84. Using the characterizations in [LLLGR18] in terms of the valuations of the Dixmier-Ohno invariants, it is straightforward to see that this is indeed the case and that the stable reduction at 13 is the hyperelliptic curve $y^{2}=z(x^{7}-z^{7})$. We apply our algorithm to find an equation for the stable model at $13$.

Starting from $F$, we get the following Aronhold system modulo $13^{2}$:

[TABLE]

[TABLE]

with

[TABLE]

where

[TABLE]

A toggle model model modulo $13^{2}$ is then $Q^{2}+\pi G=0$ where

[TABLE]

and

[TABLE]

The Dixmier-Ohno invariants of this model modulo $13^{2}$ coincide with the Dixmier-Ohno invariants of the input curve. This proves the correctness of the output with the given precision.

Remark 5.3*.*

In order to automatized completely this procedure, one would need packages to work with algebraic systems over the $p$-adic in a transparent way with control of the errors. Although there are results and fast progress on this topic [BL12, CRV14, CRV15, CRV18, Leb15], such functionalities are still not fully implemented.