The existence of supersingular curves of genus 4 in arbitrary characteristic

TL;DR

This paper proves the existence of supersingular curves of genus 4 in any characteristic, expanding understanding of supersingular algebraic curves across different fields.

Contribution

It demonstrates the existence of supersingular genus 4 curves in arbitrary characteristic and constructs such curves explicitly for p > 3.

Findings

Existence of supersingular genus 4 curves in any characteristic.

Construction of supersingular genus 4 curves via fiber products of supersingular elliptic curves for p > 3.

Verification that the desingularization process preserves supersingularity.

Abstract

We prove that there exists a supersingular nonsingular curve of genus $4$ in arbitrary characteristic $p$. For $p>3$ we shall prove that the desingularization of a certain fiber product over $\mathbb{P}^1$ of two supersingular elliptic curves is supersingular.

Equations

$$y^2$$

$$f_1(x)$$

$$f_2(x)$$

$$E_1:z^{2}y$$

$$E_2:w^{2}y$$

$${\pi }_i:E_i\to P^1=\mathrm{Proj}(K[x,y]).$$

$$f(x)=f_1(x)f_2(x)$$

$$C:u^2=f(x).$$

$$\phi :J(H)⟶E_1\times E_2\times J(C),$$

$$\psi :E_1\times E_2\times J(C)⟶J(H)$$

$$f(x{)}^{(p-1)/2}=\sum _{i=0}^{3(p-1)}{\gamma }_i{x}^i.$$

$$a={\gamma }_{p-1},b={\gamma }_{2p-1},c={\gamma }_{p-2}\text{and}d={\gamma }_{2p-2}.$$

$$M:=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right).$$

$$M{M}^{\sigma }=\left(\begin{array}{cc}{a}^{p+1}+b{c}^p& a{b}^p+b{d}^p\\ a^{p}c+c^{p}d& b^{p}c+d^{p+1}\end{array}\right)=0,$$

$$a^{p+1}+b{c}^p$$

$$a{b}^p+b{d}^p$$

$$a^{p}c+c^{p}d$$

$$b^{p}c+d^{p+1}$$

$$a^{p-1}(a{b}^{p-1}+d^p)$$

$$c^{p-1}(a{b}^{p-1}+d^p)$$

$$M^{\prime }=\left(\begin{array}{cc}{\gamma }_{p-1}^{\prime }& {\gamma }_{2p-1}^{\prime }\\ {\gamma }_{p-2}^{\prime }& {\gamma }_{2p-2}^{\prime }\end{array}\right).$$

$$M^{\prime }=P^{-1}M{P}^{\sigma }$$

$$P=\left(\begin{array}{cc}u& 0\\ uv& u^2\end{array}\right).$$

$${\gamma }_i^{\prime }=\sum _{j=i}^{3(p-1)}\left(\genfrac{}{}{0ex}{}{j}{i}\right)u^i{v}^{j-i}{\gamma }_j$$

$$f(uX+v{)}^{(p-1)/2}=\sum _{j=0}^{3(p-1)}{\gamma }_j(uX+v{)}^j=\sum _{i=0}^{3(p-1)}\left(\sum _{j=i}^{3(p-1)}\left(\genfrac{}{}{0ex}{}{j}{i}\right)u^i{v}^{j-i}{\gamma }_j\right)X^i.$$

$${\gamma }_{p-1}^{\prime }$$

$${\gamma }_{2p-1}^{\prime }$$

$${\gamma }_{p-2}^{\prime }$$

$${\gamma }_{2p-2}^{\prime }$$

$$\left(\genfrac{}{}{0ex}{}{j}{p-1}\right)=\{\begin{array}{ll}1& \text{if }j=p-1,2p-1,\\ 0& \text{otherwise},\end{array}\left(\genfrac{}{}{0ex}{}{j}{2p-1}\right)=\{\begin{array}{ll}1& \text{if }j=2p-1,\\ 0& \text{otherwise},\end{array}$$

$$\left(\genfrac{}{}{0ex}{}{j}{p-2}\right)=\{\begin{array}{ll}1& \text{if }j=p-2,2p-2\\ -1& \text{if }j=p-1,2p-1,\\ 0& \text{otherwise},\end{array}\left(\genfrac{}{}{0ex}{}{j}{2p-2}\right)=\{\begin{array}{ll}1& \text{if }j=2p-2,\\ -1& \text{if }j=2p-1,\\ 0& \text{otherwise}\end{array}$$

$$H_p(t)$$

$${\delta }_{p-1}(t)=(-1{)}^e{H}_p(t)$$

$${\delta }_{p-2}(t)=(-1{)}^{e-1}\sum _{i=1}^e(\genfrac{}{}{0ex}{}{e}{i-1}\left)\right(\genfrac{}{}{0ex}{}{e}{i})t^i.$$

$$g(x{)}^e$$

Full text

1 Introduction

Let $K$ be an algebraically closed field of positive characteristic. For a nonsingular algebraic curve $C$ over $K$ we call $C$ supersingular (resp. superspecial) if its Jacobian $J(C)$ is isogenous (resp. isomorphic) to a product of supersingular elliptic curves.

As to supersingular curves, the following is a basic problem (cf. [12, Question 2.2]).

For given $g$, does there exist a supersingular curve of genus $g$ in any characteristic $p$?

For $g\leq 3$, this problem was solved affirmatively. The case of $g=1$, i.e., elliptic curves is due to Deuring [2]. As a proof for $g=2$ with $p>3$ and for $g=3$ with $p>2$, we refer to the stronger fact that there exists a maximal curve of genus $g$ over $\mathbb{F}_{p^{2e}}$ if $g=2$ and $p^{2e}\neq 4,9$ (cf. Serre [14, Théorème 3]) and if $g=3$, $p\geq 3$ and $e$ is odd (cf. Ibukiyama [7, Theorem 1]), where we recall the general fact that any maximal curve over $\mathbb{F}_{p^{2}}$ is superspecial (and therefore supersingular). Also Ibukiyama, Katsura and Oort in [8, Proposition 3.1] proved the existence of superspecial curves of genus $2$ for $p>3$. Even in characteristic $3$, there exists a supersingular curve of genus 2: for example $y^{2}=x^{5}+1$ is supersingular (but is not superspecial), since its Cartier-Manin matrix is nilpotent, see (Eq. 7) and (Eq. 8) in Section 2 for the Cartier-Manin matrix and a criterion for the supersingularity. For the case of $p=2$, we refer to the cerebrated paper [16] by van der Geer and van der Vlugt, where they proved that there exists a supersingular curve of an arbitrary genus in characteristic $2$.

This paper focuses on the first open case, i.e., the case of $g=4$ (cf. [12, Question 3.4]). Let us recall some recent works, restricting ourselves to the case of $g=4$. In [11], Li, Montovan, Pries and Tang proved that if $p\equiv 2\bmod{3}$ or if $p\equiv 2,3,4\bmod{5}$, then there exists a supersingular curve of genus $4$, in particular for $p=3$. For odd $p\equiv 2\bmod{3}$, in [10] the first author showed that there exists a superspecial (and thus supersingular) nonsingular curve of genus $4$.

This paper aims to remove any condition on $p$ for the existence of supersingular curves of genus $4$. For this, we use curves introduced by Howe in [6], where he studied a curve of genus $4$ defined as the desingularization of the fiber product over ${\operatorname{\bf P}}^{1}$ of two elliptic curves. In this paper, we call such a curve a Howe curve, see Definition 2.1 for the precise definition of Howe curves. Our main theorem is:

Theorem 1.1**.**

For any $p>3$, there exists a supersingular Howe curve of genus $4$ in characteristic $p$.

The next corollary is deduced from Theorem 1.1 together with the existence results above by [16] for $p=2$ and by [11] for $g=4$ and $p=3$.

Corollary 1.2**.**

There exists a supersingular nonsingular curve of genus $4$ in arbitrary characteristic $p>0$.

As any supersingular Howe curve has $a$-number $\geq 3$ for odd $p$ (cf. Section 2), Theorem 1.1 is a stronger assertion than the affirmative answer for $p>3$ to the question by Pries [12, Question 3.6], which predicts that there exists a nonsingular curve of genus $4$ with $p$-rank [math] and $a$-number at least $2$.

Let us describe an outline of the proof of Theorem 1.1, with an overview of this paper. In Section 2, we review the definition of Howe curves and their properties, and show that the existence of a supersingular Howe curve of genus $4$ is equivalent to that of two supersingular elliptic curves $E_{1}:y^{2}=f_{1}$ and $E_{2}:y^{2}=f_{2}$ with coprime polynomials $f_{1}$ and $f_{2}$ of degree $3$ such that the hyperelliptic curve $C:y^{2}=f:=f_{1}f_{2}$ of genus $2$ is also supersingular. For the supersingularity of $C$, we use the fact that any curve of genus $2$ is supersingular if and only if $MM^{\sigma}=0$ holds for its Cartier-Manin matrix $M$, where $\sigma$ denotes the Frobenius map. In Section 3, for the Legendre form $y^{2}=g:=x(x-1)(x-t)$, we prove two key propositions on certain coefficients in $g^{(p-1)/2}$ by explicit computations. In Section 4, based on the two propositions, we investigate properties of entries of $MM^{\sigma}$ as polynomials, where we regard coefficients in $f_{1}$ and $f_{2}$ as variables. The properties show that we get a desired $(f_{1},f_{2})$ from a solution of a mutivariate system obtained by removing trivial factors from $MM^{\sigma}=0$. Finally, we show the existense of such a solution by proving an analogous result of the quasi-affineness of Ekedahl-Oort strata in the case of abelian varieties.

In future work, we shall enumerate the isomorphism classes of supersingular Howe curves, whereas the enumeration in the superspecial case has been done for relatively small characteristics by Senda [13]. It would be meaningful to try to apply our techniques of this paper to the case of genera higher than $4$.

Acknowledgments

This work was supported by JSPS Grant-in-Aid for Research Activity Start-up 18H05836 and JSPS Grant-in-Aid for Scientific Research (C) 17K05196.

2 Howe curves

In this section, we recall the definition of Howe curves and properties of these curves, and study the supersingularity of them.

Definition 2.1**.**

A Howe curve is a curve which is isomorphic to the desingularization of the fiber product $E_{1}\times_{{\operatorname{\bf P}}^{1}}E_{2}$ of two double covers $E_{i}\to{\operatorname{\bf P}}^{1}$ ramified over $S_{i}$, where $S_{i}$ consists of 4 points and $|S_{1}\cap S_{2}|=1$ holds.

To achieve our goal, for $p>3$ we realize a Howe curve in the following way. Let $K$ be an algebraically closed field of characteristic $p$. Let

[TABLE]

be two (nonsingular) elliptic curves, where $A_{1},B_{1},A_{2},B_{2}\in K$. Let $\lambda,\mu,\nu$ be elements of $K$ and set

[TABLE]

Consider two elliptic curves

[TABLE]

with the double covers

[TABLE]

We say that $(\lambda,\mu,\nu)$ is of Howe type if

1. (i)

$\mu\neq 0$ and $\nu\neq 0$; 2. (ii)

$f_{1}$ and $f_{2}$ are coprime.

If $(\lambda,\mu,\nu)$ is of Howe type, then the desingularization of the fiber product $E_{1}\times_{{\operatorname{\bf P}}^{1}}E_{2}$ is a Howe curve, since $E_{i}\to{\operatorname{\bf P}}^{1}$ is ramified over the set consisting of 4 points, say $S_{i}$, and $S_{1}\cap S_{2}=\{(1:0)\}$.

Suppose that $(\lambda,\mu,\nu)$ is of Howe type. Put

[TABLE]

and consider the hyperelliptic curve $C$ of genus $2$ defined by

[TABLE]

It was proven by Howe [6, Theorem 2.1] that $H$ is of genus $4$ and there exist two isogenies

[TABLE]

such that $\varphi\circ\psi$ and $\psi\circ\varphi$ are the multiplication by $2$. Hence $H$ is supersingular if and only if $E_{1}$, $E_{2}$ and $C$ is supersingular. Moreover, if $p$ is odd, the $a$-number of $H$ is equal to the sum of the $a$-numbers of $E_{1}$, $E_{2}$ and $J(C)$, whence any supersingular Howe curve is of $a$-number $\geq 3$.

Now we recall a criterion for the supersingularity of $C$. Let $\gamma_{i}$ be the $x^{i}$-coefficient of $f(x)^{(p-1)/2}$, i.e.,

[TABLE]

Put

[TABLE]

In Section 4, we shall use the fact that $\gamma_{i}$ and therefore $a,b,c$ and $d$ are homogeneous when we regard them as polynomials in $\lambda,\mu$ and $\nu$. Let $M$ be the Cartier-Manin matrix of $C$, that is a matrix representing the action of Verschiebung on $H^{0}(C,\Omega^{1})$. It is known (cf. [3, 4.1] and [18, §2]) that the Cartier-Manin matrix of $C$ is given by

[TABLE]

It follows from [1, Step 2 of the proof of Theorem 4.8] (also see [4, 4.0.3] for another proof) that a curve $D$ of genus $2$ is supersingular if and only if the Verschiebung $V$ on $H^{0}(D,\Omega_{D}^{1})$ satisfies $V^{2}=0$. Hence $C$ is supersingular if and only if

[TABLE]

where $M^{\sigma}=\begin{pmatrix}a^{p}&b^{p}\\ c^{p}&d^{p}\end{pmatrix}$.

Proposition 2.2**.**

Assume that $E_{1}$ and $E_{2}$ are supersingular. Then $H$ is supersingular if and only if $ad-bc=0$, $ab^{p-1}+d^{p}=0$ and $a^{p}+c^{p-1}d=0$.

Proof.

Since $E_{1}$ and $E_{2}$ are supersingular, $H$ is supersingular if and only if $C$ is supersingular. As explained above, $C$ is supersingular if and only if (Eq. 8) holds. If $ad-bc=0$, then we have

[TABLE]

Thus the “if”-part is true.

Suppose that (Eq. 8) holds. Then clearly we have $\operatorname{det}(M)=ad-bc=0$, which implies

[TABLE]

Also by (Eq. 9) we have $a^{p}+c^{p-1}d=0$ unless $a=c=0$ and $ab^{p-1}+d^{p}=0$ unless $b=d=0$. If $a=c=0$, by (Eq. 13) and (Eq. 14) we have $a^{p}+c^{p-1}d=0$ unless $b=d=0$. Similarly If $b=d=0$, by (Eq. 13) and (Eq. 14) we have $ab^{p-1}+d^{p}=0$ unless $a=c=0$. Obviously $(a,b,c,d)=(0,0,0,0)$ satisfies $ad-bc=0$, $ab^{p-1}+d^{p}=0$ and $a^{p}+c^{p-1}d=0$. Thus the desired assertion is true. $\Box$

For later use, we review how the Cartier-Manin matrix is changed by a linear change of variables.

Lemma 2.3**.**

Let $X$ be a new variable and consider substituting $uX+v$ for $x$, where $u,v\in K$ with $u\neq 0$. Let $\gamma^{\prime}_{i}$ be the $X^{i}$-coefficient of $f(uX+v)^{(p-1)/2}$ and set

[TABLE]

Then we have

[TABLE]

with

[TABLE]

Proof.

We have

[TABLE]

as

[TABLE]

Then we have

[TABLE]

by calculating, in characteristic $p$, the binomials

[TABLE]

[TABLE]

for $j\leq 3p-3$. $\Box$

3 Two propositions on the Legendre form

We show two propositions (Propositions 3.1 and 3.3) on the Legendre form, which play important roles in the proof of the main theorem.

Assume $p\geq 3$. Let $g(x)=x(x-1)(x-t)$ and $e=(p-1)/2$, where we regard $t$ as an indeterminate. We define a polynomial $H_{p}(t)$ by

[TABLE]

Let $\delta_{p-1}(t)$ be the $x^{p-1}$-coefficient of $g(x)^{e}$. It follows from

[TABLE]

that $y^{2}=x(x-1)(x-t_{0})$ is a supersingular elliptic curve for $t_{0}\in\overline{\mathbb{F}_{p}}$ if and only if $H_{p}(t_{0})=0$, see e.g., [15, Chap. V, Theorem 4.1].

For our purpose, we need to study the $x^{p-2}$-coefficient $\delta_{p-2}(t)$ of $g(x)^{e}$.

Proposition 3.1**.**

*We have the following: *

1. $(1)$

*The polynomial $\delta_{p-2}(t)$ is given explicitly by *

[TABLE] 2. $(2)$

$\displaystyle 2\frac{d}{dt}\delta_{p-2}(t)=2t\frac{d}{dt}\delta_{p-1}(t)+\delta_{p-1}(t)$. 3. $(3)$

$\displaystyle(e+1)\delta_{p-2}(t)=(e+1)t\delta_{p-1}(t)+t(t-1)\frac{d}{dt}\delta_{p-1}(t)$. 4. $(4)$

The polynomials $\delta_{p-1}(t)$ and $\delta_{p-2}(t)$ have no common root.

Proof.

1. (1)

It follows from the binomial theorem that

[TABLE]

Since $i+j=e-1$ if $e+i+j=p-2$, we have

[TABLE]

which is equal to (Eq. 17) by replacing the index $i$ by $i+1$. 2. (2)

The polynomial $H_{p}(t)$ is rearranged as follows:

[TABLE]

Multiplying by $(-1)^{e}t$ the both sides, we have

[TABLE]

and differentiating the both sides of (Eq. 19) yields the following equality

[TABLE]

the right hand side of which is equal to $\displaystyle(e+1)\delta_{p-1}(t)+\frac{d}{dt}\delta_{p-2}(t)$ by (Eq. 16). Here we have

[TABLE]

and by multiplying by $2$, we get

[TABLE]

whose right hand side is equal to $\displaystyle 2t\frac{d}{dt}\delta_{p-1}(t)+\delta_{p-1}(t)$ in characteristic $p$. 3. (3)

Since $\binom{e}{i}=-\binom{e}{i-1}+\binom{e+1}{i}$ for $1\leq i\leq e$, it follows from (Eq. 17) that

[TABLE]

where we factor out $t$ from the first summation of (Eq. 20), namely

[TABLE]

We have

[TABLE]

Multiplying by $e+1$ the both sides gives

[TABLE]

Changing the start of the summation in (Eq. 21) from $i=1$ to $i=0$, we have

[TABLE]

and hence the right hand side of (Eq. 21) is

[TABLE]

Thus we have

[TABLE]

and thus it suffices to show $-\sum_{i=0}^{e}\binom{e+1}{i+1}^{2}(i+1)t^{i}=(t-1)\frac{d}{dt}H_{p}(t)$, where

[TABLE]

In the following, we calculate each of the three summations in (Eq. 23). By (Eq. 16), the first term of (Eq. 23) is

[TABLE]

and the third one is

[TABLE]

Using the equality in the statement (2), we have that the second term of (Eq. 23) is

[TABLE]

It follows from (Eq. 24)–(Eq. 26) that (Eq. 23) is equal to

[TABLE]

and thus $-\sum_{i=0}^{e}\binom{e+1}{i+1}^{2}(i+1)t^{i}=(t-1)\frac{d}{dt}H_{p}(t)$, as desired. 4. (4)

By the equality in the third statement, if $\delta_{p-1}(\alpha)=\delta_{p-2}(\alpha)=0$ for some $\alpha\in\overline{\mathbb{F}_{p}}\smallsetminus\{0,1\}$, then one has $\frac{d}{dt}\delta_{p-1}(\alpha)=0$. Since $\delta_{p-1}(t)=(-1)^{e}H_{p}(t)$ by (Eq. 16), we have $H_{p}(\alpha)=\frac{d}{dt}H_{p}(\alpha)=0$, which means that $\alpha$ is a double root of $H_{p}(t)$. This contradicts the fact shown by Igusa [9] that all roots of $H_{p}(t)$ are simple (cf. [15, Chap. V, Theorem 4.1 (c)]).

$\Box$

Before we state the second proposition (Proposition 3.3), we review, for the reader’s convenience, elementary congruence relations, which is used in the proof of the second proposition.

Lemma 3.2**.**

We have

[TABLE]

*for $i=1,\ldots,p$. *

Proof.

We have

[TABLE]

where we used Wilson’s theorem $(p-1)!\equiv-1\pmod{p}$. $\Box$

Proposition 3.3**.**

Let $H_{p}(t)$ be the polynomial defined by (Eq. 15). Let $a_{i}\in\overline{\mathbb{F}_{p}}$ $(i=1,\ldots,e)$ be the roots of $H_{p}(t)=0$, i.e., $H_{p}(t)=\prod_{i=1}^{e}(t-a_{i})$. Then we have

[TABLE]

where $(g(x)^{e})^{(e)}$ denotes the $e$-th derivative of $g(x)^{e}$ with respect to $x$.

Proof.

It follows from (Eq. 18) that

[TABLE]

whose coefficient of $x^{m}t^{n}$ with $0\leq m,n\leq e$ is

[TABLE]

Putting $P:=\prod_{i=1}^{e}\{(t-a_{i})x-(1-a_{i})t\}$, we have

[TABLE]

whose right hand side is equal to $(x-t)^{e}\sum_{i=0}^{e}\binom{e}{i}^{2}\left(\frac{t(x-1)}{x-t}\right)^{i}$ by (Eq. 16). Using the binomial theorem, one obtains

[TABLE]

whose coefficient of $x^{m}t^{n}$ is

[TABLE]

From (Eq. 27) and (Eq. 28), it suffices to show

[TABLE]

Since

[TABLE]

and similarly

[TABLE]

it suffices to show that

[TABLE]

which we show in the following by rearranging the right hand into the left hand side. Putting $k=i-(m+n-e)$, one has

[TABLE]

Since $\binom{n}{k+m+n-e}=\binom{n}{n-(k+m+n-e)}=\binom{n}{(e-m)-k}$, we also have

[TABLE]

the right hand side of which is equal to

[TABLE]

by Vandermonde’s identity. By Lemma 3.2, we have $(2e-m)!=(-1)^{m+1}/m!$ and $(e-m)!=(-1)^{e+m+1}/(e+m)!$ and $e!=(-1)^{e+1}/e!$, whence (Eq. 30) is equal to

[TABLE]

modulo $p$ by Wilson’s theorem. This is equal to the left hand side of (Eq. 29). $\Box$

Lemma 3.4**.**

Let $K$ be an algebraically closed field of characteristic $p$. Let $y^{2}=g_{0}(x)$ be an elliptic curve over $K$, where $g_{0}(x)$ is a cubic polynomial in $K[x]$. Let $\epsilon_{i}$ and $\epsilon_{i}^{\prime}$ be the $x^{i}$-coefficients of $g_{0}(x)^{(p-1)/2}$ and $(rg_{0}(ux+v))^{(p-1)/2}$ respectively for $0\leq i\leq 3(p-1)/2$, where $r,u\in K^{\times}$ and $v\in K$. Then we have the following:

1. $(1)$

$\epsilon_{p-1}=0$* if and only if $\epsilon_{p-1}^{\prime}=0$.* 2. $(2)$

*If $\epsilon_{p-1}=0$ *$($or equivalently \epsilon_{p-1}^{\prime}=0$$), then $\epsilon_{p-2}\neq 0$ if and only if $\epsilon_{p-2}^{\prime}\neq 0$. 3. $(3)$

*If $\epsilon_{p-1}=0$ *$($*or equivalently \epsilon_{p-1}^{\prime}=0$$), then $\epsilon_{p-2}\neq 0$ *$($and hence \epsilon_{p-2}^{\prime}\neq 0$$).

Proof.

First we show $(1)$ and $(2)$. Similarly to the proof of Lemma 2.3, we have

[TABLE]

for $0\leq i\leq 3(p-1)/2$, and in particular $\epsilon_{p-1}^{\prime}$ and $\epsilon_{p-2}^{\prime}$ are

[TABLE]

from which (1) and (2) follows.

Next, we show (3). Assume $\epsilon_{p-1}=0$. Since $y^{2}=g_{0}(x)$ is isormorphic to a Legendre elliptic curve, there exist elements $r^{\prime}\neq 0$, $u^{\prime}\neq 0$ and $v^{\prime}$ in $K$ such that $g_{0}(x)=r^{\prime}g(X)$ with $X:=u^{\prime}x+v^{\prime}$, where $g(X)=X(X-1)(X-t)$ for some $t\in K\smallsetminus\{0,1\}$. By (1) and our assumption $\epsilon_{p-1}=0$, the $X^{p-1}$-coefficient of $g(X)^{(p-1)/2}$ is zero, and thus the $X^{p-2}$-coefficient of $g(X)^{(p-1)/2}$ is not zero by Proposition 3.1 (4). It follows from (2) that $\epsilon_{p-2}\neq 0$, and we also have $\epsilon_{p-2}^{\prime}\neq 0$. $\Box$

4 Proof of the main theorem

Assume $p>3$. Let $K$ be an algebraically closed field of characteristic $p$. We use the same notation as in Section 2, i.e., $A_{1},B_{1},A_{2},B_{2}$, $\lambda,\mu,\nu$, $f_{1},f_{2},f$, $a,b,c,d$ are as in Section 2. We choose $A_{1},B_{1},A_{2},B_{2}$ in $K$ so that (Eq. 1) and (Eq. 2) are supersingular. By Proposition 2.2, it suffices to show that there exists $(\lambda,\mu,\nu)\in K^{3}$ of Howe type such that $ad-bc=0$, $ab^{p-1}+d^{p}=0$ and $a^{p}+c^{p-1}d=0$.

From now on, we regard $\lambda$, $\mu$ and $\nu$ as intererminates, and consider $a,b,c,d$ as polynomials in $\lambda$, $\mu$, $\nu$. Note that $a,b,c$ and $d$ are homogeneous polynomials in $\lambda,\mu,\nu$ of degrees $2p-2$, $p-2$, $2p-1$ and $p-1$ respectively

Write $f_{1}^{\frac{p-1}{2}}=\sum_{k=0}^{3(p-1)/2}\alpha_{k}x^{k}$ and $f_{2}^{\frac{p-1}{2}}=\sum_{k=0}^{3(p-1)/2}\beta_{k}x^{k}$. Putting

[TABLE]

we can write

[TABLE]

and thus

[TABLE]

Moreover, we denote by $\beta_{k}^{\prime}$ the $x^{k}$-coefficient of $F_{2}^{\frac{p-1}{2}}$, i.e., $F_{2}^{\frac{p-1}{2}}=\sum_{k=0}^{3(p-1)/2}\beta_{k}^{\prime}x^{k}$ with $\beta_{3(p-1)/2}^{\prime}=1$. Note that each coefficient $\beta_{k}^{\prime}$ is a term in $K[\nu]$ of degree $3e-k$ since $F_{2}^{\frac{p-1}{2}}$ is homogeneous of degree $3e$ as a polynomial in $K[x,\nu]$. We also have

[TABLE]

Lemma 4.1**.**

The coefficient $\beta_{0}$ $($resp. $\beta_{1})$ is a homogeneous polynomial in $K[\lambda,\nu]$ of degree $3e$ $($resp. $3e-1)$ with $e=(p-1)/2$, and its highest term in $\lambda$ is $(-1)^{3e}\lambda^{3e}$ $($resp. $(3e-1)(-1)^{3e-1}\lambda^{3e-1})$.

Proof.

It follows from (Eq. 31) and (Eq. 32) that

[TABLE]

which shows that $\beta_{0}$ is homogeneous of degree $3e$, and that its highest term in $\lambda$ is $(-1)^{3e}\lambda^{3e}$. We also have

[TABLE]

which is homogeneous of degree $3e-1$ if it is not zero. It follows from $\beta_{3(p-1)/2}^{\prime}=1$ that the highest term of $\beta_{1}$ in $\lambda$ is $\frac{3}{2}(p-1)(-1)^{\frac{3}{2}(p-1)-1}\lambda^{\frac{3}{2}(p-1)-1}$, which is not zero since $p>3$. $\Box$

Lemma 4.2**.**

The coefficient $\beta_{e}$ $($resp. $\beta_{e+1})$ with $e=(p-1)/2$ is a homogeneous polynomial in $K[\lambda,\nu]$ of degree $2e$ $($resp. $2e-1)$. The highest terms of $\beta_{e}$ and $\beta_{e+1}$ in $\lambda$ are

[TABLE]

respectively.

Proof.

It follows from (Eq. 32) that

[TABLE]

which is homogeneous of degree $2e$ in $K[\lambda,\nu]$ since each $\beta_{n}^{\prime}$ is homogeneous of degree $3e-n$ in $K[\nu]$. Recall that $E_{2}$ is supersingular, and thus $\beta_{p-1}^{\prime}=0$ and $\beta_{p-2}^{\prime}\neq 0$ by Lemma 3.4. Hence the highest term of $\beta_{e}$ with respect to $\lambda$ is

[TABLE]

and it is not zero.

We also have

[TABLE]

which is homogeneous of degree $2e-1$ if it is not zero. Since $\beta_{p-2}^{\prime}\neq 0$, the highest term of $\beta_{e+1}$ with respect to $\lambda$ is

[TABLE]

which is not zero. $\Box$

Lemma 4.3**.**

We have the following:

1. (1)

${\rm ord}_{\mu}(a)=\frac{p+1}{2}$. 2. (2)

${\rm ord}_{\mu}(c)=\frac{p+1}{2}$. 3. (3)

${\rm ord}_{\mu}(ad-bc)={\rm ord}_{\nu}(ad-bc)=\frac{p+1}{2}$. 4. (4)

$\frac{ad-bc}{(\mu\nu)^{\frac{p+1}{2}}}\equiv B\lambda^{2p-4}\bmod{(\mu,\nu)}$* for some constant $B\in K^{\times}$.*

Proof.

1. (1)

We claim that there exists $\tilde{\alpha}_{k}\in K$ such that $\alpha_{k}=\mu^{3e-k}\tilde{\alpha}_{k}$ for each $0\leq k\leq 3(p-1)/2$. Indeed, we have

[TABLE]

with $e=(p-1)/2$, and thus $\alpha_{k}$ is divided by $\mu^{3e-k}$. Putting $\tilde{\alpha}_{k}:=\alpha_{k}/(\mu^{3e-k})$, we also have $\tilde{\alpha}_{k}\in K$ for $0\leq k\leq 3(p-1)/2$. Since both the elliptic curves $E_{1}:z^{2}=f_{1}$ and $E_{2}:w^{2}=f_{2}$ are supersingular, we have $\alpha_{p-1}=0$ and $\beta_{p-1}=0$, whereas $\alpha_{p-2}\neq 0$ (and thus $\tilde{\alpha}_{p-2}\neq 0$) and $\beta_{p-2}\neq 0$ by Lemma 3.4. It follows from

[TABLE]

where $\beta_{k}$ is a polynomial in $K[\lambda,\nu]$ for each $0\leq k\leq 3e$. Since $\beta_{1}\neq 0$ by Lemma 4.1, we have $\tilde{\alpha}_{3e-j}\beta_{j-e}=\tilde{\alpha}_{p-2}\beta_{1}\neq 0$ for $j=(p+1)/2$, and thus ${\rm ord}_{\mu}(a)=(p+1)/2$. 2. (2)

Similarly to the proof of (1), one has

[TABLE]

Since we have $\tilde{\alpha}_{p-2}\neq 0$, and $\beta_{0}\neq 0$ by Lemma 4.1, we also have $\tilde{\alpha}_{3e-j}\beta_{j-e-1}=\tilde{\alpha}_{p-2}\beta_{0}\neq 0$ for $j=(p+1)/2$, and thus ${\rm ord}_{\mu}(c)=(p+1)/2$. 3. (3)

Similarly to the proof of (1), remaining two entries of the Cartier-Manin matrix $M$ are written as

[TABLE]

both of which are not divided by $\mu$ since $\alpha_{3e}=1$ and since $\beta_{e},\beta_{e+1}\neq 0$ by Lemma 4.2. Thus, if the coefficient of $\mu^{(p+1)/2}$ in $ad-bc$ is not zero, then we have ${\rm ord}_{\mu}(ad-bc)=(p+1)/2$. By straightforward computation, the coefficients of $\mu^{(p+1)/2}$ in $ad$ and $bc$ are $\tilde{\alpha}_{p-2}\beta_{1}\tilde{\alpha}_{3e}\beta_{e}$ and $\tilde{\alpha}_{p-2}\beta_{0}\tilde{\alpha}_{3e}\beta_{e+1}$, respectively. Here we have

[TABLE]

where $\tilde{\alpha}_{p-2}\neq 0$ and $\tilde{\alpha}_{3e}=\alpha_{3e}=1$. If $\beta_{1}\beta_{e}-\beta_{0}\beta_{e+1}\neq 0$, we have ${\rm ord}_{\mu}(ad-bc)=(p+1)/2$. By Lemmas 4.1 and 4.2, the highest term of $\beta_{1}\beta_{e}$ in $\lambda$ is

[TABLE]

and that of $\beta_{0}\beta_{e+1}$ is

[TABLE]

Since $\beta_{p-2}^{\prime}\neq 0$ by Lemma 3.4, it suffices to show

[TABLE]

If the equality holds, we have

[TABLE]

and thus $1\equiv 2\pmod{p}$. This is a contradiction.

Next, we show ${\rm ord}_{\nu}(ad-bc)=(p+1)/2$. To show this, we consider the transformation $X=x-\lambda$. The polynomials $f_{1}(x)$ and $f_{2}(x)$ are rewritten as

[TABLE]

respectively. Let $\gamma_{l}^{\prime}$ denote the coefficient of $X^{l}$ in $(f_{1}^{\prime}(X)f_{2}^{\prime}(X))^{(p-1)/2}$. Putting $a^{\prime}=\gamma^{\prime}_{p-1}$, $c^{\prime}=\gamma^{\prime}_{p-2}$, $b^{\prime}=\gamma^{\prime}_{2p-1}$ and $d^{\prime}=\gamma^{\prime}_{2p-2}$, we have $ad-bc=a^{\prime}d^{\prime}-b^{\prime}c^{\prime}$ by Lemma 2.3. By the same argument as the proof of ${\rm ord}_{\mu}(ad-bc)=(p+1)/2$ for $f_{1}(x)$ and $f_{2}(x)$, we have ${\rm ord}_{\nu}(a^{\prime}d^{\prime}-b^{\prime}c^{\prime})=(p+1)/2$ for $f_{1}^{\prime}(X)$ and $f_{2}^{\prime}(X)$, and thus ${\rm ord}_{\nu}(ad-bc)=(p+1)/2$. 4. (4)

From the first part of the proof of (3), the coefficient of $(\mu\nu)^{\frac{p+1}{2}}$ in $ad-bc$ is $B\lambda^{2p-4}$ with

[TABLE]

which is not zero. Recall from the proof of (1) that $\tilde{\alpha}_{p-2}$ and $\tilde{\alpha}_{3e}$ are non-zero constants in $k$. Recall also from the beginning of this section that $\beta_{p-2}^{\prime}$ is a term in $K[\nu]$ of degree $3e-(p-2)=(p+1)/2$, and thus $(\beta_{p-2}^{\prime}/\nu^{\frac{p+1}{2}})\in K^{\times}$. Thus the claim holds.

$\Box$

Let $R:=K[\lambda,\mu,\nu]$ and put

[TABLE]

which belong to $R$. Since $a,b,c$ and $d$ are homogeneous polynomials in $\lambda,\mu,\nu$ of degrees $2p-2$, $p-2$, $2p-1$ and $p-1$ respectively, we have that $h_{0}$, $h_{1}$ and $h_{2}$ are homogeneous of degrees $2p-4$, $p(p-1)$ and $2p(p-1)$ respectively. Consider the homogeneous ideal

[TABLE]

of $K[\lambda,\mu,\nu]$.

Lemma 4.4**.**

Assume $p>3$. For any point $(\lambda:\mu:\nu)$ on $V(h_{0})$ in ${\operatorname{\bf P}}^{2}$, we have $\mu\neq 0$ or $\nu\neq 0$.

Proof.

It suffices to show that $\mu=\nu=0$ implies $\lambda=0$ for $(\lambda,\mu,\nu)\in K^{3}$ satisfying $h_{0}=0$. This immediately follows from Lemma 4.3 (4). $\Box$

Consider the point with $\nu\neq 0$. From now on, we substitute $1$ for $\nu$ and consider $a,b,c,d$ as polynomials in $\lambda,\mu$. Set

[TABLE]

and let $J$ be the ideal of $S$ obtained from $I$ by substituting $1$ for $\nu$. Set $a^{\prime}=a/\mu^{(p+1)/2}$ and $c^{\prime}=c/\mu^{(p+1)/2}$. Let $a^{\prime}_{0},c^{\prime}_{0},b_{0},d_{0}$ be the constant terms of $a^{\prime},c^{\prime},b,d$ as polynomials in $\mu$, which are polynomials in $\lambda$.

Lemma 4.5**.**

As polynomials in $\lambda$, we have

1. (1)

$c^{\prime}_{0}$* and $d_{0}$ are coprime.* 2. (2)

$b_{0}$* and $d_{0}$ are coprime.*

Proof.

(1) Since

[TABLE]

and $\tilde{\alpha}_{p-2}$ and $\tilde{\alpha}_{3e}$ are non-zero constant as polynomials in $\lambda$, it suffices to see that $\beta_{0}$ and $\beta_{e}$ are coprime as polynomials in $\lambda$. As

[TABLE]

and $F_{2}$ is a separable polynomial, $\beta_{0}$ and $\beta_{e}$ are coprime.

(2) Since

[TABLE]

and $\tilde{\alpha}_{3e}$ is a non-zero constant, it suffices to see that $\beta_{e}$ and $\beta_{e+1}$ are coprime. As

[TABLE]

it suffices to show that $(F_{2}^{e})^{(e)}(-\lambda)$ is a separable polynomial. A linear coordinate change makes $F_{2}$ a Legendre form $g(x)=x(x-1)(x-a_{1})$, where $a_{1}$ is a solution of $H_{p}(t)=0$ as $y^{2}=F_{2}$ is a supersingular elliptic curve. Let $H_{p}(t)=\prod_{i=1}^{e}(t-a_{i})$ be a factorization of $H_{p}(t)$. By Proposition 3.3, up to multiplication by a non-zero constant, $(g(x)^{e})^{(e)}$ is factored as

[TABLE]

Here we note that $a_{i}\neq 0,1$ for $i=1,2,\ldots,e$ as $H_{p}(0)=1$ and $H_{p}(1)=(-1)^{e}$, see the proof of [15, Chap. V. Theorem 4.1 (c)]. Then it is clear that (Eq. 35) is a separated polynomial, and therefore $\beta_{e}$ is separated. $\Box$

Proposition 4.6**.**

Suppose that $p>3$. Then any point $(\lambda:\mu:\nu)$ on $V(h_{0},h_{1},h_{2})$ is of Howe type.

Proof.

Let $(\lambda:\mu:\nu)$ be a point of $V(h_{0},h_{1},h_{2})$. By Lemma 4.4, we have $\mu\neq 0$ or $\nu\neq 0$. Suppose $\nu\neq 0$. Substututing $1$ for $\nu$ and consider everything as a polynomial in $\mu$ and $\lambda$. It follows from $h_{0}=0$ and $h_{1}=0$ that

[TABLE]

This and Lemma 4.5 imply that $1\equiv 0\quad({\rm mod}\ \mu)$, whence $\mu$ is unit in $S/J$. Thus we have shown that $\nu\neq 0$ implies $\mu\neq 0$ for any point $(\lambda:\mu:\nu)$ on $V(h_{0},h_{1},h_{2})$. A similar argument shows that $\mu\neq 0$ implies that $\nu\neq 0$. Hence we conclude that both of $\mu$ and $\nu$ are not zero.

It remains to show that $f_{1}$ and $f_{2}$ defined in (Eq. 3) and (Eq. 4) are coprime. As $\mu$ and $\nu$ are not zero, $f_{1}$ and $f_{2}$ are separated polynomials. Suppose that $f_{1}$ and $f_{2}$ were not coprime. After taking a linear coordinate change, they are written as $f_{1}=x(x-1)(x-t_{1})$ and $f_{2}=x(x-t_{2})(x-t_{3})$. Then

[TABLE]

Then $M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ becomes a upper triangular matrix (i.e., $c=0$) with $a=(t_{1}t_{2}t_{3})^{e}$. Since $f_{1}$ and $f_{2}$ are separated, we have $t_{i}\neq 0$ for $i=1,2,3$ and therefore $h_{2}\neq 0$. This is a contradiction. $\Box$

Finally we show that $V(h_{0},h_{1},h_{2})$ is not empty. The fact is reminiscent of the quasi-affineness (cf. [17, (6.5). Theorem]) of Ekedahl-Oort strata in the case of the moduli space of principally polarized abelian varieties.

Proposition 4.7**.**

$V(h_{0},h_{1},h_{2})\neq\emptyset$.

Proof.

We suppose that $V(h_{0},h_{1},h_{2})=\emptyset$ were true. Set $X:=V(h_{0})=V(h_{0})\smallsetminus V(h_{0},h_{1},h_{2})$. Note that $(a,d)=(0,0)$ is not allowed on $X$, as $(a,d)=(0,0)$ implies $h_{1}=h_{2}=0$. Consider the morphism

[TABLE]

sending $(\lambda:\mu:\nu)$ to $(a^{p}+c^{p-1}d:c^{p-1}d)$ for the part of $a\neq 0$ and $(ab^{p-1}+d^{p}:d^{p})$ for the part of $d\neq 0$, which is well-defined by (Eq. 13) and (Eq. 14). Moreover the image of $\varphi$ consists of infinitely many points from the following two facts. Firstly $X$ is connected since it is a hypersurface in ${\operatorname{\bf P}}^{2}$. Secondly $(1:1)$ (resp. $(1:0)$) is an image of $\varphi$ since there exists $(\lambda:\mu:\nu)\in{\operatorname{\bf P}}^{2}$ such that $h_{0}=0$ and $a=0$ (resp. $h_{0}=0$ and $d=0$), by [5, Chap.1, Theorem 7.2]. The existence of such a morphism as $\varphi$ implies that $X$ is not projective. This is a contradiction. $\Box$