# The existence of supersingular curves of genus 4 in arbitrary characteristic

## Abstract

We prove that there exists a supersingular nonsingular curve of genus 4 in arbitrary characteristic $$p>0$$. For $$p>3$$ we shall prove that the desingularization of a certain fiber product over $$\mathbf{P }^1$$ of two supersingular elliptic curves is supersingular.

## 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. [19, Question 2.2]).

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

For $$g\le 3$$, this problem was solved affirmatively. The case of $$g=1$$, i.e., elliptic curves is due to Deuring [4]. Ibukiyama, Katsura and Oort in [11, Proposition 3.1] proved the existence of superspecial curves of genus 2 for $$p>3$$. For the existence of supersingular curves of genus 3 in any characteristic $$p>0$$, see Oort [17, Theorem 5.12]. Also, 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 $$p^{2e}\ne 4,9$$ (cf. Serre [20, Théorème 3]) and if $$g=3$$, $$p\ge 3$$ and e is odd (cf. Ibukiyama [10, Theorem 1]), where we recall the general fact that any maximal curve over $$\mathbb {F}_{p^2}$$ is superspecial (and therefore supersingular). Even for $$(g,p)=(2,3)$$, there exists a supersingular curve: for example $$y^2=x^5+1$$ is supersingular (but is not superspecial), since its Cartier-Manin matrix is nilpotent, see (9) and (10) in Sect. 2 for the Cartier–Manin matrix and a criterion for the supersingularity. For the case of $$p=2$$, we refer to the celebrated paper [22] 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. [19, Question 3.4]). Let us recall some recent works, restricting ourselves to the case of $$g=4$$. According to [16, Remark 7.2] and [15, Theorem 7.1] by Li, Mantovan, Pries and Tang, there exists a supersingular curve of genus 4 if $$p \equiv 5 \bmod {6}$$ or if $$p \equiv 2, 3, 4 \bmod {5}$$. Among them, we review the existence for $$(g,p)=(4,3)$$ for the reader’s convenience. Indeed $$y^{10} = x(1-x)$$ is supersingular, as this is a quotient of the Hermitian curve $$X^{q+1}+Y^{q+1}=1$$ with $$q=3^2$$, where the quotient map is given by $$x=X^{q+1}$$ and $$y=XY$$. For odd $$p \equiv 2 \bmod {3}$$, in [14] 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 [9], where he studied a curve of genus 4 defined as the desingularization of a fiber product over $$\mathbf{P }^1$$ 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 in characteristic p. The number of isomorphism classes of supersingular Howe curves over an algebraically closed field is finite.

The next corollary is deduced from Theorem 1.1 together with the existence results for $$p=2$$ and for $$(g,p)=(4,3)$$ explained above.

### 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 $$\ge 3$$ for odd p (cf. Sect. 2), Theorem 1.1 is a stronger assertion than the affirmative answer for $$p>3$$ to the question by Pries [19, Question 3.6], which predicts that there exists a nonsingular curve of genus 4 with p-rank 0 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 Sect. 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 pair $$(f_1, f_2)$$ of separable polynomials of degree 3 such that the hyperelliptic curve $$C : y^2 = f:=f_1 f_2$$ of genus 2 is also supersingular. Here $$E_1$$ and $$E_2$$ can be isomorphic. For the supersingularity of C, we use the fact that any curve of genus 2 is supersingular if and only if $$M M^{\sigma } = 0$$ holds for its Cartier-Manin matrix M, where $$\sigma$$ denotes the Frobenius map. The moduli space $${\mathcal M}_2$$ (resp. $$\overline{{\mathcal M}_2}$$) of nonsingular (resp. stable) curves of genus two is of dimension 3 and it is known that the supersingular locus on $$\overline{{\mathcal M}_2}$$ (naturally extending that of $${\mathcal M}_2$$) is one. On the other hand, C’s constructed as above from supersingular $$E_1$$ and $$E_2$$ make a two-dimensional family, say $$\mathcal P$$, in $${\mathcal M}_2$$. Then, the Zariski closure $$\overline{{\mathcal P}}$$ of $${\mathcal P}$$ in $$\overline{{\mathcal M}_2}$$ must inersect the supersingular locus in $$\overline{{\mathcal M}_2}$$. The difficulty lies in showing that not all of these intersection points lie outside $${\mathcal M}_2$$. Almost all of the remaining sections are devoted to overcoming the difficulty. In the proof, instead of $$\overline{{\mathcal P}}$$, we use a more concrete space, see Sect. 3 for our moduli-theoretic framework. In Sect. 4, we prove two assertions of cubic polynomials (Corollaries 4.4 and 4.7) by reducing the problems to two propositions (Propositions 4.3 and 4.6) for the Legendre form $$y^2=g:=x(x-1)(x-t)$$. In Sect. 5, based on the two assertions, we investigate properties of entries of $$M M^{\sigma }$$ 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 multivariate system obtained by removing trivial factors from $$M M^{\sigma } = 0$$, where “desired” means that $$f_1$$ and $$f_2$$ are separable and are coprime (in other words, the corresponding point belongs to the interior $${\mathcal M}_2$$). Finally, we show the existence of such a solution by proving an analogous result of the quasi-affineness of Ekedahl-Oort strata in the case of abelian varieties.

## 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 $$E_1 \times _{\mathbf{P }^1} E_2$$ of two double covers $$E_i\rightarrow \mathbf{P }^1$$ 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

\begin{aligned} y^2= & {} x^3 + A_1 x + B_1, \end{aligned}
(1)
\begin{aligned} y^2= & {} x^3 + A_2 x + B_2 \end{aligned}
(2)

be two (nonsingular) elliptic curves, where $$A_1,B_1,A_2,B_2\in K$$. Here two elliptic curves (1) and (2) can be isomorphic. Let $$\lambda , \mu , \nu$$ be elements of K and set

\begin{aligned} f_1(x)= & {} x^3 + A_1 \mu ^2 x + B_1\mu ^3, \end{aligned}
(3)
\begin{aligned} f_2(x)= & {} (x-\lambda )^3 + A_2 \nu ^2(x-\lambda ) + B_2 \nu ^3. \end{aligned}
(4)

Consider two elliptic curves

\begin{aligned} E_1:\quad z^2y= & {} y^3f_1(x/y) = x^3 + A_1 \mu ^2 xy^2 + B_1\mu ^3y^3, \end{aligned}
(5)
\begin{aligned} E_2:\quad w^2y= & {} y^3f_2(x/y) = (x-\lambda y)^3 + A_2 \nu ^2(x-\lambda y)y^2 + B_2 \nu ^3 y^3 \end{aligned}
(6)

with the double covers

\begin{aligned} \pi _i : E_i \rightarrow \mathbf{P }^1={\text {Proj}}(K[x,y]). \end{aligned}

Note that the isomorphism classes of $$E_1$$ and $$E_2$$ are independent of the choice of $$(\lambda ,\mu ,\nu )$$ provided $$\mu \ne 0$$ and $$\nu \ne 0$$. We say that $$(\lambda ,\mu ,\nu )$$ is of Howe type if

1. (i)

$$\mu \ne 0$$ and $$\nu \ne 0$$;

2. (ii)

$$f_1$$ and $$f_2$$ are coprime.

If $$(\lambda ,\mu ,\nu )$$ is of Howe type, then the desingularization of the fiber product $$E_1\times _{\mathbf{P }^1}E_2$$ is a Howe curve, since $$E_i\rightarrow \mathbf{P }^1$$ is ramified over the set consisting of 4 points, say $$S_i$$, and $$S_1\cap S_2 = \{(1:0)\}$$. Conversely, any Howe curve is realized in this way. Indeed, let $$E_i$$ for $$i=1,2$$ be two elliptic curves with double covers $$E_i\rightarrow \mathbf{P }^1$$, defining a Howe curve. An automorphism of $$\mathbf{P }^1$$ moves the ramified point of both of $$E_i\rightarrow \mathbf{P }^1$$ ($$i=1,2$$) to (1 : 0). By a usual coordinate-change, we make $$E_1$$ to be of the form (5) if $$p>3$$. Note that any elliptic curve ramified at (1 : 0) is written as in (6).

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

\begin{aligned} f(x) = f_1(x)f_2(x) \end{aligned}

and consider the hyperelliptic curve C of genus 2 defined by

\begin{aligned} C: u^2 = f(x). \end{aligned}
(7)

It was proven by Howe [9, Theorem 2.1] that H is of genus 4 and there exist two isogenies

\begin{aligned} \varphi : J(H) \longrightarrow E_1 \times E_2 \times J(C),\\ \psi : E_1 \times E_2 \times J(C) \longrightarrow J(H) \end{aligned}

such that $$\varphi \circ \psi$$ and $$\psi \circ \varphi$$ are the multiplication by 2, see [13, Theorem C] for a more general result. Hence H is supersingular if and only if $$E_1$$, $$E_2$$ and C are supersingular. For an abelian variety Z, let Z[p] denote the kernel of the multiplication-by-p map on Z. If p is odd, then $$\psi \circ \varphi$$ (resp. $$\varphi \circ \psi$$) is an automorphism of J(H)[p] (resp. $$(E_1 \times E_2 \times J(C))[p]$$), whence J(H)[p] and $$E_1[p] \times E_2[p] \times J(C)[p]$$ are isomorphic, see [5, Corollary 2] for a more general result. Hence 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 $$\ge 3$$.

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.,

\begin{aligned} f(x)^{(p-1)/2} = \sum _{i=0}^{3(p-1)} \gamma _i x^i. \end{aligned}
(8)

Put

\begin{aligned} a=\gamma _{p-1},\quad b=\gamma _{2p-1},\quad c=\gamma _{p-2}\quad \text {and}\quad d=\gamma _{2p-2}. \end{aligned}

In Sect. 4, we shall use the fact that $$\gamma _i$$ and therefore abc 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 Cartier operator on $$H^0(C,\Omega ^1_C)$$ (referred to as the modified Cartier operator in [23]). It is known (cf. [6, 4.1] and [23, §2]) that the Cartier–Manin matrix of C is given by

\begin{aligned} M := \begin{pmatrix} a &{}\quad b \\ c &{}\quad d \end{pmatrix}. \end{aligned}
(9)

Recall the fact (cf. [11, Lemma 1.1, (i)]) that a curve D of genus 2 is supersingular if and only if the Cartier operator V on $$H^0(D,\Omega _D^1)$$ satisfies $$V^2=0$$. Hence C is supersingular if and only if

\begin{aligned} M M^\sigma = \begin{pmatrix} a^{p+1}+bc^p &{}\quad ab^p+bd^p\\ a^pc+c^pd &{}\quad b^pc+d^{p+1} \end{pmatrix} = 0, \end{aligned}
(10)

where $$M^\sigma = \begin{pmatrix}a^p &{}\quad b^p \\ c^p &{}\quad d^p\end{pmatrix}$$. See [2, Step 2 of the proof of Theorem 4.8] and also [7, 4.0.3] for a sufficient condition for supersingularity, generalized in the higher genus case.

### Remark 2.2

As Achter and Howe suggested in [1], we need to be careful when we use Cartier–Manin matrices. A notation different from (9) for Cartier–Manin matrices exists: the transpose A of M, see [23]. In the notation, the supersingularity for genus-2 curves is equivalent to $$A^\sigma A=0$$, as in [11, Lemma 1.1, (i)].

### Proposition 2.3

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 (10) holds. First remark that $$ad-bc=0$$ implies

\begin{aligned} a^{p+1}+bc^p= & {} a(a^p+c^{p-1}d), \end{aligned}
(11)
\begin{aligned} ab^p+bd^p= & {} b(ab^{p-1}+d^p),\end{aligned}
(12)
\begin{aligned} a^pc+c^pd= & {} c(a^p+c^{p-1}d),\end{aligned}
(13)
\begin{aligned} b^pc+d^{p+1}= & {} d(ab^{p-1}+d^p). \end{aligned}
(14)

To prove the “if”-part, suppose that $$ad-bc=0$$, $$ab^{p-1}+d^p=0$$ and $$a^p+c^{p-1}d=0$$. It follows from $$ad-bc=0$$ that (11)–(14) hold. This means that every entry of the matrix in (10) is a multiple of $$ab^{p-1}+d^p$$ or $$a^p+c^{p-1}d$$. Thus the “if”-part is true.

Conversely, suppose that (10) holds. Since $${\text {det}}(M M^\sigma ) = {\text {det}}(M)^{p+1}$$, we have $${\text {det}}(M)=ad-bc=0$$, which implies

\begin{aligned} a^{p-1}(ab^{p-1}+d^p)= & {} b^{p-1}(a^p+c^{p-1}d),\end{aligned}
(15)
\begin{aligned} c^{p-1}(ab^{p-1}+d^p)= & {} d^{p-1}(a^p+c^{p-1}d). \end{aligned}
(16)

Also by (11)–(14) 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 (15) and (16) we have $$a^p+c^{p-1}d=0$$ unless $$b=d=0$$. Similarly, if $$b=d=0$$, by (15) and (16) 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 “only if”-part is true. $$\square$$

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

### Lemma 2.4

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

\begin{aligned} M' = \begin{pmatrix}\gamma '_{p-1}&{}\quad \gamma '_{2p-1}\\ \gamma '_{p-2}&{}\quad \gamma '_{2p-2}\end{pmatrix}. \end{aligned}

Then we have

\begin{aligned} M' = P^{-1}MP^\sigma \end{aligned}

with

\begin{aligned} P = \begin{pmatrix}u &{}\quad 0 \\ uv &{}\quad u^2\end{pmatrix}. \end{aligned}

In particular, we have $$\det (M') = \det (M)$$ if $$u=1$$.

### Proof

We have

\begin{aligned} \gamma '_i = \sum _{j = i}^{3(p-1)} \left( {\begin{array}{c}j\\ i\end{array}}\right) u^iv^{j-i}\gamma _j \end{aligned}

as

\begin{aligned} 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( {\begin{array}{c}j\\ i\end{array}}\right) u^iv^{j-i}\gamma _j \right) X^i. \end{aligned}

Then we have

\begin{aligned} \gamma '_{p-1}= & {} u^{p-1}(\gamma _{p-1} + v^p\gamma _{2p-1}),\\ \gamma '_{2p-1}= & {} u^{2p-1}\gamma _{2p-1},\\ \gamma '_{p-2}= & {} u^{p-2}(\gamma _{p-2}-v\gamma _{p-1} +v^p\gamma _{2p-2} - v^{p+1}\gamma _{2p-1}),\\ \gamma '_{2p-2}= & {} u^{2p-2}(\gamma _{2p-2} - v\gamma _{2p-1}) \end{aligned}

by calculating, in characteristic p, the binomials

\begin{aligned} \left( {\begin{array}{c}j\\ p-1\end{array}}\right)= & {} {\left\{ \begin{array}{ll}1 &{} \text {if } j=p-1,2p-1,\\ 0 &{} \text {otherwise},\end{array}\right. } \qquad \left( {\begin{array}{c}j\\ 2p-1\end{array}}\right) ={\left\{ \begin{array}{ll}1 &{} \text {if } j=2p-1,\\ 0 &{} \text {otherwise},\end{array}\right. }\\ \left( {\begin{array}{c}j\\ p-2\end{array}}\right)= & {} {\left\{ \begin{array}{ll}1 &{} \text {if } j=p-2,2p-2\\ -1 &{} \text {if } j=p-1,2p-1,\\ 0 &{} \text {otherwise},\end{array}\right. }\qquad \left( {\begin{array}{c}j\\ 2p-2\end{array}}\right) ={\left\{ \begin{array}{ll}1 &{} \text {if } j=2p-2,\\ -1 &{} \text {if } j=2p-1,\\ 0 &{} \text {otherwise}\end{array}\right. } \end{aligned}

for $$j\le 3p-3$$. These are obtained by applying a general formula

\begin{aligned} \left( {\begin{array}{c}kp+\ell \\ k'p+\ell '\end{array}}\right) = \left( {\begin{array}{c}k\\ k'\end{array}}\right) \left( {\begin{array}{c}\ell \\ \ell '\end{array}}\right) \end{aligned}
(17)

for $$0\le \ell ,\ell '\le p-1$$ to the case of $$k'=0,1$$ and $$\ell '=p-1,p-2$$ with $$\left( {\begin{array}{c}\ell \\ \ell '\end{array}}\right) =0$$ for $$\ell '>\ell$$. The formula (17) follows from

\begin{aligned} (x+1)^{kp+\ell }=(x^p+1)^k(x+1)^\ell = \sum _{k'}\sum _{\ell '}\left( {\begin{array}{c}k\\ k'\end{array}}\right) \left( {\begin{array}{c}\ell \\ \ell '\end{array}}\right) x^{pk'+\ell '} \end{aligned}

in characteristic p. $$\square$$

## A moduli-theoretic observation and ingredients of the proof

From the next section onward, we go into the details of the proof of the main theorem. Before that, let us give an outline of the proof with a moduli-theoretic observation. We use the same notation as in Sect. 2. We choose $$A_1$$, $$B_1$$, $$A_2$$ and $$B_2$$ so that the elliptic curves (1) and (2) are supersingular. Let $${\tilde{C}}$$ be the desingularizaion only at the infinity of C defined in (7). Note that $${\tilde{C}}$$ is nonsingular if and only if the corresponding point $$(\lambda : \mu : \nu )$$ is of Howe type. Considering $$\lambda$$, $$\mu$$ and $$\nu$$ as indeterminates, we have a family of curves $${\tilde{C}}$$ over $$\mathbf{P }^2={\text {Proj}}K[\lambda , \mu , \nu ]$$. To prove the existence of a supersingular Howe curve, it suffices to find a nonsingular supersingular fiber from this family. The possible Newton polygons of nonsingular $${\tilde{C}}$$’s are supersingular, p-rank-one and ordinary. Note that a nonsingular $${\tilde{C}}$$ has p-rank $$\le 1$$ if and only if its Cartier–Manin matrix $$M=\begin{pmatrix}a&{}\quad b\\ c&{}\quad d\end{pmatrix}$$ has determinant zero. Since the determinant $$ad-bc$$ turns out to be a multiple of a power of $$\mu \nu$$ and every fiber $${\tilde{C}}$$ over the locus $$\mu \nu =0$$ is singular, it would be better to divide $$ad-bc$$ by $$\mu \nu$$ as much as possible. We shall prove in Lemma 5.3 that the order of $$ad-bc$$ with respect to $$\mu \nu$$ is exactly $$(p+1)/2$$. The exactness is important and its proof requires an assertion (Corollary 4.4) on cubic polynomials from the next section.

Put

\begin{aligned} h_0:= & {} (ad-bc)/(\mu \nu )^{(p+1)/2},\\ h_1:= & {} ab^{p-1}+d^p,\\ h_2:= & {} a^p+c^{p-1}d. \end{aligned}

We consider

\begin{aligned} V(h_0,h_1,h_2) \subset V(h_0) \subset \mathbf{P }^2 \end{aligned}

as the Newton polygon stratification for this family over $$\mathbf{P }^2$$, where $$V(h_0,h_1,h_2)$$, $$V(h_0)\smallsetminus V(h_0,h_1,h_2)$$ and $$\mathbf{P }^2\smallsetminus V(h_0)$$ are called the supersingular locus, the p-rank-one locus and the ordinary locus respectively. The expected dimensions of the loci would be 0, 1 and 2 respectively. Indeed, this is proved in Proposition 5.7. (A priori, on the open subscheme where $${\tilde{C}}$$ is nonsingular, the dimension of each locus is greater than or equal to the expected one by the purity theorem [3, 4.1] due to de Jong and Oort.) If the p-rank-one locus is proved not to be projective, then $$V(h_0,h_1,h_2)\ne \emptyset$$ follows from it (cf. Proposition 5.7). Our proof is along this line. However, an obstruction is that $$V(h_0,h_1,h_2)$$ may be contained in the locus consisting of points which are not of Howe type (equivalently, over which $${\tilde{C}}$$ is singular). The main part of our proof is to show that there is no obstruction, i.e., the condition $$h_i=0$$ for $$i=0,1,2$$ implies that $${\tilde{C}}$$ is nonsingular. This is proven in Proposition 5.6, where the proof needs another assertion (Corollary 4.7) on cubic polynomials from the next section.

## Two assertions on cubic polynomials

In this section, we prove two assertions on cubic polynomials (Corollaries 4.4 and 4.7), which play important roles in the proof of the main theorem. The proof is done by reducing the problems to those of the Legendre form (Propositions 4.3 and 4.6).

Assume $$p\ge 3$$. 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

\begin{aligned} H_p (t):= & {} \sum _{i=0}^{e} \left( {\begin{array}{c}e\\ i\end{array}}\right) ^2 t^i. \end{aligned}
(18)

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

\begin{aligned} \delta _{p-1}(t) = (-1)^e H_p(t) \end{aligned}
(19)

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., [21, Chap. V, Theorem 4.1].

For our purpose, we need to study the $$x^{p-2}$$-coefficient $$\delta _{p-2}(t)$$ of $$g(x)^e$$. Specifically, we shall show Proposition 4.3 below, which asserts that $$\delta _{p-1}(t)$$ and $$\delta _{p-2}(t)$$ are coprime. We start with giving an explicit formula of $$\delta _{p-2}(t)$$.

### Lemma 4.1

We have

\begin{aligned} \delta _{p-2}(t) = (-1)^{e-1} \sum _{i=0}^{e-1} \left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ i+1\end{array}}\right) t^{i+1} . \end{aligned}
(20)

### Proof

By the binomial theorem, $$g(x)^e = x^e(x-1)^e(x-t)^e$$ is equal to

\begin{aligned} x^e \sum _{i=0}^e \left( {\begin{array}{c}e\\ i\end{array}}\right) x^i(-1)^{e-i} \sum _{j=0}^e \left( {\begin{array}{c}e\\ j\end{array}}\right) x^{e-j}(-t)^{j} = \sum _{i=0}^e\sum _{j=0}^e (-1)^{e-i+j}\left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ j\end{array}}\right) x^{2e+i-j}t^{j}. \end{aligned}
(21)

The $$x^{p-2}$$-coefficient of this is the sum over (ij) with $$j=i+1$$, since $$j=i+1$$ if $$2e+i-j=p-2$$. Thus, we have the desired equation. $$\square$$

The next lemma is the essential part of the proof of the coprimeness of $$\delta _{p-1}(t)$$ and $$\delta _{p-2}(t)$$.

### Lemma 4.2

We have the equality

\begin{aligned} \delta _{p-2}(t) = t \delta _{p-1}(t) + 2 t (t-1) \frac{d}{dt}\delta _{p-1}(t). \end{aligned}

### Proof

The lemma follows from the equality

\begin{aligned} \left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ i+1\end{array}}\right) = - \left( {\begin{array}{c}e\\ i\end{array}}\right) ^2 - 2\left( {\begin{array}{c}e\\ i\end{array}}\right) ^2 i + 2\left( {\begin{array}{c}e\\ i+1\end{array}}\right) ^2(i+1) \end{aligned}
(22)

in characteristic p for $$i=0,1,\ldots ,e$$ with $$\left( {\begin{array}{c}e\\ e+1\end{array}}\right) =0$$. Indeed, multiplying both sides of (22) by $$(-1)^{e-1}t^{i+1}$$ and taking the sum over $$i=0,1,\ldots ,e$$, we have

\begin{aligned} \delta _{p-2}(t)= t\delta _{p-1}(t) + 2t^2 \frac{d}{dt}\delta _{p-1}(t) - 2t \frac{d}{dt}\delta _{p-1}(t). \end{aligned}

Since $$2(e-i) \equiv -1-2i\ \mathrm{mod}\ p$$, we have

\begin{aligned} 1=2(i+1)-1-2i=\frac{(-1-2i)(i+1)}{e-i}+2(e-i) =-\frac{i+1}{e-i}-2i\frac{i+1}{e-i}+2(e-i). \end{aligned}

The equation (22) is obtained by multiplying both sides by $$\displaystyle \left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ i+1\end{array}}\right)$$. $$\square$$

Now we prove

### Proposition 4.3

The polynomials $$\delta _{p-1}(t)$$ and $$\delta _{p-2}(t)$$ are coprime, i.e., have no common root in $$\overline{\mathbb {F}_p}$$.

### Proof

Note that $$\delta _{p-1}(0)$$ and $$\delta _{p-1}(1)$$ are not zero, since $$\delta _{p-1}(t) = (-1)^e H_p (t)$$ by (19) and $$H_p(0)=1$$ and $$H_p(1)=(-1)^e$$, see the proof of [21, Chap. V, Theorem 4.1 (c)]). If $$\delta _{p-1}(\alpha ) =\delta _{p-2}(\alpha ) = 0$$ for some $$\alpha \in \overline{\mathbb {F}_p} \smallsetminus \{ 0, 1 \}$$, then one has $$\left( \frac{d}{dt} \delta _{p-1}\right) (\alpha ) = 0$$ by Lemma 4.2. We have $$H_{p}(\alpha )=\left( \frac{d}{dt} H_{p}\right) (\alpha ) = 0$$, which means that $$\alpha$$ is a double root of $$H_p (t)$$. This contradicts the fact shown by Igusa [12] that all roots of $$H_p(t)$$ are simple (cf. [21, Chap. V, Theorem 4.1 (c)]). $$\square$$

The following corollary derived from Proposition 4.3 will be used in the next section.

### Corollary 4.4

Let K be an algebraically closed field of characteristic p. Let $$E_0$$ be an elliptic curve $$y^2 = g_0(x)$$ over K, where $$g_0(x)$$ is a cubic separable polynomial in K[x]. Assume that $$E_0$$ is supersingular, i.e., the $$x^{p-1}$$-coefficient of $$g_0(x)^e$$ is zero. Then $$x^{p-2}$$-coefficient of $$g_0(x)^e$$ is not zero.

### Proof

There exist $$u \ne 0$$ and v in K such that $$g_0 (x) = g (X)$$ with $$X := ux + v$$, where $$g(X)=X(X-1)(X-t)$$ for some $$t \in K \smallsetminus \{ 0, 1 \}$$. Let $$\epsilon _i$$ (resp. $$\epsilon _i'$$) be the $$X^i$$-coefficients (resp. $$x^i$$-coefficients) of $$g(X)^{(p-1)/2}$$ and $$g_0(x)^{(p-1)/2}$$ respectively for $$0 \le i \le 3 (p-1)/2$$. Similarly to the proof of Lemma 2.4, we have

\begin{aligned} \epsilon _{i}' = \sum _{j=i}^{3 (p-1)/2} \left( {\begin{array}{c}j\\ i\end{array}}\right) u^i v^{j-i} \epsilon _{j} \end{aligned}

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

\begin{aligned} \epsilon _{p-1}'= & {} u^{p-1} \epsilon _{p-1} , \end{aligned}
(23)
\begin{aligned} \epsilon _{p-2}'= & {} u^{p-2} (\epsilon _{p-2} - v \epsilon _{p-1} ). \end{aligned}
(24)

It follows from $$\epsilon '_{p-1}=0$$ and (23) that $$\epsilon _{p-1}=0$$. By Proposition 4.3, $$\epsilon _{p-1}=0$$ implies $$\epsilon _{p-2}\ne 0$$. By (24) we obtain $$\epsilon '_{p-2}\ne 0$$. $$\square$$

The second aim of this section is to show that the e-th derivative $$(g_0(x)^e)^{(e)}$$ of $$g_0(x)^e$$ is separable for any cubic separable polynomial $$g_0(x)$$. We start with reviewing an elementary congruence relation for the reader’s convenience.

### Lemma 4.5

We have

\begin{aligned} (p-1-n)! \equiv \frac{(-1)^{n+1}}{n!} \pmod {p} \end{aligned}

for $$n=0,1,\ldots ,p-1$$.

### Proof

We have

\begin{aligned} (p - 1 - n )! = \frac{(p-1)!}{(p-1)(p-2) \cdots \left( p -n\right) } \equiv \frac{-1}{(-1)(-2) \cdots (-n)} = \frac{(-1)^{n+1}}{n!} \pmod {p}, \end{aligned}

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

The next proposition gives a factorization of $$(g(x)^e)^{(e)}$$ for $$g(x)=x(x-1)(x-t)$$. Let $$H_p(t)$$ be the polynomial defined by (18). Let $$a_i \in \overline{\mathbb {F}_p}$$ $$(i=1,\ldots ,e)$$ be the roots of $$H_p(t)=0$$, i.e.,

\begin{aligned} H_p(t) = \prod _{i=1}^{e}(t-a_i). \end{aligned}
(25)

### Proposition 4.6

We have

\begin{aligned} \frac{1}{e!}\cdot (g(x)^e)^{(e)} = \prod _{i=1}^e\left( (t-a_i)x-(1-a_i)t\right) , \end{aligned}
(26)

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

### Proof

Replacing j by $$e-j$$ in (21) and taking its e-th derivative, we have

\begin{aligned} (g(x)^e)^{(e)}=\sum _{i=0}^e\sum _{j=0}^e (-1)^{i+j}\frac{(e+i+j)!}{(i+j)!}\left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ j\end{array}}\right) x^{i+j}t^{e-j}. \end{aligned}
(27)

Hence $$x^{i+j}t^{e-j}$$-coefficient of the left side of (26) is $$(-1)^{i+j}\left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ j\end{array}}\right)$$ times

\begin{aligned} \frac{(e+i+j)!}{e!(i+j)!}. \end{aligned}
(28)

Put $$P:=\prod _{i=1}^e\left( (t-a_i)x-(1-a_i)t\right)$$. We have

\begin{aligned} P= & {} (x-t)^e\prod _{i=1}^e\left( \frac{t(x-1)}{x-t}-a_i\right) = (x-t)^e H_p\left( \frac{t(x-1)}{x-t}\right) \\= & {} (x-t)^e\sum _{k=0}^e \left( {\begin{array}{c}e\\ k\end{array}}\right) ^2 \left( \frac{t(x-1)}{x-t}\right) ^k \end{aligned}

by (25) and (18). Using the binomial theorem, one obtains

\begin{aligned} P= & {} \sum _{k=0}^e \left( {\begin{array}{c}e\\ k\end{array}}\right) ^2 (x-t)^{e-k}t^k(x-1)^k \nonumber \\= & {} \sum _{k=0}^e \left( {\begin{array}{c}e\\ k\end{array}}\right) ^2 t^k \sum _{j=0}^{e} \left( {\begin{array}{c}e-k\\ j\end{array}}\right) x^j(-t)^{e-k-j} \sum _{i=0}^{e} \left( {\begin{array}{c}k\\ i\end{array}}\right) x^i(-1)^{k-i}\nonumber \\= & {} \sum _{k=0}^e\sum _{j=0}^{e}\sum _{i=0}^{e} (-1)^{e-i-j}\left( {\begin{array}{c}e\\ k\end{array}}\right) ^2\left( {\begin{array}{c}e-k\\ j\end{array}}\right) \left( {\begin{array}{c}k\\ i\end{array}}\right) x^{i+j}t^{e-j}, \end{aligned}
(29)

where we used $$\left( {\begin{array}{c}k\\ i\end{array}}\right) =0$$ for $$i>k$$ and $$\left( {\begin{array}{c}e-k\\ j\end{array}}\right) =0$$ for $$j>e-k$$ in the second equality. Since

\begin{aligned} \left( {\begin{array}{c}e\\ k\end{array}}\right) \left( {\begin{array}{c}e-k\\ j\end{array}}\right) =\left( {\begin{array}{c}e\\ j\end{array}}\right) \left( {\begin{array}{c}e-j\\ k\end{array}}\right) \quad \text {and}\quad \left( {\begin{array}{c}e\\ k\end{array}}\right) \left( {\begin{array}{c}k\\ i\end{array}}\right) =\left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e-i\\ e-k\end{array}}\right) , \end{aligned}

the $$x^{i+j}t^{e-j}$$-coefficient of P is

\begin{aligned} (-1)^{e-i-j}\left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ j\end{array}}\right) \sum _{k=0}^e\left( {\begin{array}{c}e-j\\ k\end{array}}\right) \left( {\begin{array}{c}e-i\\ e-k\end{array}}\right) . \end{aligned}
(30)

By applying Vandermonde’s identity to the sum, (30) is equal to $$(-1)^{i+j}\left( {\begin{array}{c}e\\ i\end{array}}\right) \left( {\begin{array}{c}e\\ j\end{array}}\right)$$ times

\begin{aligned} (-1)^{e}\left( {\begin{array}{c}2e-i-j\\ e\end{array}}\right) . \end{aligned}
(31)

If $$i+j > e$$, then both of (28) and (31) are zero. Otherwise, applying Lemma 4.5 to $$n=i+j$$ and $$n=e-i-j$$, we have that (28) is equal to (31). Thus the equation (26) holds. $$\square$$

As a corollary of Proposition 4.6, we have

### Corollary 4.7

Let $$g_0(x)$$ be a separable cubic polynomial. Then $$(g_0(x)^e)^{(e)}$$ is a separable polynomial.

### Proof

A linear coordinate change makes $$g_0$$ a Legendre form $$g(x)=x(x-1)(x-t)$$ with $$t\ne 0,1$$. It is clear that $$(g_0(x)^e)^{(e)}$$ is separable if and only if $$(g(x)^e)^{(e)}$$ is separable. Let $$a_1,\ldots ,a_e$$ be the roots of $$H_p(t)$$ as above. Note that $$a_i\ne 0,1$$ for $$i=1,2,\ldots ,e$$ as $$H_p(0)=1$$ and $$H_p(1)=(-1)^e$$ (cf. the proof of [21, Chap. V, Theorem 4.1 (c)]). If $$t \not \in \{a_1,\ldots ,a_e\}$$, the roots of $$(g(x)^e)^{(e)}$$ are $$(1-a_i)t/(t-a_i)$$ for $$i=1, \ldots , e$$. Otherwise, we may assume $$t=a_1$$ without loss of generality, and then the roots of $$(g(x)^e)^{(e)}$$ are $$(1-a_i)a_1/(a_1-a_i)$$ for $$i=2,\ldots ,e$$. In both cases, it is straightforward to see that the roots are mutually distinct, using the fact that $$H_p(t)$$ is separable, i.e., $$a_1,\ldots ,a_e$$ are mutually distinct (cf. [12] and [21, Chap. V, Theorem 4.1 (c)]). $$\square$$

## 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 Sect. 2, i.e., $$A_1,B_1,A_2,B_2$$, $$\lambda , \mu , \nu$$, $$f_1, f_2, f$$, abcd are as in Sect. 2. We choose $$A_1,B_1,A_2,B_2$$ in K so that (1) and (2) are supersingular. By Proposition 2.3, 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 indeterminates, and consider abcd as polynomials in $$\lambda$$, $$\mu$$, $$\nu$$. Note that abc and d are homogeneous polynomials in $$\lambda ,\mu ,\nu$$ of degrees $$2p-2$$, $$p-2$$, $$2p-1$$ and $$p-1$$ respectively.

We start with studying the factors of $$\mu$$ and $$\nu$$ in $$ad-bc$$. More precisely, we prove that $$ad-bc$$ is a multiple of $$(\mu \nu )^{(p+1)/2}$$ in Lemma 5.3 below. Set $$e := (p-1)/2$$ as in Section 3. Let $$\alpha _k$$ and $$\beta _k$$ be the $$x^k$$-coefficients of $$f_1(x)$$ and $$f_2(x)$$ respectively, i.e.,

\begin{aligned} f_1(x)^{e} = \sum _{k=0}^{3e} \alpha _k x^k \quad \text {and} \quad f_2(x)^{e} = \sum _{k=0}^{3e} \beta _k x^k. \end{aligned}

Let $$\gamma _i$$ be the $$x^i$$-coefficient of $$f(x)^{(p-1)/2}$$ with $$f(x)=f_1(x)f_2(x)$$ as in (8). We have

\begin{aligned} \gamma _i = \sum _{k=0}^{3e} \alpha _k\beta _{i-k} \end{aligned}

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

\begin{aligned} F_2 (x):= & {} x^3 + A_2 \nu ^2 x + B_2 \nu ^3. \end{aligned}
(32)

Note $$f_2(x) = F_2(x - \lambda )$$. We denote by $$\beta _n^{\prime }$$ the $$x^n$$-coefficient of $$F_2(x)^{e}$$, i.e.,

\begin{aligned} F_2(x)^{e} = \sum _{n=0}^{3e} \beta _n^{\prime } x^n \end{aligned}
(33)

with $$\beta _{3e}^{\prime }=1$$. Looking at the $$x^k$$-coefficients of both sides of $$f_2(x)^{e} = F_2(x - \lambda )^{e}$$, we have

\begin{aligned} \beta _k = \sum _{n=k}^{3e} \left( {\begin{array}{c}n\\ k\end{array}}\right) \beta _n'\cdot (-\lambda )^{n-k}. \end{aligned}
(34)

From this, we find the highest terms in $$\lambda$$ of $$\beta _k$$ for some k, making use of $$\left( {\begin{array}{c}n\\ k\end{array}}\right) = 0$$ in $$\mathbb {F}_p$$ if $$0\le k \le p-1$$ and $$p\le n \le p+k-1$$:

### Lemma 5.1

The coefficient $$\beta _k$$ is a homogeneous polynomial in $$K[\lambda ,\nu ]$$ of degree $$3e-k$$. Moreover

1. (1)

the highest term in $$\lambda$$ of $$\beta _0$$ is $$(-\lambda )^{3 e}$$,

2. (2)

the highest term in $$\lambda$$ of $$\beta _1$$ is $$3e(-\lambda )^{3 e - 1}$$,

3. (3)

the highest term in $$\lambda$$ of $$\beta _e$$ is $$\left( {\begin{array}{c}p-2\\ e\end{array}}\right) \beta _{p-2}^{\prime }\cdot (-\lambda )^{\frac{p-3}{2}}$$ and

4. (4)

the highest term in $$\lambda$$ of $$\beta _{e+1}$$ is $$\left( {\begin{array}{c}p-2\\ e+1\end{array}}\right) \beta _{p-2}^{\prime }\cdot (-\lambda )^{\frac{p-5}{2}}$$.

### Proof

Note that $$\beta '_n$$ is a monomial in $$K[\nu ]$$ of degree $$3 e - n$$, since $$F_2(x)^{e}$$ is homogeneous of degree 3e as a polynomial in $$K[x, \nu ]$$. Hence, by (34) the coefficient $$\beta _k$$ is a homogeneous polynomial in $$K[\lambda ,\nu ]$$ of degree $$3e-k$$.

(1) and (2): For $$k=0,1$$, the highest term in $$\lambda$$ of $$\beta _k$$ is the term of $$n=3e$$ of (34), which is

\begin{aligned} \left( {\begin{array}{c}3e\\ k\end{array}}\right) (-\lambda )^{3e-k}. \end{aligned}

(3) and (4): Let k be e or $$e+1$$. Since $$\left( {\begin{array}{c}n\\ k\end{array}}\right) = 0$$ for $$p\le n \le p+k-1$$ with $$p+k-1 \ge 3e$$, it follows from (34) that

\begin{aligned} \beta _k = \sum _{n=k}^{p-1} \left( {\begin{array}{c}n\\ k\end{array}}\right) \beta _n'\cdot (-\lambda )^{n-k}. \end{aligned}
(35)

Since $$E_2$$ is supersingular, we have $$\beta '_{p-1}=0$$ and $$\beta _{p-2}^{\prime } \ne 0$$ by Corollary 4.4. Hence, the highest term in $$\lambda$$ of (35) is

\begin{aligned} \left( {\begin{array}{c}p-2\\ k\end{array}}\right) \beta _{p-2}'\cdot (-\lambda )^{p-2-k}. \end{aligned}

Note that $$p-2\ge k$$ for $$k=e$$ and $$e+1$$ if $$p\ge 5$$. $$\square$$

### Remark 5.2

The assumption $$p>3$$ is used in Lemma 5.1 (2) and (4). This is the reason why the proof of the main theorem does not work for $$p=3$$. In fact, there is no superspecial Howe curve in characteristic 3, see Remark 5.8.

In the next lemma, we study the order $$\mathrm{ord}_\mu (a)$$ of a with respect to $$\mu$$ and so on, where $$\mathrm{ord}_\mu (a)$$ is the largest number k such that a is a mulpliple of $$\mu ^k$$ in $$K[\lambda ,\mu ,\nu ]$$.

### Lemma 5.3

We have the following:

1. (1)

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

2. (2)

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

3. (3)

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

4. (4)

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

### Proof

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

\begin{aligned} \displaystyle \alpha _{k}= & {} \sum _{3 n_0 + n_1=k} \frac{e!}{{n_0}! {n_1}! (e - n_0 - n_1)!} (A_1 \mu ^2)^{n_1} (B_1 \mu ^3)^{e - n_0 - n_1} \\= & {} \left( \sum _{3 n_0 + n_1=k} \frac{e!}{{n_0}! {n_1}! (e - n_0 - n_1)!} A_1^{n_1} B_1^{e - n_0 - n_1} \right) \mu ^{3 e - k}, \end{aligned}

and thus $$\alpha _{k}$$ is a multiple of $$\mu ^{3 e - k}$$. Putting $$\tilde{\alpha }_{k} := \alpha _{k} / (\mu ^{3 e - k})$$, we also have $$\tilde{\alpha }_k \in K$$ for $$0 \le k \le 3e$$. Since both the elliptic curves $$E_1 : z^2 = f_1(x)$$ and $$E_2 : w^2 = f_2(x)$$ are supersingular, we have $$\alpha _{p-1} = 0$$ and $$\beta _{p-1} = 0$$, whereas $$\alpha _{p-2} \ne 0$$ (and thus $$\tilde{\alpha }_{p-2} \ne 0$$) and $$\beta _{p-2} \ne 0$$ by Corollary 4.4. It follows that

\begin{aligned} \displaystyle a = \gamma _{p-1} = \sum _{k=1}^{p-2} \alpha _{k} \beta _{p-1-k} = \sum _{k=1}^{p-2} \mu ^{3 e - k} \tilde{\alpha }_{k} \beta _{p-1-k} = \sum _{j=\frac{p+1}{2}}^{\frac{3 p -5}{2}} \mu ^{j} \tilde{\alpha }_{3 e - j} \beta _{j - e} , \end{aligned}
(36)

where $$\beta _k$$ is a polynomial in $$K [\lambda , \nu ]$$ for each $$0 \le k \le 3 e$$. Since $$\beta _1 \ne 0$$ by Lemma 5.1 (2), we have $$\tilde{\alpha }_{3 e - j} \beta _{j-e} = \tilde{\alpha }_{p-2} \beta _{1} \ne 0$$ for $$j = (p+1)/2$$, and thus $$\mathrm{ord}_{\mu } (a) = (p+1)/2$$.

(2) Similarly to the proof of (1), one has

\begin{aligned} \displaystyle c = \gamma _{p-2} = \sum _{k=0}^{p-2} \alpha _{k} \beta _{p-2-k} = \sum _{k=0}^{p-2} \mu ^{3e - k} \tilde{\alpha }_{k} \beta _{p-2-k} = \sum _{j=\frac{p+1}{2}}^{\frac{3 p - 3}{2}} \mu ^{j} \tilde{\alpha }_{3 e - j} \beta _{j - e - 1}. \end{aligned}
(37)

Since we have $$\tilde{\alpha }_{p-2}\ne 0$$, and $$\beta _0 \ne 0$$ by Lemma 5.1 (1), we also have $$\tilde{\alpha }_{3 e - j} \beta _{j-e-1} = \tilde{\alpha }_{p-2} \beta _{0} \ne 0$$ for $$j = (p+1)/2$$, and thus $$\mathrm{ord}_{\mu } (c) = (p+1)/2$$.

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

\begin{aligned} b= & {} \gamma _{2p-1} = \sum _{k=\frac{p+1}{2}}^{\frac{3}{2}(p-1)} \alpha _{k} \beta _{2p-1-k} = \sum _{k=\frac{p+1}{2}}^{\frac{3}{2}(p-1)} \mu ^{3e - k} \tilde{\alpha }_{k} \beta _{2p-1-k} = \sum _{j=0}^{p-2} \mu ^{j} \tilde{\alpha }_{3 e - j} \beta _{j + e + 1}, \end{aligned}
(38)
\begin{aligned} d= & {} \gamma _{2p-2} = \sum _{k=\frac{p-1}{2}}^{\frac{3}{2}(p-1)} \alpha _{k} \beta _{2p-2-k} = \sum _{k=\frac{p-1}{2}}^{\frac{3}{2}(p-1)} \mu ^{3e - k} \tilde{\alpha }_{k} \beta _{2p-2-k} =\sum _{j=0}^{p-1} \mu ^{j} \tilde{\alpha }_{3e - j} \beta _{j+e}, \end{aligned}
(39)

both of which are not a multiple of $$\mu$$ since $$\alpha _{3 e} = 1$$ and since $$\beta _{e}, \beta _{e+1} \ne 0$$ by Lemma 5.1 (3) and (4). Thus, if the coefficient of $$\mu ^{(p+1)/2}$$ in $$ad-bc$$ is not zero, then we have $$\mathrm{ord}_{\mu }(a d - b c) = (p+1)/2$$. By straightforward computation, the coefficients of $$\mu ^{(p+1)/2}$$ in ad and bc are $$\tilde{\alpha }_{p-2} \beta _1 \tilde{\alpha }_{3 e} \beta _e$$ and $$\tilde{\alpha }_{p-2} \beta _0 \tilde{\alpha }_{3 e} \beta _{e+1}$$, respectively. Here we have

\begin{aligned} \tilde{\alpha }_{p-2} \beta _1 \tilde{\alpha }_{3 e} \beta _e - \tilde{\alpha }_{p-2} \beta _0 \tilde{\alpha }_{3 e} \beta _{e+1} = \tilde{\alpha }_{p-2} \tilde{\alpha }_{3 e} ( \beta _1 \beta _e - \beta _0 \beta _{e+1}), \end{aligned}

where $$\tilde{\alpha }_{p-2} \ne 0$$ and $$\tilde{\alpha }_{3 e} = \alpha _{3 e} = 1$$. If $$\beta _1 \beta _e - \beta _0 \beta _{e+1} \ne 0$$, we have $$\mathrm{ord}_{\mu }(ad - b c)= (p+1)/2$$. By Lemma 5.1, the highest term of $$\beta _1 \beta _{e}$$ in $$\lambda$$ is

\begin{aligned} 3e(-\lambda )^{3e-1}\cdot \left( {\begin{array}{c}p-2\\ e\end{array}}\right) \beta '_{p-2}\cdot (-\lambda )^{e-1} =3e\left( {\begin{array}{c}p-2\\ e\end{array}}\right) \beta '_{p-2} \cdot (-\lambda )^{2p-4} \end{aligned}

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

\begin{aligned} (-\lambda )^{3e}\cdot \left( {\begin{array}{c}p-2\\ e+1\end{array}}\right) \beta '_{p-2}\cdot (-\lambda )^{e-2} =\left( {\begin{array}{c}p-2\\ e+1\end{array}}\right) \beta '_{p-2}\cdot (-\lambda )^{2p-4}. \end{aligned}

Since $$\beta _{p-2}' \ne 0$$ by Corollary 4.4, it suffices to show

\begin{aligned} 3e\left( {\begin{array}{c}p-2\\ e\end{array}}\right) \ne \left( {\begin{array}{c}p-2\\ e+1\end{array}}\right) \end{aligned}

in $$\mathbb {F}_p$$. If the equality held, we would have $$3e/(p-2-e) = 1/(e+1)$$ in $$\mathbb {F}_p$$. Multiplying the denominators and the numerators of both sides by 2, we get $$3(p-1)/(p-3)= 2/(p+1)$$ in $$\mathbb {F}_p$$. But the left hand side is equal to 1 and the right hand side is equal to 2. This is a contradiction.

Next, we show $$\mathrm{ord}_{\nu } (a d - b c) = (p+1)/2$$. To show this, we consider the transformation $$X = x - \lambda$$. The polynomials $$f_1 (x)$$ and $$f_2 (x)$$ are rewritten as

\begin{aligned} F_1 (X)= & {} (X + \lambda )^3 + A_1 \mu ^2 (X + \lambda )+ B_1\mu ^3, \end{aligned}
(40)
\begin{aligned} F_2 (X)= & {} X^3 + A_2 \nu ^2X + B_2 \nu ^3 \end{aligned}
(41)

respectively. Let $$\gamma _{l}^{\prime }$$ denote the coefficient of $$X^{l}$$ in $$(F_1(X) F_2(X))^{(p-1)/2}$$. 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 the second assertion of Lemma 2.4. By the same argument as in the proof of $$\mathrm{ord}_{\mu }(ad - bc) = (p+1)/2$$ for $$f_1(x)$$ and $$f_2(x)$$, we have $$\mathrm{ord}_{\nu }(a' d' - b' c') = (p+1)/2$$ for $$F_2 (X)$$ and $$F_1(X)$$, and thus $$\mathrm{ord}_{\nu }(ad-bc) = (p+1)/2$$.

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

\begin{aligned} B:= \tilde{\alpha }_{p-2} \tilde{\alpha }_{3 e} ( \beta _{p-2}^{\prime } / \nu ^{\frac{p+1}{2}} ) \left( 3e\left( {\begin{array}{c}p-2\\ e\end{array}}\right) -\left( {\begin{array}{c}p-2\\ e+1\end{array}}\right) \right) (-1)^{2 p - 4} , \end{aligned}

which is not zero. Recall from the proof of (1) that $$\tilde{\alpha }_{p-2}$$ and $$\tilde{\alpha }_{3 e}$$ are non-zero constants in k. Recall also from the beginning of this section that $$\beta _{p-2}^{\prime }$$ is a monomial in $$K[\nu ]$$ of degree $$3 e - (p-2) = (p+1)/2$$, and thus $$( \beta _{p-2}^{\prime } / \nu ^{\frac{p+1}{2}} ) \in K^\times$$. Thus, the assertion of (4) holds. $$\square$$

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

\begin{aligned} h_0:= & {} (ad-bc)/(\mu \nu )^{(p+1)/2},\\ h_1:= & {} ab^{p-1}+d^p,\\ h_2:= & {} a^p+c^{p-1}d, \end{aligned}

which belong to R. Since abc 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. We have the stratification

\begin{aligned} V(h_0,h_1,h_2) \subset V(h_0) \subset \mathbf{P }^2 \end{aligned}

as mentioned in Sect.3. Our aim is to prove that any point $$(\lambda : \mu : \nu )$$ of $$V(h_0,h_1,h_2)$$ is of Howe type. In particular, we need to prove that $$\mu$$ and $$\nu$$ are not zero. First we show:

### Lemma 5.4

Assume $$p>3$$. For any point $$(\lambda _0:\mu _0:\nu _0)$$ on $$V(h_0)$$ in $$\mathbf{P }^2$$, we have $$\mu _0\ne 0$$ or $$\nu _0\ne 0$$.

### Proof

It suffices to show that $$\mu _0=\nu _0=0$$ implies $$\lambda _0=0$$ for $$(\lambda _0: \mu _0: \nu _0)\in V(h_0)$$. This immediately follows from Lemma 5.3 (4). $$\square$$

Thanks to Lemma 5.4, it suffices to consider the set of points with $$\mu \ne 0$$ or $$\nu \ne 0$$. We study only the case of $$\nu \ne 0$$, since a similar argument works also for $$\mu \ne 0$$ (exchange the roles of $$\mu$$ and $$\nu$$ by the coordinate change $$X=x-\lambda$$ as in (40) and (41)). From now on, we substitute 1 for $$\nu$$ and consider abcd as polynomials in $$\lambda , \mu$$, i.e., we work over the ring

\begin{aligned} S:=K[\lambda ,\mu ]. \end{aligned}

Set $$a'=a/\mu ^{(p+1)/2}$$ and $$c'=c/\mu ^{(p+1)/2}$$, which belong to S by Lemma 5.3 (1) and (2). Let $$a'_0,c'_0, b_0,d_0$$ be the constant terms of $$a',c', b,d$$ as polynomials in $$\mu$$, which are polynomials in $$\lambda$$.

### Lemma 5.5

As polynomials in $$\lambda$$, we have

1. (1)

$$c'_0$$ and $$d_0$$ are coprime.

2. (2)

$$b_0$$ and $$d_0$$ are coprime.

### Proof

Since $$\beta _k$$ is the $$x^k$$-coefficient of “the Maclaurin series expansion” of $$f_2(x)^{e} = F_2(x - \lambda )^{e}$$, we have

\begin{aligned} \beta _{k}= & {} \frac{1}{k !} \left( F_2^{e} \right) ^{(k)} ( - \lambda ) \end{aligned}
(42)

if $$k < p$$. We use the fact that $${\tilde{\alpha }}_k := \alpha _k/\mu ^{3e-k}$$ belongs to K for $$0\le k \le 3e$$ with $$\tilde{\alpha }_{p-2}\ne 0$$ and $${\tilde{\alpha }}_{3e}\ne 0$$, as seen in the proof of Lemma 5.3 (1).

(1) By (37) and (39), we have

\begin{aligned} c'_0={\tilde{\alpha }}_{p-2}\beta _0\quad \text {and}\quad d_0={\tilde{\alpha }}_{3e}\beta _e. \end{aligned}

As $${\tilde{\alpha }}_{p-2}$$ and $${\tilde{\alpha }}_{3e}$$ are non-zero constants, it suffices to see that $$\beta _0$$ and $$\beta _e$$ are coprime as polynomials in $$\lambda$$. Since

\begin{aligned} \beta _0 = F_2^{e}(-\lambda ),\qquad \beta _e = \frac{1}{e!}(F_2^{e})^{(e)}(-\lambda ) \end{aligned}

and $$F_2$$ is a separable polynomial, $$\beta _0$$ and $$\beta _e$$ are coprime by the next claim. For any separable $$Q(\lambda )\in K[\lambda ]$$ and for any integer n with $$0\le n <p$$, we claim that $$Q(\lambda )^n$$ and $$(Q(\lambda )^n)^{(n)}$$ are coprime. Indeed, let r be any root of $$Q(\lambda )$$ and write $$Q(\lambda )=(\lambda -r)Q_0(\lambda )$$ with $$Q_0(r)\ne 0$$. By the generalized Leibniz rule, we have

\begin{aligned} (Q(\lambda )^n)^{(n)} = \sum _{i=0}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) \frac{n!}{i!}(\lambda -r)^{i}(Q_0(\lambda )^n)^{(i)}. \end{aligned}

Hence $$(Q(\lambda )^n)^{(n)}|_{\lambda =r}=n!Q_0(r)^n \ne 0$$.

(2) By (38) and (39), we have

\begin{aligned} b_0 = {\tilde{\alpha }}_{3e}\beta _{e+1}\quad \text {and}\quad d_0={\tilde{\alpha }}_{3e}\beta _e. \end{aligned}

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

\begin{aligned} \beta _e = \frac{1}{e!}(F_2^{e})^{(e)}(-\lambda ),\qquad \beta _{e+1} = \frac{1}{(e+1)!}(F_2^{e})^{(e+1)}(-\lambda ), \end{aligned}

since $$(F_2^{e})^{(e)}(-\lambda )$$ is separable by Corollary 4.7. $$\square$$

Now let us show that if $$(\lambda _0,\mu _0,\nu _0)$$ satisfies $$h_i=0$$ for $$i=0,1,2$$, then $$(\lambda _0,\mu _0,\nu _0)$$ is of Howe type. This is a key ingredient of this paper, as we explained in Sect. 3.

### Proposition 5.6

Suppose that $$p>3$$. Then any point $$(\lambda _0:\mu _0:\nu _0)$$ on $$V(h_0,h_1,h_2)$$ is of Howe type.

### Proof

Let $$(\lambda _0:\mu _0:\nu _0)$$ be a point of $$V(h_0,h_1,h_2)$$. By Lemma 5.4, we have $$\mu _0\ne 0$$ or $$\nu _0 \ne 0$$. Consider the case of $$\nu _0\ne 0$$. We study the open affine subscheme of $$V(h_0,h_1,h_2)$$ defined by $$\nu \ne 0$$. Substituting 1 for $$\nu$$, it is isomorphic to $${\text {Spec}}S/J$$, where $$S=K[\lambda ,\mu ]$$ and J is the ideal $$\langle h_0(\lambda ,\mu ,1), h_1(\lambda ,\mu ,1), h_2(\lambda ,\mu ,1) \rangle _S$$ of S. On S/J, we have

\begin{aligned} a'_0 d_0 - b_0 c'_0\equiv & {} h_0(\lambda ,\mu ,1)=0 \quad (\mathrm{mod}\ \mu ),\\ d_0^p\equiv & {} h_1(\lambda ,\mu ,1)=0 \quad (\mathrm{mod}\ \mu ). \end{aligned}

By Lemma 5.5, $$a'_0 d_0 - b_0 c'_0$$ and $$d_0^p$$ are coprime as polynomials in $$\lambda$$, and therefore there exist polynomials uv in $$\lambda$$ such that $$u(a'_0 d_0 - b_0 c'_0)+v d_0^p = 1$$. Hence we have $$1\equiv 0 \quad (\mathrm{mod}\ \mu )$$. This means that $$\mu$$ is unit in S/J. Thus we have shown that $$\nu _0\ne 0$$ implies $$\mu _0\ne 0$$ for any point $$(\lambda _0:\mu _0:\nu _0)$$ on $$V(h_0,h_1,h_2)$$. A similar argument shows that $$\mu _0\ne 0$$ implies that $$\nu _0\ne 0$$ (exchange the roles of $$\mu$$ and $$\nu$$ by the coordinate change $$X=x-\lambda$$ as in (40) and (41)). Hence we conclude that both of $$\mu _0$$ and $$\nu _0$$ are not zero.

It remains to show that $$f_1(x)$$ and $$f_2(x)$$ defined in (3) and (4) for $$(\lambda _0,\mu _0, \nu _0)$$ are coprime. As $$\mu _0$$ and $$\nu _0$$ are not zero, $$f_1(x)$$ and $$f_2(x)$$ are separable polynomials. Suppose that $$f_1(x)$$ and $$f_2(x)$$ were not coprime. After taking a linear coordinate change, they are written as $$f_1(x) = x(x-1)(x-t_1)$$ and $$f_2(x) = x(x-t_2)(x-t_3)$$. Then

\begin{aligned} \left( f_1(x)f_2(x)\right) ^e = x^{p-1}\left( (x-1)(x-t_1)(x-t_2)(x-t_3)\right) ^e. \end{aligned}

Then the Cartier-Manin matrix $$M=\begin{pmatrix}a &{}\quad b\\ c &{}\quad d\end{pmatrix}$$ of the curve $$y^2=f_1(x)f_2(x)$$ becomes a upper triangular matrix (i.e., $$c=0$$) with $$a = (t_1t_2t_3)^e$$. Since $$f_1(x)$$ and $$f_2(x)$$ are separable, we have $$t_i\ne 0$$ for $$i=1,2,3$$ and therefore $$h_2=a^p+c^{p-1}d \ne 0$$. This is a contradiction. $$\square$$

Finally we show that $$V(h_0,h_1,h_2)$$ is not empty. The fact is reminiscent of the quasi-affineness (cf. [18, (6.5). Theorem]) of Ekedahl-Oort strata in the case of the moduli space of principally polarized abelian varieties.

### Proposition 5.7

Assume $$p>3$$. Then $$V(h_0,h_1,h_2)$$ is non-empty and finite.

### 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 (ad) can not be (0, 0) on X, since $$(a,d)=(0,0)$$ implies $$h_1=h_2=0$$. Consider the morphism

\begin{aligned} \varphi : X \longrightarrow \mathbf{P }^1\smallsetminus \{(0:1)\}\simeq \mathbf{A }^1 \end{aligned}

sending $$(\lambda :\mu :\nu )$$ to $$(a^p+c^{p-1}d:c^{p-1}d)$$ for the part of $$a\ne 0$$ and $$(ab^{p-1}+d^p:d^p)$$ for the part of $$d\ne 0$$, which is well-defined, i.e., $$(a^p+c^{p-1}d,c^{p-1}d)\ne (0,0)$$ for $$a\ne 0$$ and $$(ab^{p-1}+d^p,d^p)\ne (0,0)$$ for $$d\ne 0$$ with the equality $$(a^p+c^{p-1}d:c^{p-1}d)=(ab^{p-1}+d^p:d^p)$$ for $$a\ne 0$$ and $$d\ne 0$$ deduced from (16). Let Y be the scheme-theoretic image of $$\varphi$$ ( [8, Chap. II, Exercise 3.11, (d)]). Note that Y is affine, as it is a closed subscheme of $$\mathbf{A }^1$$. Since the scheme-theoretic image of a proper scheme is proper (cf. [8, Chap. II, Exercise 4.4]), Y has to be finite. But, this is absurd, since the image of $$\varphi$$ is not finite by the following two facts. Firstly X is connected, since it is a hypersurface in $$\mathbf{P }^2$$ (cf. [8, Chap. III, Cor. 7.9]). Secondly the image of $$\varphi$$ contains two distinct points. Indeed, (1 : 1) (resp. (1 : 0)) is in the image of $$\varphi$$, since there exists a point of $$\mathbf{P }^2$$ where $$h_0=0$$ and $$a = 0$$ (resp. $$h_0=0$$ and $$d = 0$$), by [8, Chap. I, Theorem 7.2].

In the proof of Proposition 5.6, we have seen that $$V(h_0,h_1,h_2)$$ is contained in $$\mathbf{P }^2\smallsetminus V(\mu ) \simeq \mathbf{A }^2$$, whence $$V(h_0,h_1,h_2)$$ is affine. Thus, $$V(h_0,h_1,h_2)$$ is finite, as it is projective. $$\square$$

Finally, we look at the case of $$p=3$$.

### Remark 5.8

Assume $$p=3$$. There exists a unique supersingular elliptic curve $$y^2=x^3-x$$. We use the notation in Sect. 2. Remark that we assumed $$p>3$$ in Sect. 2 only to assure that any elliptic curve can be written as $$y^2=x^3+Ax+B$$ and that all arguments in Sect. 2 work for $$A_1=A_2=-1$$ and $$B_1=B_2=0$$ with $$p=3$$. It is straightforward to see that the (1, 1)-entry of $$MM^\sigma$$ is equal to $$(\mu \nu )^8$$. Hence the supersingularity of C associated to $$(\lambda ,\mu ,\nu )$$ implies that $$\mu$$ or $$\nu$$ is zero, whence $$(\lambda ,\mu ,\nu )$$ is not of Howe type. Thus, we conclude that there is no supersingular Howe curve for $$p=3$$.

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## Acknowledgements

The authors thank Everett Howe, Tomoyoshi Ibukiyama and Rachel Pries for their valuable helpful comments to earlier versions of this paper. The authors are also grateful to the anonymous referees for their careful reading of this article and their comments. This work was supported by JSPS Grant-in-Aid for Research Activity Start-up Grant Nos. 18H05836, 19K21026, 20K14301 and JSPS Grant-in- Aid for Scientific Research (C) 17K05196.

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