Assume \(p>3\). Let K be an algebraically closed field of characteristic p. We use the same notation as in Sect. 2, i.e., \(A_1,B_1,A_2,B_2\), \(\lambda , \mu , \nu \), \(f_1, f_2, f\), a, b, c, d 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 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.
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)
the highest term in \(\lambda \) of \(\beta _0\) is \((-\lambda )^{3 e}\),
-
(2)
the highest term in \(\lambda \) of \(\beta _1\) is \(3e(-\lambda )^{3 e - 1}\),
-
(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)
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)
\(\mathrm{ord}_\mu (a) = \frac{p+1}{2}\).
-
(2)
\(\mathrm{ord}_\mu (c) = \frac{p+1}{2}\).
-
(3)
\(\mathrm{ord}_\mu (ad-bc)=\mathrm{ord}_\nu (ad-bc)=\frac{p+1}{2}\).
-
(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 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. 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 a, b, c, d 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)
\(c'_0\) and \(d_0\) are coprime.
-
(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 u, v 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 (a, d) 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 \(p=3\).