My question is regarding Elkies' paper on "The existence of infinitely many supersingular primes for every elliptic curve over $\mathbb{Q}$".
In the section "Nuts and Bolts", Elkies has the following proposition:
Proposition. Modulo $\ell$, $P_\ell(X)$ and $P_{4\ell}(X)$ factor into $(X-12^3)R(X)^2$ and $(X-12^3)S(X)^2$ for some polynomials $R(X)$ and $S(X)$ respectively.
Here $P_D(X)$ refers to the Hilbert class (or ring class) polynomial for the imaginary quadratic order $O_D$ of discriminant $-D$, and $\ell$ is a prime congruent to $-1$ mod $4$. He goes on to prove the following lemmas:
Lemma 1. $P_\ell(12^3) \equiv P_{4\ell}(12^3) \equiv 0$ mod $\ell$.
The proof of Lemma 1 is easy to understand. Next, he says that the proofs of the $P_{\ell}$ part and the $P_{4\ell}$ part of the proposition proceed in the same way (by proving Lemma 2), and he does it for $P_\ell$. However, it is unclear to me if Lemma 2 can be proven similarly for $P_{4\ell}$, and I will explain it below.
Lemma 2. Let $D = \ell$ or $4\ell$. If $x_0$ is any root of $P_D(X)$, then there exists a unique prime $\lambda_0$ lying above $\ell$ in the splitting field $K_D$ of $P_D$ such that $x_0 \equiv 12^3$ mod $\lambda_0$.
Proof of Lemma 2. The existence claim in Lemma 2 follows from Lemma 1. For uniqueness, he assumes for a contradiction that there exists another prime $\lambda_1$ lying above $\ell$ such that $x_0 \equiv 12^3$ mod $\lambda_1$. Since some $\sigma \in Gal(K_D/\mathbb{Q})$ carries $\lambda_1$ to $\lambda_0$, we obtain another root $x_1 = \sigma(x_0) \neq x_0$ of $P_D$ such that $x_1 \equiv 12^3$ mod $\lambda_0$ (as well).
If $E_0$ and $E_1$ are elliptic curves of $j$-invariants $x_0$ and $x_1$, then $E_0$ and $E_1$ both reduce (mod $\lambda_0$) to elliptic curves which are isomorphic to the reduction of \begin{equation} \mathscr{E} \colon Y^2=X^3-X \end{equation} modulo $\ell$, which we shall denote by $\mathscr{E}_\ell$.
Some facts about $\mathscr{E}_\ell$:
- $\mathscr{E}_\ell$ is supersingular, whose Frobenius $\ell^{\text{th}}$-power isogeny $F$ satisfies $F^2 = [-\ell]$. Since $ker(1+F) \supseteq \ker[2]$, $\mathscr{E}_\ell$ has an endomorphism $\frac{1+F}{2}$.
- $\mathscr{E}$ has CM by $\sqrt{-1}$, given by $(x,y) \mapsto (-x,iy)$. We shall denote the reduction of this isogeny modulo $\ell$ by $I$.
- $(IF)^2 = [-\ell]$.
- $End(\mathscr{E}_\ell) = \mathbb{Z}[I,\frac{1+F}{2}] = \mathbb{Z} \oplus \mathbb{Z}I \oplus \mathbb{Z}\frac{1+F}{2} \oplus \mathbb{Z}\frac{I+IF}{2}$.
Therefore, we get a degree-preserving embedding \begin{equation} \iota \colon Hom(E_0,E_1) \hookrightarrow End(\mathscr{E}_\ell). \end{equation}
Where the proof diverges for the cases $D=\ell$ and $D=4\ell$:
For $D=\ell$: Elkies says that it can be shown that there exists a $\mathbb{Z}$-basis $\lbrace \psi_1,\psi_2 \rbrace$ for $Hom(E_0,E_1)$ such that $\deg(\psi_i) <\frac{1+\ell}{4}$ for each $i$ (this is okay for me --- I've managed to obtain a sharper upper bound of $\deg(\psi_i) \leq \frac{\ell}{6}$). For each $i$, let $\theta_i = \iota(\psi_i)$, and Fact #4 above allows us to write $\theta_i = a+bI+c\frac{1+F}{2}+d\frac{I+IF}{2}$ for some $a,b,c,d \in \mathbb{Z}$. Then \begin{equation} \frac{\ell}{6} \geq \deg(\theta_i) = (a+bI+c\frac{1+F}{2}+d\frac{I+IF}{2})(a-bI+c\frac{1-F}{2}-d\frac{I+IF}{2}) = (a+\frac{c}{2})^2 + (b+\frac{d}{2})^2 + (c^2+d^2)\frac{\ell}{4} \geq (c^2+d^2)\frac{\ell}{4}, \end{equation} which forces $c=d=0$ (Elkies' upper bound works as well). Thus, for each $i=1,2$, $\theta_i \in \mathbb{Z}[I]$. In other words, $\iota$ is the following embedding \begin{equation} \iota \colon Hom(E_0,E_1) \hookrightarrow \mathbb{Z}[I] \subseteq End(\mathscr{E}_\ell). \end{equation} It can be shown that the image of $\iota$ is a rank $2$ lattice whose period parallelogram has Lebesgue area $\frac{\sqrt{\ell}}{2}$, but all the sublattices of $\mathbb{Z}[I]$ have unit parallelograms of integral area --- which culminates in a contradiction.
For $D=4\ell$: Elkies doesn't do this case in his paper, but I've managed to show that there exists a $\mathbb{Z}$-basis $\lbrace \psi_1,\psi_2 \rbrace$ for $Hom(E_0,E_1)$ such that $\deg(\psi_i) \leq \frac{4\ell}{6} = \frac{2\ell}{3}$ for each $i$. Once again, for each $i$, let $\theta_i = \iota(\psi_i)$, and write $\theta_i = a+bI+c\frac{1+F}{2}+d\frac{I+IF}{2}$ for some $a,b,c,d \in \mathbb{Z}$. By the same computation, we get \begin{equation} \frac{2\ell}{3} \geq \deg(\theta_i) = (a+\frac{c}{2})^2 + (b+\frac{d}{2})^2 + (c^2+d^2)\frac{\ell}{4} \geq (c^2+d^2)\frac{\ell}{4}, \end{equation} which implies $c^2+d^2 \leq \frac{8}{3}$. This is where the proof seems to fall apart for $D=4\ell$, since $c^2+d^2 \leq \frac{8}{3}$ does not imply $c=d=0$ --- and hence does not produce a contradiction. (Note that there are clearly solutions $(a,b,c,d)$ such that $c \neq 0$ or $d \neq 0$, e.g. $(0,0,1,1)$.)
What I have tried so far: (1) I don't believe that I can improve my upper bound of $\frac{2\ell}{3}$ any further (if this is possible, please enlighten me). (2) I tried to change the $\mathbb{Z}$-basis $\lbrace 1,I,\frac{1+F}{2},\frac{I+IF}{2} \rbrace$ for $End(\mathscr{E}_\ell)$ and (possibly) get some quadratic form which works, but all of them (so far) arrive at the same inequality $c^2+d^2 \leq \frac{8}{3}$ (i.e. no contradiction).
I'm aware that Elkies has proven a more general version of the proposition (at the start) in his PhD thesis, but I feel like that proof is out of reach for me at the moment. Therefore, I'm hoping that someone who has worked out this proof for the specific case $D=4\ell$ can enlighten me on this issue. Thank you.
