# Homomorphism ring of two elliptic curves with the same supersingular reduction.

$\textbf{Motivation:}$ I am studying the article "The existence of infinitely many supersingular primes for every elliptic curve over $\mathbb{Q}$", by Noam Elkies, and there is a part that I do not fully understand and my advisors have been unable to help me.

$\textbf{What I have understood:}$ Let $K$ be a quadratic imaginary field and define $\mathcal{ELL}(\mathcal{O}_{D})$ as the set of elliptic curves with complex multiplication by the order $\mathcal{O}_{D}$ of discriminant $-D$, up to isomorphism. We know that every elliptic curve in $\mathcal{ELL}(\mathcal{O}_{D})$ is determined by its $j$-invariant and all the $j$-invariants of the curves in $\mathcal{ELL}(\mathcal{O}_{D})$ are conjugated algebraic integers.

Let $l$ be a prime congruent to $3$ mod $4$. We start by considering $P_{l}$ as the minimal polynomial of the $j$-invariants of elliptic curves in $\mathcal{ELL}(\mathcal{O}_{l})$. The objective is to prove: $$P_{l}(X)\equiv(X-12^3)R^2(X)\ \text{mod}\ l$$ for some polynomial $R$.

The first step is to obtain the factor $12^3$. It is not difficult if we look at the curve $$E : y^2=x^3-x$$ over the finite field $\mathbb{F}_{l}$. The $j$-invariant is $12^3$ and it has complex multiplication by the Frobenius automorphism, which verifies $F^2=-l$. We can factorise $1+F$ by the multiplication by $2$ obtaining $\mathbb{Z}[\frac{1+F}{2}]\subset \text{End}(E)$ (it is not difficult to see that End$(E)=\mathbb{Z}[\frac{1+F}{2},I]$).

By Deuring Lifting's Theorem, there is an elliptic curve over a number field with complex multiplication by $\mathcal{O}_{l}$ and $j$-invariant equal to $x_{0}$, and there is a place $\lambda_{0}$ lying over $l$ such that $x_{0}\equiv 12^3$ mod $\lambda_{0}$. $x_{0}$ is a root of $P_{l}$ in the splitting field $K_l$ and so, $P_{l}(12^3)\equiv 0$ mod $l$.

Now, we want to prove that in fact, $\lambda_{0}$ is the only prime lying over $l$ in the splitting field such that $x_{0}\equiv 12^3$ mod $\lambda_{0}$. Assume that there is another prime $\lambda_{1}$ lying over $l$ with $x_{0}\equiv 12^3$ mod $\lambda_{l}$. Then there is an element of $Gal(K_{l}/\mathbb{Q})$ which takes $\lambda_{1}$ to $\lambda_{0}$ and $x_{0}$ to another root $x_{1}$ of $P_{l}$ such that $x_{1}\equiv 12^3$ mod $\lambda_{0}$. So there are two elliptic curves $E_{0}$ and $E_{1}$ which reduce to $E$ mod $\lambda_{0}$ with $j$-invariants $x_{0}$ and $x_{1}$ respectively. Then we can consider Hom$(E_{0},E_{1})$ and its degree-preserving embedding into End$(E)$.

$\textbf{Where it goes wrong:}$ At this point, Elkies says that Hom$(E_{0},E_{1})$ is a 2-rank lattice of $\mathcal{O}_{l}$ whose period parallelogram has area $\frac{\sqrt{-l}}{2}$ (Shouldn't it be $\frac{\sqrt{l}}{2}$? An imaginary area seems odd.) and which is spanned over $\mathbb{Z}$ by two elements of degree less than $\frac{1+l}{4}$ but from the explicit form of End$(E)$, the only such endomorphisms are in $\mathbb{Z}[I]$, all of whose sublattices have unit parallelograms of integer area and that leads us to contradiction.

$\textbf{Questions:}$ How does Elkies arrives to give an explicit description of Hom$(E_{0},E_{1})$? How does he know that all such endomorphisms must be in $\mathbb{Z}[I]$?

I would be most grateful for any help whether explanations or references.

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By choosing an embedding of $K_l$ into $\mathbb{C}$, then looking at the parameterising tori and their lattices, and after taking a suitable homothety, the lattices are ideals in $\mathcal{O}_l$, which cannot lie in the same class (in the class group) since they have different j-invariants. The Hom space is now a fractional ideal, that cannot be principal. That the ideal is spanned by two elements of reduced norm less than $\frac{l+1}{4}$, for $l\ge 7$ is an exercise. For $l=3$, there should be a sperate computation. – Dror Speiser May 25 '12 at 8:44
Finally, elements of $\mathbb{Z}[\frac{1+F}{2},I]$ which aren't in $\mathbb{Z}[I]$, quite clearly have degree at least $\frac{l+1}{4}$, the smallest element being $\frac{1+F}{2}$ with that degree. – Dror Speiser May 25 '12 at 8:47
* When $l=3$ the class number is one and the unique j-invariant is 0, so that $P_3(X)=X$, and indeed $X=X-12^3\pmod{3}$. – Dror Speiser May 25 '12 at 8:59
Thank you for the reponse. I agree with what you said, but indeed the point that is blocking me is precisely how to determine the degree of the elements of this fractional ideal. I know that this degree is going to be N$(x)$/N$(\mathfrak{a})$ for every $x\in \mathfrak{a}$ ($\mathfrak{a}$ being the Hom space.) but since I don't explicitely know $\mathfrak{a}$, I don't know how to calculate the degree of its elemtents. – MPI May 25 '12 at 9:46
This is an elementary exercise. Remember that you're not calculating the norm of arbitrary elements, but of generators. My favorite solution involves the terms quadratic forms and fundamental domain. I will point out that it is crucial the discriminant is prime. If after trying a bit more you still haven't solved this problem, I suggest you post it on math.stackexchange. – Dror Speiser May 25 '12 at 11:07