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Let $A$ be a supersingular elliptic curve over $\mathbb{Z}/p\mathbb{Z}$ and $\mathcal{O}$ an order in an imaginary quadratic field contained in the quaternion algebra $\operatorname{End}(A)$, then by Deuring's lifting lemma, $A$ lifts to an elliptic curve $E$ over $\bar{\mathbb{Q}}$ with complex multiplication by an order $\mathcal{O}'$ containing $\mathcal{O}$. For a reference, see "supersingular j-invariants as singular moduli mod-p" by M. Kaneko. My question is about whether this sort of assertion about lifting can be made quantitative in the following sense.

Let $\mathcal{S}$ be the set of all imaginary quadratic fields $K=\mathbb{Q}(\sqrt{-D})$, and $\mathcal{S}_A$ the subset of $\mathcal{S}$ consisting of all imaginary quadratic fields $K$ such that there exists an elliptic curve $E$ over $\bar{\mathbb{Q}}$ with complex multiplication by $\mathcal{O}_K$ such that $E$ is a lift of $A$. Given a real number $x>0$, set $\mathcal{S}(x)$ to consist of $K=\mathbb{Q}(\sqrt{-D})$ such that $D$ is positive, squarefree and $D\leq x$; set $\mathcal{S}_A(x):=\mathcal{S}_A\cap \mathcal{S}(x)$. The sets $\mathcal{S}(x)$ and $\mathcal{S}_A(x)$ are finite, set $\#\mathcal{S}(x)$ (resp. $\#\mathcal{S}_A(x)$) to denote the cardinality of $\mathcal{S}(x)$ (resp. $\mathcal{S}_A(x)$). Then, what can be said about the relative upper density $\limsup_{x\rightarrow \infty} \frac{\#\mathcal{S}_A(x)}{\#\mathcal{S}(x)}$? Is it positive?

There are only finitely many elliptic curves $A$ over $\mathbb{Z}/p\mathbb{Z}$, perhaps these densities are the same for all such $A$. This would be some sort of statement about the equi-distibution of CM j invariants with respect to mod-$p$ reduction. Working explicitly with quarterion algebras (as is done in loc. cit.), it seems to be difficult to derive such a statement. However, perhaps there are tools from arithmetic geometry to prove such a statement? In fact, I'm looking for asymptotic lower bounds for $\#\mathcal{S}_A(x)$ as $x\rightarrow \infty$. I'm hoping to do better than simply show that the set $\mathcal{S}_A$ is infinite.

The reason why I'm interested in this sort of statement has to do with the average behavior of the classical Iwasawa $\lambda$-invariant of an imaginary quadratic field. A recent preprint by M. Stokes https://arxiv.org/abs/2302.09594 establishes the connection between the $\lambda$-invariant of the cyclotomic $\mathbb{Z}_p$-extension of an imaginary quadratic field $K$ in which $p$ splits, and the number of points on the mod-$p$ reduction of any CM elliptic curve $E$ with complex multiplication by $\mathcal{O}_K$.

EDIT: Let $\mathcal{S}$ be the set of all imaginary quadratic fields $K=\mathbb{Q}(\sqrt{-D})$ such that $p$ is inert in $K$. The set $\mathcal{S}(x)$ consists of $K=\mathbb{Q}(\sqrt{-D})$ in $\mathcal{S}$ such that $D$ is positive squarefree and $D\leq x$. This seems to be setting in which this question is nontrivial, although it doesn't quite square with the context in which Stokes' result is proven.

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  • $\begingroup$ This doesn't exactly imply that the upper density is positive when $p\geq 5$, or does it? $\endgroup$
    – Anwesh Ray
    Commented Apr 7, 2023 at 0:12
  • $\begingroup$ Writing $K = \mathbb{Q}(\sqrt{-D})$, the prime $p$ does not split in $K$ if and only if $\left( \tfrac{-D}{p} \right) = -1$. This happens for half of all discriminants $D$ by Dirichlet's theorem on primes in arithmetic progressions. For each such $K$, the curves with CM by $\mathcal{O}_K$ are supersingular at $p$. I think that, combined with user491858's comment that all supersingular curves appear when any of them do, this means that the limit of your fraction is $\tfrac{1}{2}$. $\endgroup$ Commented Apr 7, 2023 at 2:39
  • $\begingroup$ @DavidESpeyer The paper I cited mentions a result of Elkies, which would be vacuously true if $\mathcal{S}_A$ contained all such $D$. In fact the paper goes into some detail in finding a $D$ with $D\leq 4/\sqrt{3} \sqrt{p}$. So something about the assertion following user491858's comment doesn't quite add up. In any case, I am actually interested in this question for all $A$, without the supersingular hypothesis and also the family of imaginary quadratic fields for which I would like to prove my result are those in which $p$ splits. $\endgroup$
    – Anwesh Ray
    Commented Apr 7, 2023 at 3:16
  • $\begingroup$ You're right, I am confused. Well, I am sure of the statement that $p$ doesn't split in $\mathbb{Q}(\sqrt{-D})$ for $1/2$ of $D$, and I'm pretty sure that this forces the reduction to be supersingular, so my remaining point of confusion is user491858's statement. $\endgroup$ Commented Apr 7, 2023 at 3:42
  • $\begingroup$ Thinking more directly about the isogeny claim, if $E_1$ has CM by $\mathcal{O}_K$, and $E_2$ is isogenous to $E_1$, then we can only conclude that $E_2$ has CM by an order in $\mathcal{O}_K$. So, even if there are no issues lifting the isogeny from char. $p$ to char. $0$, you'd still need to be careful about exactly what CM ring you get. $\endgroup$ Commented Apr 7, 2023 at 3:47

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Let $E$ be a supersingular elliptic curve in characteristic $p$. Then $A = \mathrm{End}(E)$ is a maximal order in the quaternion algebra $D/\mathbf{Q}$ ramified exactly at $p$ and $\infty$. If $\alpha \in A \smallsetminus \mathbf{Z}$, then Deuring's theorem guarantees that the pair $(E,\alpha)$ can be lifted to a characteristic zero, and the lift will necessarily be a CM elliptic curve. If $K$ is an imaginary quadratic field then $\mathcal{O}_K = \mathbf{Z}[\alpha]$ for some $\alpha$, so the question as to whether $E$ can be lifted to an elliptic curve with CM by the full ring of integers $\mathcal{O}_K$ is equivalent to asking that there is an inclusion

$$\mathcal{O}_K \rightarrow A.$$

A necessary condition is that $K \hookrightarrow D$. This is equivalent to the condition that $K$ splits $D$, which is equivalent to the condition that $p$ does not split in $K$. So from now on restrict to such $p$.

Consider for convenience of exposition the case when $K$ has even discriminant so one can choose $\alpha = \sqrt{-\Delta_K}$. Then $\alpha \in A$ if and only if there exists a trace zero element in $A$ with norm $\Delta_K$. The trace zero elements form a rank three module (over $\mathbf{Z}$), and so the condition that $\Delta_K$ be a norm is equivalent to asking that $\Delta_K$ is represented by the corresponding ternary quadratic form $P_A(x,y,z)$. Understanding what numbers are represented by ternary quadratic forms is a tricky problem in general. In this setting, the ternary quadratic forms $P_A$ depend on $A$. However, these forms are all locally the same for all finite places. So the easy thing to compute is how many ways an integer is represented by $P_A(x,y,z)$ for some $A$. But this is just the number of CM elliptic curves with endomorphisms by $\mathcal{O}_K$ which is the class number $h_K$. This is why things are much simpler when there is only one supersingular point which occurs for only finitely many $p$. When $p=2$, for example, then $D$ is the Hamilton quaternions and $P_A(x,y,z) = x^2 + y^2 + z^2$, and the number of ways of representing a prime in this form is directly related to class numbers, as was understood by Gauss. For larger $p$, there will be a similar formula for the number of ways of representing numbers in terms of some $P_A$, but understanding the individual factors is more complicated.

In more general settings of ternary definite quadratic forms, The best one can hope for is that an integer $\Delta_K$ is represented by such a form as long as it is locally representable and $D$ is sufficiently large. (Small $D$ are always going to be difficult.) Even this is too much to ask for general ternary quadratic forms (a good introductory reference is here https://personal.math.ubc.ca/~cass/siegel/hanke-ternary.pdf). Fortunately it turns out for this particular case that we are in the best situation, and the conclusion is indeed that, fixing $p$, lifts exist for any $K$ where $p$ does not split as long as $\Delta_K \gg_p 1$. But even more is true. We know there are $h_K \gg D^{1/2 - \varepsilon}$ CM elliptic curves by the ring of integers of $K$. It turns out that there reductions are distributed uniformly among the supersingular points as $\Delta_K \rightarrow \infty$. This follows from Theorem 3 of Michel's paper (https://annals.math.princeton.edu/wp-content/uploads/annals-v160-n1-p05.pdf)

Some Variations There are a number of other variations that could be asked. The question (somewhat artificially) considered supersingular elliptic curves over $\mathbf{F}_p$. One could insist on finding CM lifts which are themselves defined over $\mathbf{Z}_p$. To be precise, since $p$ is inert in $K$, these lifts do not actually have CM over the base field, but they are lifts over $\mathbf{Z}_p$ which have CM over some extension. For convenience let us also restrict to $K/\mathbf{Q}$ of prime discriminant. Then one can ask:

Problem: For a fixed $p$, let $K/\mathbf{Q}$ range over imaginary quadratic fields of prime discriminant. What is the distribution of which supersingular elliptic curves over $\mathbf{F}_p$ lift to a (potentially) CM elliptic curve over $\mathbf{Z}_p$?

To be more specific, we know that when $K$ has prime discriminant then $h_K$ is odd. Let $H_K$ be the class field. The decomposition group at $p$ of $\mathrm{Gal}(H_K/\mathbf{Q})$ surjects onto $\mathrm{Gal}(K/\mathbf{Q})$ because $p$ is inert in this field; since $h_K$ is odd, this implies that the decomposition group in $\mathrm{Gal}(H_K/\mathbf{Q})$ is (any of the) reflections. Since the $j$-invariant of any CM elliptic curve is fixed by exactly one reflection, it follows that there will be a unique elliptic curve over $\mathbf{Z}_p$ which has (potential) CM by $\mathcal{O}_K$. The mod-$p$ reduction of this curve then gives a unique supersingular elliptic curve over $\mathbf{F}_p$ determined by $\Delta_K$. How are these reductions distributed as $\Delta_K$ ranges over all admissible primes?

To make this even more explicit, let $p=11$, so there are two supersingular elliptic curves with $j=0$ and $j = 1728$. Let $p \equiv 3 \bmod 4$ be a prime such that $11$ is inert in $\mathbf{Q}(\sqrt{-p})$, so $p \not\equiv 2,6,7,8,10 \bmod 11$. There is a class polynomial $P(x)$ of degree $h_K$, and this will have a unique root defined over $\mathbf{Z}_{11}$. For what primes $p$ is this root $0 \bmod 11$ and for what primes is it $1728 \bmod 11$? (This question came up naturally from discussions with David Speyer in the comments). For example, with $\Delta_K = -3, -23, -31, -47$, the reductions are $j = 0,0,0,1728$. The first case is obvious, the modular equations in the next two cases are

$$t^3 + 3491750t^2 - 5151296875t + 12771880859375,$$ $$t^3 + 39491307t^2 - 58682638134t + 1566028350940383,$$

and the last case is computed in the comments). The default guess in this case is presumably that they each occur roughly half the time.

The question of understanding supersingular isogenies is something that seems to be intensively studied for cryptography, so these questions may well be known.

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  • $\begingroup$ For $A$ the quaternion algebra of endomorphisms of an elliptic curve and $\alpha \in A$ such that $p$ is inert in the field $\mathbb Q(\alpha)$, a necessary condition for $\alpha$ to lift to an endomorphism of an elliptic curve over $\mathbb Z_p$ is that $\operatorname{Frob}_p \alpha \operatorname{Frob}_p^{-1} \in \mathbb Q(\alpha)$ and thus $\operatorname{Frob}_p \alpha \operatorname{Frob}_p^{-1} = \operatorname{tr}(\alpha)-\alpha$. This forces $\alpha$ to lie in a subspace of $A$. $\endgroup$
    – Will Sawin
    Commented May 12 at 14:45
  • $\begingroup$ If we also have $\operatorname{tr}(\alpha)=0$ then $\alpha$ lies in a two-dimensional subspace, so your Problem now involves looking for primes represented by binary quadratic form. The two-dimensional subspace is locally free over $\mathbb Z[\sqrt{-p}]$ so we're looking for primes $\ell$ that split and represent a fixed ideal class (the fact that $p$ doesn't split in $\mathbb Q(\sqrt{-\ell})$ and $\ell\equiv 3 \bmod 4$ combine to give splitness of $\ell$ in $\mathbb Q(\sqrt{-p})$). The number of supersingular elliptic curves is the class number so each class comes from one curve. $\endgroup$
    – Will Sawin
    Commented May 12 at 14:52
  • $\begingroup$ So the necessary condition is sufficient and the equidistribution follows from the usual Hecke equidistribution of split primes among ideal classes. $\endgroup$
    – Will Sawin
    Commented May 12 at 14:53
  • $\begingroup$ @WillSawin nice argument! $\endgroup$
    – user491858
    Commented May 12 at 22:54

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