Are supersingular primes and supersingular elliptic curves related?
(this was essentially a subquestion in my earlier question, but still looks sufficiently different to me to deserve a separate post)
Are supersingular primes and supersingular elliptic curves related?
(this was essentially a subquestion in my earlier question, but still looks sufficiently different to me to deserve a separate post)
Let F be the finite field with p elements.
A supersingular elliptic curve is an elliptic curve E/F with the property that the endomorphism ring (ring of homomorphisms from E to E) of E over the algebraic closure of F_p is has rank 4 as a Z-module.
It is a theorem that End(E/F) has rank 2 or rank 4 (and that in the former case, End(E/F) is isomorphic to an order in an imaginary quadratic number ring whereas in the latter case, End(E/F) is isomorphic to an order in a quaternion algebra over Q). Hence, a supersingular elliptic curve over F_p can be thought of as an elliptic curve over F with a "big" endomorphism ring.
It is a theorem that another characterization supersingular elliptic curves is that E/F is supersingular is if and only if the number of points on E/F is exactly p + 1 (edit: for p > 3, see Voloch's comment below).
I believe that the etymology of the term "supersingular" is as follows: if you start with an elliptic curve E/Q over Q, for all but finitely many primes p, reduction (mod p) gives an elliptic curve E/F. For a generic E/Q (specifically, one without "complex multiplication") then the set of primes such that reduction (mod p) turns E/Q into a supersingular E/F has asymptotic density 0. Such primes are called "supersingular primes for E/Q" - supersingular refers to "really unusual." The reductions for such primes are then called supersingular elliptic curves over F. I'm pretty sure that every elliptic curve over F is a (mod p) reduction of an elliptic curve over Q so that all supersingular elliptic curves arise in this way.
I'll remark (following Silverman) that a supersingular elliptic curve over F is not "singular" in the sense of algebraic geometry - by definition all elliptic curves are nonsingular.
I do not know much about supersingular primes in the context of monstrous moonshine. According to Wikipedia, a supersingular prime is a prime that divides the order of the monster group; and there are 15 such primes. Given E/Q, by a theorem of Elkies there will be infinitely many super singular primes p for E/Q. So it's difficult to imagine how the list of 15 supersingular (with respect to moonshine) primes could emerge from the notion of "supersingular elliptic curve." I imagine that the etymology of the term "supersingular" in the context of moonshine is again that that supersingular primes are special - but that they special in a completely different way from the supersingular primes for a elliptic curve over Q.
There are many equivalent ways to define supersingularity for an elliptic curve over a characteristic $p$ field. One of them is that the $p$-torsion of the curve is connected, i.e., it is a purely infinitesimal group scheme of order $p^2$. As Jonah mentioned, supersingular means very special, and is not a statement about smoothness. There is a theorem of Deuring that implies the j-invariant of a supersingular elliptic curve always lies in $\mathbb{F}_{p^2}$, and as a consequence, all such curves are defined over a finite degree extension of $\mathbb{F}_p$.
There are two notions of supersingular prime: one is relative to a fixed elliptic curve over $\mathbb{Q}$, and one is absolute. For any elliptic curve $E/\mathbb{Q}$, a prime $p$ is supersingular for $E$ if $E$ has good supersingular reduction at $p$. Such primes are known to be asymptotically density zero, but infinite in number (by a theorem of Elkies). Lang and Trotter conjectured that the number of supersingular primes less than $N$ limits to a constant times $\sqrt{N}/\log(N)$ as $N$ gets large.
Supersingular primes in the absolute sense are those primes $p$ for which all supersingular elliptic curves over an algebraic closure of $\mathbb{F}_p$ have $j$-invariant in $\mathbb{F}_p$ instead of just $\mathbb{F}_{p^2}$. These happen to be the primes that divide the order of the monster simple group, and they are also the primes for which the normalizer of $\Gamma_0(p) \in SL_2(\mathbb{R})$ acts on the complex upper half plane with a genus zero quotient. For general $p$, this normalizer contains $\Gamma_0(p)$ as an index 2 subgroup, with the nontrivial coset called the "Fricke involution" (a special case of Atkin-Lehner involultion). There is a standard order 2 representative, taking $z \mapsto -1/pz$. The quotient curve classifies unordered pairs of elliptic curves with dual degree p isogenies between them. I do not know any canonical relations between these characterizations of supersingularity.
Edit: Thanks to Emerton for pointing out the connection. I'll try to expand on it a bit. The moduli problem of generalized elliptic curves with $\Gamma_0(p)$-structure has a coarse moduli space that is a smooth irreducible curve away from p, but has mod p fiber given by taking a disjoint union of two copies of $X(1)$ (a genus zero curve) and gluing along supersingular points (this description is more or less in Katz-Mazur, chapter 13). A geometric point describing an elliptic curve with j-invariant $\alpha \in \mathbb{F}_{p^2}$ is glued to a geometric point on the other irreducible component describing an elliptic curve with j-invariant $\alpha^p$. The Fricke involution switches the components, so the quotient of $X_0(p)$ by this involution is a genus zero curve glued to itself at finitely many supersingular points. The quotient has arithmetic genus zero if and only if all supersingular geometric points are glued to themselves - otherwise, the flat modular deformation to characteristic zero yields a smooth curve of higher genus. In other words, it is necessary and sufficient that all supersingular geometric points have no Frobenius conjugates, i.e., that the j-invariants of all supersingular curves lie in $\mathbb{F}_p$.
More Edit: I should give a more honest reply to Mariano's question, which was originally raised by Ogg in the mid 1970s (and he famously offered a bottle of Jack Daniels to anyone who could solve it). Half of the question has an answer. If we combine the results of Borcherds's paper Monstrous moonshine and monstrous Lie superalgebras with the results of the paper Modular equations and the genus zero property of moonshine functions by Cummins and Gannon, we get the following fact:
Let G be a finite group acting faithfully on a conformal vertex algebra V by conformal symmetries, and suppose V has central charge 24 and character $\operatorname{Tr}(q^{L_0-1}|V) = j(\tau)-744$. Then for any element $g \in G$, the series $\operatorname{Tr}(gq^{L_0-1}|V)$ is the q-expansion of a modular function that is holomorphic on the upper half plane, invariant under a discrete group $\Gamma \subset PSL_2(\mathbb{R})$ satisfying $\Gamma \supset \Gamma_0(N)$ for some $N$, and generates the function field of the quotient curve $\Gamma \backslash \mathbf{H}$. In particular, the quotient curve is genus zero.
The monster simple group arises in this context because I. Frenkel, Lepowsky, and Meurman constructed a conformal vertex algebra satisfying the above hypotheses, whose group of conformal automorphisms is the monster. The fact given above implies that for each prime p dividing the order of the monster, the quotient $X_0^+(p) = X_0(p)/\langle w_p \rangle$ is genus zero, where $w_p$ is the Fricke involution. In particular, each prime dividing the order of the monster is necessarily supersingular.
I know some people who would like there to be a conceptual (read: non-enumerative) explanation for why all supersingular primes divide the order of the monster. So far, the best I've heard is that the monster is really big, while there aren't that many supersingular primes.