(The setup)
Let \(X = V(f) \subseteq \PP^2_{/\ZZ}\) with \(f \in \ZZ[x,y,z]\) a cubic whose base change to \(\CC\) is smooth. For all but finitely many primes the reduction \(X_{(p)} \subseteq \PP^2_{/\FF_p}\) is again smooth, and one can ask for \[ B \da \ts{ p \st X_{(p)} \text{ is smooth over } \bar\FF_p \text{ and has Hasse invariant } 0 } . \] The answer depends entirely on whether \(X_{/\CC}\) has complex multiplication:
With CM by an order in \(K\): for \(p\) of good reduction, \(p \in B\) exactly when \(p\) does not split in \(K\), that is when \(p\) is inert or ramified. Half the primes split, so \(B\) has density \(\tfrac{1}{2}\). This is Deuring’s theorem.
Without CM: \(B\) has density \(0\), and \(B\) is nevertheless infinite. Both of these are theorems, the second Elkies’. A finer count is not known: the expected asymptotic \(\abs{\ts{p \in B \st p \leq x}} \sim c\,\tfrac{\sqrt x}{\log x}\) is the Lang–Trotter conjecture and is open.
(What is proved and what is not)
The three claims have different logical status and should remain separate. Deuring’s criterion is a clean equivalence and gives the density \(\tfrac12\) immediately from Chebotarev, or from Dirichlet in the cases where the splitting condition is a congruence. Density zero in the non-CM case follows from the distribution of Frobenius traces. Infinitude in the non-CM case is much harder and was open for a long time; it is a theorem only for elliptic curves over \(\QQ\). The \(\sqrt x/\log x\) growth is a conjecture and should be labelled as one.
($y^2 = x^3 - x$, worked out)
Here \(\lambda = -1\), \(j = 1728\), and the curve has CM by \(\ZZ[i]\). Apply the criterion directly: with \(m = \tfrac{p-1}{2}\) and \(g(x) = x(x^2-1)\), \[ g(x)^m = x^m (x^2-1)^m , \] so the coefficient of \(x^{p-1} = x^{2m}\) is the coefficient of \(x^m\) in \((x^2-1)^m\). Only even powers appear there, so that coefficient is \(0\) when \(m\) is odd, and \((-1)^{m/2}\binom{m}{m/2}\) when \(m\) is even — which is prime to \(p\), since \(m < p\). Hence \[ X_{(p)} \text{ supersingular} \iff m = \tfrac{p-1}{2} \text{ is odd} \iff p \equiv 3 \bmod 4 . \] And \(p \equiv 3 \bmod 4\) is exactly the condition that \(p\) stays prime in \(\ZZ[i]\), matching the general statement. By Dirichlet these have density \(\tfrac{1}{2}\) and there are infinitely many.
($y^2 = x(x-1)(x-3)$, the contrast)
Here \(\lambda = 3\) and \[ j = 2^8 \cdot \frac{(9-3+1)^3}{9 \cdot 4} = \frac{2^6 \cdot 7^3}{3^2} \notin \ZZ . \] Computing \(h_p(3) \bmod p\) for the primes of good reduction, the only supersingular prime below \(120\) is \(p = 23\). The density-zero statement follows from the non-CM theorem above, and \(j\notin\overline{\ZZ}\) rules out complex multiplication entirely.
The density \(\tfrac12\) in the CM case comes from the splitting behavior of primes in the CM field. For a CM curve the reduction inherits the CM order, and the splitting behaviour of \(p\) in \(K\) decides whether the reduced endomorphism ring stays commutative. When \(p\) splits, Frobenius generates the old imaginary quadratic order and the curve is ordinary; when \(p\) is inert or ramified it cannot, the endomorphism ring jumps to a quaternion order, and the curve is supersingular. So a single congruence condition governs every \(p\) at once.
Without CM there is no such rigid structure and the supersingular primes are governed instead by how often the Frobenius trace \(a_p\) vanishes, a much thinner condition; density zero, infinitude, and the \(\sqrt x / \log x\) heuristic all belong to that side.
Conversely, integrality of the \(j\)-invariant gives an obstruction to CM: a non-integral rational \(j\) rules out CM, and more generally a \(j\)-invariant that is not an algebraic integer cannot be a CM \(j\)-invariant.
Dependencies
T-CRVCMComplex multiplication, and $\operatorname{End}(E,p_0)$ as an order in an imaginary quadratic fieldusesT-CRVHASSEThe Hasse invariant criterion and the Hasse polynomial $h_p(\lambda)$uses