The Hasse invariant criterion and the Hasse polynomial $h_p(\lambda)$

Supersingularity is a coefficient test: extract $(xyz)^{p-1}$ from $f^{p-1}$, or solve $h_p(\lambda)=0$ in Legendre form.Suggest a slogan

(The coefficient criterion)

Let $\operatorname{ch} k = p > 0$ and let $E = V(f) \subseteq \PP^2$ be a smooth cubic, $f$ homogeneous of degree $3$. Then $E$ has Hasse invariant zero, that is $E$ is supersingular, if and only if \[ \text{the coefficient of } (xyz)^{p-1} \text{ in } f^{p-1} \text{ is } 0 . \]

(The Legendre family)

For $p \neq 2$ and $E \colon y^2 = x(x-1)(x-\lambda)$ with $\lambda \neq 0,1$, put $m \da \tfrac{p-1}{2}$ and \[ h_p(\lambda) \da \sum_{i=0}^{m} \binom{m}{i}^2 \lambda^i . \] Then $E$ is supersingular exactly when $h_p(\lambda) = 0$. Consequently, for each $p$ there are at most $\left\lfloor p/12 \right\rfloor + 2$ supersingular elliptic curves over $\bar k$ up to isomorphism.

(Where the criterion comes from)

Each step in the coefficient criterion is forced by the cohomological realization below.

The ideal sheaf of a plane cubic is $\OO_{\PP^2}(-3)$, so \[ 0 \to \OO_{\PP^2}(-3) \xrightarrow{\ f\ } \OO_{\PP^2} \to \OO_E \to 0 . \] Since $h^1(\OO_{\PP^2}) = h^2(\OO_{\PP^2}) = 0$, the long exact sequence gives $H^1(\OO_E) \iso H^2(\PP^2; \OO(-3))$. That group is one-dimensional with basis the Čech class $\tfrac{1}{xyz}$ — it is the unique monomial of degree $-3$ with all three exponents negative. So the line whose endomorphism is being tested has a canonical monomial basis, and the test becomes a computation with monomials.

Frobenius on $\PP^2$ sends $\tfrac{1}{xyz} \mapsto \tfrac{1}{(xyz)^p}$. Transporting along the diagram that compares $E = V(f)$ with $V(f^p)$ multiplies by $f^{p-1}$, so the image is the class of \[ \frac{f^{p-1}}{(xyz)^{p}} . \] In $H^2(\PP^2; \OO(-3))$ every monomial with a non-negative exponent is zero, so only the term of $f^{p-1}$ equal to $(xyz)^{p-1}$ survives, and it contributes that coefficient times $\tfrac{1}{xyz}$. The criterion is then the definition read off: $F^* = 0$ exactly when that coefficient vanishes.

The degree bookkeeping explains why the distinguished monomial can occur: both $f^{p-1}$ and $(xyz)^{p-1}$ have degree $3(p-1)$.

(Using the Legendre form)

Dehomogenising, the criterion for $y^2 = g(x)$ with $\deg g = 3$ becomes: supersingular exactly when the coefficient of $x^{p-1}$ in $g(x)^{(p-1)/2}$ vanishes. Expanding $\big(x(x-1)(x-\lambda)\big)^m$ and collecting gives $(-1)^m h_p(\lambda)$, so the sign is irrelevant and the vanishing is $h_p(\lambda) = 0$.

Two low-characteristic checks are:

$p = 3$: $m = 1$, $h_3(\lambda) = 1 + \lambda$, so $\lambda = -1$, and there $\lambda^2 - \lambda + 1 = 3 = 0$, giving $j = 0$ — the one supersingular curve in characteristic $3$.
$p = 5$: $m = 2$, $h_5(\lambda) = \lambda^2 + 4\lambda + 1$, whose roots lie in $\FF_{25}$ and satisfy $\lambda^2 - \lambda + 1 = -5\lambda = 0$, again giving $j = 0$.

The count follows because $h_p$ has degree $m = \tfrac{p-1}{2}$ and the map $\lambda \mapsto j$ is generically six-to-one, so the supersingular $j$-values number about $p/12$; the exact count is $\left\lfloor p/12 \right\rfloor$ adjusted by $0, 1, 1, 2$ according to $p \equiv 1, 5, 7, 11 \bmod 12$, which the stated bound covers. Supersingular curves therefore exist for every $p$ and are always finite in number; $h_p$ has simple roots, so no value of $\lambda$ is counted twice.

Given a plane cubic in characteristic $p$, how do you compute its Hasse invariant?

Write down the Hasse polynomial for the Legendre family.

Why does the criterion involve $f^{p-1}$ and the monomial $(xyz)^{p-1}$?

How many supersingular curves are there in characteristic $p$?

Dependencies

D-CRVHASSEThe Hasse invariant, and ordinary versus supersingularuses
T-IJW1KThe cohomology of $\OO_{\PP^n}(d)$uses

Backlinks

FE-CRVSSPRIMESFixing the curve and varying $p$, and the density of supersingular primesuses

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Elliptic curves