1 in which p is not split. Let π = ±pζ when p is inert in Z, and let π generate the
unique prime over p in Z when p is ramified in Z, so π is a Weil
Z = Q(π). (Note that π = ±2

−1 = (1 ±

since 2 is ramified in Q(

By Corollary (also see Example, a simple abelian variety
E0 over κ with
equal to π must have endomorphism algebra Z and
dimension 1. The elliptic curve E0 is supersingular because p is not split in Z. The
isogeny class of E0 contains no member that is the scalar extension of an elliptic
curve over Fp, as otherwise π would have a square root π0 Z, which is visibly
absurd by inspection since π = ±2

The abelian surface A0 := Resκ/Fp (E0) satisfies (A0)κ E0 ×
E0p), (
so (A0)κ is
not simple. But A0 is simple, as otherwise there would be a non-zero homomorphism
E0 A0 from an elliptic curve E0 over Fp and hence (by the universal property of
Weil restriction) a non-zero homomorphism (E0)κ E0, contrary to what we just
saw concerning the isogeny class of E0. (Note that (A0)κ is isotypic, since
E0p) (
isogenous to E0 via the relative Frobenius morphism E0
Taking K/Q to be a quadratic field in which p is inert, we can lift E0 over
OK,(p) to get an elliptic curve E over K having good reduction E0 at pOK. Then
A := ResK/Q(E) is an abelian surface over Q having good reduction Resκ/Fp (E0)
at p that is simple over Fp, so (via consideration of eron models over Z(p)) A
is simple over Q. However, AK E × E where E is the twist σ∗(E) by the
non-trivial automorphism σ of K over Q, so AK is not simple.
1.6.4. Example. Pushing the end of Example 1.6.3 further over Q, we now prove
that if π = ±pζ and p −1 (mod 4) with
+ 1 = 0 (resp. p −1 (mod 3) with
+ ζ + 1 = 0) then E and E over K are not isogenous (so AK is not isotypic, in
contrast with its reduction (A0)κ). Suppose that there were an isogeny ψ : E E ,
and choose it with minimal degree. In particular, ψ is not divisible by [p]E. We
claim that ordp(deg ψ) is odd (and in particular, is positive). Suppose otherwise,
so deg ψ = mp2n with n 0 and p m. Consider the reduction ψ0 : E0
of ψ, also an isogeny with degree
In particular, ker(ψ0) is a finite subgroup
scheme of E0 with order
so its p-part has order
But E0 is supersingular,
so it has a unique subgroup scheme of each p-power order. Hence, the p-part of
ker(ψ0) is E0[pn], so ψ0 = ψ0 [pn]E0 with ψ0 : E0
E0p) (
of degree m.
Consider the composite isogeny

( (
= E0
using the Frobenius isogeny of
E0p). (
This is an endomorphism of E0 with degree
pm. Since End(E0) is an order in Z[ζ] on which the degree is computed as the
norm to Z, we get an element of Z[ζ] whose norm in Z is divisible exactly once by
p. That is impossible since p is prime in Z[ζ], and so completes the verification that
deg ψ has p-part
for some odd j.
We conclude that the finite K-subgroup N := ker(ψ) E has non-trivial p-
part, and this p-part has cyclic geometric fiber (as otherwise it would contain E[p],
contradicting that we arranged ψ to not be divisible by [p]E). By cyclicity, N[p]
is a K-subgroup of E with order p. Consider its scheme-theoretic closure G in the
eron model of E at pOK,(p). This is a finite flat group scheme over OK,(p) of
order p, and its special fiber is an order-p subgroup scheme of the supersingular
Previous Page Next Page