fact F , R ) always consists of a single element when it is non-empty, even when
R is not noetherian. The key point is that
= 0 for some n 1, so we can
apply: Lemma. Let R be a ring in which a prime p is nilpotent and let J R
be an ideal such that Jn = 0 with n 1. Let f : X Y be a homomorphism
between p-divisible groups over R. If f vanishes modulo J then f = 0.
Proof. The case n = 1 is trivial, and by induction on n we may assume n = 2.
Our problem is comparing two R-homomorphisms with the same reduction mod-
ulo J. Equipping J with trivial divided powers, this problem will be solved used
Grothendieck–Messing theory. Consider the Lie algebras of the universal vec-
tor extensions E(X) and E(Y ), equipped with their respective Hodge subbundles
Lie(Xt)∨ and Lie(Y t)∨. For any R-homomorphism u : X Y , Grothendieck–
Messing theory shows that the map Lie(E(u)) respecting Hodge subbundles uniquely
determines u and only depends on u0 := u mod J. Hence, if u0 = 0 then u = 0.
We now apply Proposition to the functor F defined above whose value
on any , R ) InfΛ is empty or a singleton. The established part (1) of Theorem (with varying Λ) implies the pro-representability hypothesis in Proposition Thus, we just have to verify conditions (i) and (ii) in Proposition
Condition (ii) is immediate via Proposition To establish (i) for , R ) =

, Ri), we just have to show that if F , R ) is non-empty then so is F , Ri)
for some i. We may rename Λ as Λ, R as R, and Ri as Ri for simplicity of notation.
Our problem is to show that if f0 lifts to an R-homomorphism f : XR YR then
f descends to an Ri-homomorphism XRi YRi for some large i. We shall induct
on the integer n 1 such that the maximal ideal m of R and maximal ideal mi of
every Ri have vanishing nth power, the case n = 1 being trivial.
Let R = R/mn−1 and Ri = Ri/mi
so lim

Ri = R. We may assume n 2
and (by induction) that the lift f := fR of f0 descends to a (necessarily unique) lift
: XRi0 YRi0 of f0. For i i0 let f
= f
⊗Ri0 Ri. For i i0, if there is a
lift fi : XRi YRi of f
then fi ⊗Ri R is a lift of f0 and thus coincides with f (by
Lemma Hence, it is necessary and sufficient to find i i0 so that f
over Ri.
By Grothendieck–Messing theory (see [75, IV, 2.5; V, 1.6]), for every i i0
there is a canonical map
Li : Lie(E(XRi )) Lie(E(YRi ))
that only depends on f
and has reduction Lie(E(f
)) modulo the square-zero ideal
Ri, and moreover Li respects the Hodge subbundles if and only if f
lifts to an Ri-homomorphism XRi YRi . Thus, it is necessary and sufficient to
prove that Li respects the Hodge subbundles for large i. Compatibility with base
change ensures that Li = Li ⊗Ri Ri whenever i i i0 and that Li ⊗Ri R =
Lie(E(f)). But this latter map respects the Hodge subbundles since it arises from
an R-homomorphism f lifting f. Hence, by standard limit arguments (and the
compatibility of the Hodge subbundles with respect to base change) it follows that
Li respects the Hodge subbundles for sufficiently large i. This completes the proof
of Theorem
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