is injective and the target is free of rank 1 over L . Hence, the finitely generated
torsion-free OL-module M is invertible. Also, the injective map
Hom((A , i ), (A, i)) Hom((AKs , i ), (AKs , i))
between finitely generated OL-modules has image with finite index since it becomes
an equality after applying Q ⊗Z (·) (for L-dimension reasons); let n be the index.
It follows that all L-linear Ks-homomorphisms f : AKs AKs are Gal(Ks/K)-
invariant because nf is defined over K. This shows that the formation of M is
unaffected by ground field extension to Ks, and hence by any ground field extension
(due to Lemma
Now we assume that char(K) = 0 and seek to prove that M ⊗OL
A A is
an isomorphism. We may assume K is finitely generated, and then that K = C.
The OL-modules H1(A(C), Z) and H1(A (C), Z) are each invertible (due to being
Z-flat of rank [L : Q]). By Example 1.5.3, we get OL-linear isomorphisms A(C) =
(R ⊗Q L)Φ/a and A (C) = (R ⊗Q L)Φ/a for non-zero ideals a, a OL. Hence,
elements of M are precisely multiplication on (R ⊗Q L)Φ by those c L such that
ca a. We conclude that M = HomOL (a , a) = aa
, with M ⊗OL A A given
by the evident evaluation pairing on C-points. This is an isomorphism because the
induced map on homology lattices is the natural map M ⊗OL a a that is clearly
an isomorphism.
The isomorphism property for the map M ⊗OL A A in Example fails
away from characteristic 0, even for elliptic curves over finite fields. For example, if
L is imaginary quadratic with class number 1 then the relative Frobenius isogeny
provides counterexamples (using elliptic curves whose j-invariant is not in the prime
field). More explicitly, for p −1 mod 4 and κ := Z[i]/(p) Fp2 with p 3,
consider the elliptic curves = E = {y2 = x3 x} over κ with CM by OL
via the actions [i](x, y) = (−x, ±iy). These are not OL-linearly isomorphic (since
Aut(E) = μ4 OL
, as p 3) but the Frobenius isogeny E
is an OL-linear
isogeny E+ E−. Thus, the module M of OL-linear homomorphisms from E+ to
E− is non-zero but the OL-linear map M ⊗OL E+ E− cannot be an isomorphism
(since M OL as OL-modules). Example. Let (A, i, L) be as in Example over a field K, so OL
End(A). For an invertible OL-module M, another such abelian variety is given by
M ⊗OL A. Any m M defines an OL-linear map em : A M ⊗OL A via x m⊗x.
For = char(K), the map T (em) induced by em on -adic Tate modules is the map
T (A) M ⊗OL T (A) given by v m v, so em = 0 when m = 0. In particular,
the module HomOL (A, M ⊗OL A) of OL-linear homomorphisms is non-zero and
therefore invertible (by Example
The natural map of invertible OL-modules
eA,M : M HomOL (A, M ⊗OL A)
is injective, hence of finite index. We shall now show that this map is an isomor-
phism. It suffices to check the result after applying Z ⊗Z (·) for every prime
(allowing = char(K)). This scalar extension is the first map in the diagram
M Z ⊗Z HomOL (A, M ⊗OL A) HomOL, (A[

],M ⊗OL, A[

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