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Let $k$ be a field. We say that the isogeny theorem holds over $k$ if, for any abelian variety $A$ over $k$, there are only finitely many $k$-isomorphism classes of abelian varieties $B$ over $k$ which are $k$-isogenous to $A$.

Here are some examples of fields for which the isogeny theorem hold.

  1. $k$ is a finite field.
  2. $k$ is a number field (Faltings)

Does the isogeny theorem hold for $k$ a function field over an algebraically closed field of characteristic zero?

Does the isogeny theorem hold for $k$ a function field over a finite field?

In the number field case, by results of Serre-Tate, the Shafarevich conjecture for abelian varieties implies the isogeny theorem.

Does a similar implication hold over $\mathbf C(t)$? (Of course, the naive analogue of the Shafarevich conjecture is false, but Faltings proves a "correct" version in his paper: Arakelov's theorem for abelian varieties.)

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A related question is mathoverflow.net/questions/53014/… –  Mike Lowrey Dec 20 '12 at 20:00
    
I should also mention that this question was inspired by mathoverflow.net/questions/108543/… –  Mike Lowrey Dec 20 '12 at 20:00
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For a function field over a finite fields, Zarhin proves the isogeny theorem, if one restricts the discussion to isogenies of degree prime to $p$ (otherwise, compositions of relative Frobenius morphisms give a counterexample). See "Isogenies of abelian...", Math USSR Sbornik Vol. 24 (1974), no. 3. For a finitely generated field over a number field the theorem holds by work of Faltings, if I am not mistaken. See the book by Faltings-Wüstholz on the Mordell conjecture (I don't have it at hand). –  Damian Rössler Dec 20 '12 at 21:15
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I think I understand how to deduce the isogeny theorem for fields of finite type over $\mathbf{Q}$ by a standard specialization argument. The question still remains open for $\mathbf{C}(t)$, I believe. Do note that we have Mordell (Manin's theorem) and "Shafarevich" (Faltings-Arakelov-Parshin) for this field. –  Mike Lowrey Dec 20 '12 at 23:25
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If you want a sensible answer over $\mathbb{C}(t)$ you'll need some non-isotriviality hypothesis. –  Felipe Voloch Dec 21 '12 at 0:16

1 Answer 1

Over $\mathbb{C}(T)$ the Isogeny theorem doesn't hold, even if you throw away isotrivial components.

In his paper "Arakelov's theorem for Abelian Varieties" Faltings proves the following analogue of Shafarevich:

Let $B$ be complete smooth complex curve and $S\subset B$ a finite subset of points, and $f:X\rightarrow B\backslash{S}$ a family of Abelian varieties. Let $G$ be the fundamental group of $B\backslash{S}$. Fix a point $p\in B\backslash S$ so that we get an action of $G$ on $H_1(X_p,\mathbb{Z})$ by monodromy. Now, say $X$ satisfies (A) if $$End(X)=End_{G}(H_1(X_p,\mathbb{Z})).$$ Note that the LHS also injects into the RHS.

Faltings proves there are finitely many isomorphisms classes $X$ satisfying (A) of a fixed dimension, and he also gives an example of an 8-dimensional family $X$ which doesn't satisfy (A). Now we'll use this $X$ to give a negative answer to the Isogeny theorem:

By a standard argument (see Milne's Abelian Varieties online notes, Chapter IV, Theorem 2.5) if both $X$ and $X\times X$ have at most finitely many isogenous abelian varieties up to isomorphism, by an isogeny of degree a power of some prime $l$, then $$End(X)\otimes\mathbb{Q}_l=End_G(H_1(X_p,\mathbb{Z})\otimes\mathbb{Q}_l)$$ where the second endomorphism ring is as $\mathbb{Q}_l$ vector spaces. However, the latter is false for our $X$ by assumption, since taking group invariants commutes with base change over fields, hence either $X$ or $X\times X$ don't satisfy the isogeny theorem.

$\textbf{However},$ if $X$ is a family of abelian varieties over the generic point of $B$ which extends to a family away from $S$ (in other words, it only has bad reduction at points of $S$) then every abelian varietiy isogenous to $X$ shares this property, since good reduction is an isogeny invariant. Hence the isogeny therem is true for all $X$ satisfying (A), and this includes all $X$ of dimension at most 3 by a theorem of Deligne (see Hodge II, 4.4.13), as long as you insist $X$ has no isotrivial subvarieties.

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