Finiteness theorems for abelian varieties
Posted by Martin Orr on Monday, 19 September 2011 at 16:34
Faltings famously proved the Mordell, Shafarevich and Tate conjectures in 1983. In this post I will discuss the relationships between the Tate and Shafarevich conjectures and some other finiteness theorems for abelian varieties.
Everything which I call a conjecture in this post is known to be true: they all follow from Finiteness Theorem I. Proving Finiteness Theorem I was the bulk of Faltings' work, but I am not going to talk about that today.
Finiteness Theorem I. Given a number field
and an abelian variety
defined over
, there are only finitely many isomorphism classes of abelian varieties defined over
and isogenous to
.
Overview
Here is a diagram of the standard route by which these conjectures are proved (a little simpler than the path taken by Faltings).
At the centre of these results is the Shafarevich conjecture for abelian varieties.
Shafarevich conjecture for abelian varieties. Given a number field
, a finite set
of primes of
, and an integer
, there are only finitely many isomorphism classes of abelian varieties of dimension
, defined over
and having good reduction outside
.
The Shafarevich conjecture for curves (Shafarevich's original conjecture) is the same thing, but with "abelian varieties of dimension " replaced by "nonsingular complete curves of genus
", with the condition that if
then we only consider curves having a
rational point (elliptic curves).
The Mordell conjecture, that a curve of genus
over a number field
has finitely many
points, is outside the scope of this post.
The Shafarevich conjecture for abelian varieties is equivalent to the following pair of theorems.
Finiteness Theorem I. Given a number field
and an abelian variety
defined over
, there are only finitely many isomorphism classes of abelian varieties defined over
and isogenous to
.
Finiteness Theorem II. Given a number field
, a finite set
of primes of
, and an integer
, there are only finitely many isogeny classes of abelian varieties of dimension
, defined over
and having good reduction outside
.
The surprising thing, which I will sketch in a little more detail below, is that Finiteness Theorem I implies Finiteness Theorem II, passing through the Tate conjecture and semisimplicity theorem.
The semisimplicity theorem and the Tate conjecture
I will now sketch the proof that Finiteness Theorem I implies the semisimplicity theorem and the Tate conjecture.
Semisimplicity theorem. Let
be an abelian variety over a number field
. Then
is semisimple as a representation of
(that is, it is a direct sum of irreducible representations).
Tate conjecture. Let
be an abelian variety over a number field
. Then
.
Any endomorphism of induces an endomorphism of
, so there is a homomorphism of
modules
.
It is injective because an endomorphism of
which vanishes at all
torsion points, for all
, must be
.
So Tate's conjecture is really the claim that this map is surjective.
To prove semisimplicity, we need to show that if and
is a Galoisinvariant subspace of
, then there exists an idempotent
such that
. Then
splits as the direct sum
.
The key step is the following lemma:
Lemma. Let
be a Galoisinvariant subspace. Then there exists
such that
.
In this lemma, lies not just in
but in the apparently smaller
.
This is important because:

We know that
is a product of matrix rings of division algebras. This allows us to upgrade our
to an idempotent
satisfying
, as required to prove semisimplicity.

The Tate conjecture is all about showing that things lie in
.
To prove the lemma, we construct abelian varieties and isogenies
(defined over
) such that
By Finiteness I, the belong to only finitely many isomorphism classes so there is an infinite set
such that all the
(for
) are isomorphic.
We use the
(
) to construct elements
such that some subsequence converges (in the
adic topology) to an element
which has the desired property
.
Once the lemma is proved, the semisimplicity theorem follows as in point 1 above.
To prove the Tate conjecture, let .
We use the double centraliser theorem from algebra and apply the lemma to the graph
of
, a Galoisinvariant subspace of
.
Finiteness Theorem II
Finally I will sketch the proof that semisimplicity and the Tate conjecture imply Finiteness Theorem II.
The Tate conjecture has some important corollaries:
by applying it to , we deduce that if
and
are two abelian varieties over
,
then
.
We further deduce that if
and
are two abelian varieties such that
as Galois representations, then
and
are isogenous.
Hence to prove Finiteness Theorem II it suffices to show that the adic Tate modules of abelian varieties of dimension
over
, with good reduction outside
, fall into finitely many isomorphism classes.
The idea here is that a semisimple representation of a group
is determined by the traces of
for all
.
In the case of a Galois representation, Frobenius elements are dense in
, so it suffices to use traces of Frobenius elements.
A clever argument shows that we only need to look at finitely many Frobenius elements.
Precisely, given and
and an integer
,
there is a finite set
of primes of
such that a semisimple
representation of
of dimension
unramified outside
is determined by the traces of Frobenius elements of primes in
.
This argument ultimately relies on Hermite's theorem that there are finitely many extensions of
of bounded degree unramified outside
.
The Weil conjectures imply that for each prime , the trace of the Frobenius at
can take only finitely many values.
So we are done.
Great Post Martin! I especially like that diagram, it looks like it took you a long time to TeX/HTML.
The proof of your lemma is rather sketchy. How does one construct the
and
satisfying that property? And how about using
to construct
satisfying the subsequence property? Neither of these two steps is clear to me.
I was also confused by your "To prove the Tate conjecture" paragraph. Applying the lemma to the graph of
furnishes a
such that
But where does the
come from, whose graph on
is the same as the graph of
?
How do you upgrade your
to an idempotent
?
Because this is an important point: how does
as Galois modules imply that
and
are isogenous?
In your 'Clever Argument' paragraph, does Chebotarev's theorem come in? Does Weil Restriction of Scalars come in?
I made the diagram in LibreOffice Draw. I agree that my proofs are rather sketchy, as I thought it would be too long if I wrote more. For this morning I will just deal with your question 3.
We use that every right ideal of a product of matrix rings of division algebras is generated by an idempotent. This is because any right ideal of a matrix ring of a division algebra consists of those matrices which are zero in some specified rows and anything in the other rows. Such an ideal is generated by the matrix with 1s in the appropriate places on the diagonal and zeroes elsewhere. Apply this to the right ideal
Now I will answer question 1. Let
.
(In the original text of the blog, I made a typo in the statement of the lemma, putting
instead of
. This is now fixed.)
For the construction of
and
, we use the following general lemma.
Then to construct
, let
be the smallest element of
. For each
, we can choose an isomorphism
Then
is an isogeny
satisfying
.
To get an element of
, we compose this with
. Of course
is not usually invertible as an isogeny, but it has an inverse in
. We set
and this satisfies
.
In particular
so
. Now
is compact (it is a finite rank
module) so some subsequence of
converges, say to
.
Taking the limit of
gives
, and multiplying by
gives
.
For your question 2: in proving the Tate conjecture, we do not actually construct a
inducing
. Instead we use the double centraliser theorem. I tend to remember the theorem like this, though this is not the most general possible statement:
We apply this theorem with
and
. By the theorem, it will suffice to show that
, i.e. that all
commute with all
.
Note that
. Since
,
commutes with all of
. In particular, it commutes with the
such that
. Hence
.
So for any
,
. But
contains only one element of the form
, so this implies that
. So
commutes with
as required.
Your question 4 is indeed important, and it looks like it should be quite easy. You can deduce
from the Tate conjecture applied to
. So an isomorphism
comes from something in
. The only problem is the tensoring with
.
The injections in
form an open set, and by hypothesis this set is nonempty. By the Tate conjecture,
is dense in
. Hence there is an element
which induces an injection
.
By multiplying by an integer, we may assume that
. Now the injectivity of
implies that
has finite kernel, and the surjectivity of
implies that
is surjective.
And for question 5, yes the clever argument uses the Chebotarev density theorem. But I think that the clever bit is the use of Nakayama’s lemma to allow you to look at the mod
representation, which is finite, instead of the
adic representation.
More precisely: let
be a semisimple
representation of
of dimension
, with
being the
module on which it acts. Let
be the sub
module of
generated by the image of
.
Let
be the representation obtained by reducing
mod
. The image of
is finite, so we can take a finite set
which hits all of the image of
. Then
generates
as a
module, so by Nakayama's lemma
generates
as a
module.
However the set
may vary from one representation to another. This is not a problem because the representation
factors through
for some field
of degree at most
over
, and by Hermite's theorem there are only finitely many such fields unramified outside
. Let
be the composite of all these fields, and let
be a finite subset of
which surjects onto
. (Using Chebotarev we can find primes whose Frobenii give us
.)
Then for any
,
is a
linear combination of
. This is still not enough to prove the theorem, because the coefficients of this combination may depend on
.
To solve this we work with two representations
and
simultaneously, and apply the above argument to their direct sum
. Let
be the sub
module of
generated by the image of
. The mod
representation
factors through
so by the previous argument,
generates
as a
module. This implies that for every
, we can write
as a linear combination of
and
as a linear combination of
with the same coefficients. So if the traces of
and
agree on
, then they agree everywhere, as required.
Thanks a lot for those clarifications Martin!
Regarding your first rebuttal, about my question 3, your argument gives us an
such that
, and I couldn't see why it had to be equality, until I realised that this follows from the existence of the
in the right ideal with
.
Also, there is a very small typo in Comment number 3, where you construct
; it should read
``
is an isogeny such that."