1
$\begingroup$

Let $X, Y$ be abelian varieties over $k$. Let $l$ be a prime not equal to the characteristic of $k$. Then one shows that $\text{Hom}_k(X, Y)\to \text{Hom}_{\mathbb{Z}_l}(T_l X, T_l Y)$ is injective. Moreover, it is shown that the $\mathbb{Z}_l$ linear map $\text{Hom}_k(X, Y)\otimes_{\mathbb{Z}} \mathbb{Z}_l \to \text{Hom}_{\mathbb{Z}_l}(T_l X, T_l Y), f\otimes c\mapsto c\cdot T_l(f)$ is injective with torsion-free cokernel and that $T_l X$ and $T_l Y$ are non-canonically isomorphic to free $\mathbb{Z}_l$ modules of finite rank.

Then it is concluded that $\text{Hom}_k(X,Y)$ is a free abelian group of finite rank. I can not understand how that can be concluded from the above. (See for example Ben Moonen's draft for a book on abelian varieties chapter 15).

At first, this seemed rather obvious but check out the second answer to the question Checking that a torsion-free abelian group has finite rank here on math-exchange.

$\endgroup$
4
  • 1
    $\begingroup$ You could vary the prime $\ell$, but I guess that still does not rule out the possibility $\hom_k(X,Y)=\mathbb Z[1/p]$ where $char(k)=p$. $\endgroup$
    – Kenta S
    Commented Jan 20, 2023 at 16:29
  • $\begingroup$ Yes, varying $l$ doesn't work by the linked question and the second answer there $\endgroup$ Commented Jan 20, 2023 at 17:01
  • $\begingroup$ @FabioNeugebauer: "Then it is concluded that [...]" - maybe tell us where it is concluded that [...]? $\endgroup$
    – Alex M.
    Commented Jan 29, 2023 at 19:13
  • $\begingroup$ @Alex M. Its concluded in Ben Moonens book on Abelian Varieties or in Milne's 1986 text on Abelian Varieties in theorem 12.5 $\endgroup$ Commented Jan 29, 2023 at 19:23

1 Answer 1

0
$\begingroup$

Mb, I rushed through the problem and accidentally made a mistake, that your third problem highlighted. Revised answer below:

The statement that HomZl(TlX,TlY) is a free abelian group of finite rank was incorrect. Instead, HomZl(TlX,TlY) is a free Zl-module of finite rank.

Despite this, the conclusion that Homk(X,Y) is a free abelian group of finite rank still holds. This is because of the following reasons:

The homomorphism Homk(X,Y) → HomZl(TlX,TlY) being injective implies that the image of Homk(X,Y) in HomZl(TlX,TlY) is a subgroup of finite index.

The Zl-linear map Homk(X,Y) ⊗ Zl → HomZl(TlX,TlY), f ⊗ c → c ⋅ Tl(f) being injective with torsion-free cokernel implies that Homk(X,Y) is a direct summand of HomZl(TlX,TlY), which is a free Zl-module of finite rank.

As a direct summand of a free Zl-module of finite rank, Homk(X,Y) must also have finite rank as a abelian group.

Therefore, we can conclude that Homk(X,Y) is a free abelian group of finite rank.

$\endgroup$
3
  • 2
    $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Commented Jan 29, 2023 at 17:41
  • $\begingroup$ I do not understand the following: For the first fact, why has it finite index. For the second fact, I think I see that $\text{Hom}_k(X, Y)\otimes \mathbb{Z}_l$ will be a direct summand of a free $\mathbb{Z_l}$ module of finite rank, how does the rest follow? Also, I don't get why $\text{Hom}_{\mathbb{Z}_l}(T_l X, T_l Y)$ should have finite rank as abelian group, I think it has finite rank as $\mathbb{Z}_l$ module, but $\mathbb{Z}_l$ has infinite rank as abelian group? $\endgroup$ Commented Jan 29, 2023 at 18:11
  • $\begingroup$ I see that $\text{Hom}_k(X,Y)$ must have finite rank as torsion free abelian group. But e.g. $\mathbb{Q}$ satisfies the same and is far from being free. $\endgroup$ Commented Jan 29, 2023 at 19:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .