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Waterhouse in his thesis (Abelian varieties over finite fields, Ann. scient. \'Ec. Norm. Sup., t. 2, 1969, p 521-560) seems to use without comments the following fact:

Let $k$ be a finite field, and let $A$, $B$ be two abelian varieties over $k$ that are $k$-isogenous. Consider the set $I(A,B)$ of all the $k$-isogenies from $A$ to $B$. Then for any finite, non-trivial, subgroup $H$ of $A$, there is $\varphi\in I(A,B)$ that does not vanish identically on $H$.

This fact is used implicitly in lines 8-9 page 533, right after the definition of kernel ideal.

Does anyone have an argument to see it? Also, I do not know what role the assumption that the base field $k$ is finite should play. Thanks.

[EDIT: Actually what Waterhouse uses is that, under the assumption of the second paragraph above, there is a ${\it morphism}$ $\varphi:A\rightarrow B$ that does not vanish identically on $H$]

[EDIT 2: I report here Waterhouse's statement. Let $A$ be an abelian variety over a finite field $k$, and let $R$ be its $k$-endomorphism ring. Let $I$ be a left ideal of $R$ that contains an isogeny of $A$. Define $H(I)$ to be the finite subgroup of $A$ given by the intersection of all ker($\varphi$), as $\varphi$ ranges in $I$.

By definition, $I$ is a kernel ideal if $I=$ { $r\in R: r\cdot H(I)=0$ }.

Here comes the line I can't verify:

"Every $I$ is contained in a kernel ideal $J$ with $H(J)=H(I)$, namely $J=$ { $r\in R: r\cdot H(I)=0$ }."

The question is "how do we know that $J$, as just defined, is a kernel ideal?" I think this question is just a reformulation of the main question I asked above.]

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I do not see why your statement is true in the following example. Take $\psi$ to be the $2$-isogeny from $E : y^2=x^3+x$ to the non-isomorphic $E': y^2 = x^3+x+3$ killing the point $(2,0)$. Now the image under the dual $\hat\psi$ of $I(E,E')$ in $End(E)$ is the ideal generated by $[2]$. I believe that one concludes from this that any isogeny $\varphi:E\to E'$ has to factor through $\psi$. So $H=<(2,0)>$ vanishes for all $\varphi$ – Chris Wuthrich Mar 2 '11 at 12:53
I took $k=\mathbf{F}_5$. Am I doing something wrong, here ? Sorry I do not have Waterhouse at hand. – Chris Wuthrich Mar 2 '11 at 12:54
Thanks for your comment. I am not fully able to verify your counterexample. You are saying that the map from Hom(E,E') to End(E) sending x to $\hat\varphi$x gives an iso between Hom(E,E') and the ideal (2) as right End(E) modules, right? If this is the case then shouldn't E be isomorphic to E', because of the principality of that ideal? Anyway I am going to edit my question and post what Waterhouse actually says... – Tommaso Centeleghe Mar 2 '11 at 13:09
They have different $j$-invariant. Yes, as right modules. I don't see why the fact that this ideal is principal will imply that they should be isomorphic. – Chris Wuthrich Mar 2 '11 at 13:17
up vote 7 down vote accepted

Let me describe a natural straightforward generalization of Chris Wutrich's counterexample.

Let $B$ be a $g$-dimensional abelian variety over a field $k$ and assume that $End_k(B)$ is a principal ideal domain. Let $A$ be another abelian variety over $k$ that is not $k$-isomorphic to $B$ but $k$-isogenous to it. Then the group $Hom_k(A,B)$ becomes a free $End_k(B)$-module of rank 1. This means that there exists a (generator) isogeny $\lambda:A \to B$ such that every $k$-homomorphism $v: A\to B$ is a composition $u\lambda$ of $\lambda$ and a certain $u \in End_k(B)$. In particular, $ker(v)$ always contains $\ker(\lambda)$. Since $A$ and $B$ are not isomorphic over $k$, the isogeny $\lambda$ is not an isomorphism and therefore $H:=ker(\lambda)\subset A$ is nontrivial but is killed by every $k$-homomorphism from $A$ to $B$.

In order to construct explicit examples (over finite fields) pick any imaginary quadratic field $K$ of discriminant 1 amd let $O$ be the ring of integers in $K$, which is PID. Then for a ``half" of the primes $p$ there exist a finite field $k$ of characteristic $p$ and an ordinary elliptic curve $B$ over $k$ with $End_k(B)=End(B)=O$. Enlarging (if necessary) $k$, we may find an elliptic curve $A$ over $K$ that is not $k$-isomorphic to $A$ but $k$-isogenous to it. For example, if a prime $\ell$ is different from $p$ and inert in $O$ then we may pick a cyclic order $\ell$ subgroup $C$ in $B(k)$ (enlarging $k$ if necessary) and put $A=B/C$. Then the cyclic order $\ell$ subgroup $H=B_{\ell}/C\subset A$ is killed by every homomorphism $A \to B$.

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Thanks, this helps and, together with Chris' example, convince me that what I was asking cannot be true. Let me just point out that your endomorphism ring End_k(B) better do not have too many units, for otherwise you may be able to pick several generators for your principal ideal, whose associated kernels need not coincide. right? What about Waterhouse's actual claim, posted in my EDIT2 above? Can you sketch a proof of it? (I realized that what it says is actually weaker than what I was asking) – Tommaso Centeleghe Mar 2 '11 at 14:03
You are welcome. Well, units in $End_k(B)$ are actually automorphisms of $B$, so they allow us to vary $\lambda$'s while $\ker(\lambda)$ remains the same. As for Waterhouse actual claim, please notice that at this point he deals not with arbitrary finite group subschemes of $A$ but only with those that (slightly reformulating) are kernels of homomorphisms $A \to A^m$. He does not claim that every finite group subscheme $H \subset A$ coincides with $H(J)$ for a certain ideal $J \subset \End_k(B)$. – Yuri Zarhin Mar 2 '11 at 14:16
OOPS! Sorry. In the last sentence one should read $J\subset End_k(A)$. – Yuri Zarhin Mar 2 '11 at 14:31
You are right. Now I understand, I answered my own question with the proof of Waterhouse's original claim. – Tommaso Centeleghe Mar 2 '11 at 16:29

As the comments and the answer above explain, the answer to my original question is NO. For completeness, I include the (very easy) proof of the statement in the second edit of my question (whose notation will be adopted) which is a weaker statement than that I was asking in the first place.

The group of homomorphisms Hom$_k(A/H(I),A)$ has the property that the intersection of all the kernels of its elements is the trivial subgroup of $A/H(I)$. To see this, one can just observe that the ideal $I$ can be identified with a subset of Hom$_k(A/H(I),A)$, and this subset already has the property that the intersection of all the kernels of its elements is trivial.

It now immediately follows that $H(I)=H(J)$.

I would like to thank the authors of the comments and answer above for their interest.

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So the statement Waterhouse uses is correct and fairly obvious, and the false statement is your misunderstanding of what Waterhouse wrote? Perhaps you should make this clear. – mephisto Mar 2 '11 at 20:24
Yes. I thought this is what I just did in my answer above! – Tommaso Centeleghe Mar 3 '11 at 11:07

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