Visible Elements of $H^1(K,A)$

In this section, we produce a map $B(K)/n B(K)\rightarrow \Vis_J(H^1(K,A))$ with bounded kernel.

Lemma 3.7   Let $A$ and $B$ be abelian subvarieties of an abelian variety $J$ over a number field $K$ such that $A\cap B$ is finite. Suppose $n$ is a natural number such that

$$\gcd\left(n,\,\,\char93 (J/B)(K)_{\tor}\cdot\char93 B(K)_{\tor}\right) = 1$$

and $B[n] \subset A$ as subgroup schemes of $J$. Then there is a natural map

$$\varphi :B(K)/nB(K)\rightarrow \Vis_J(H^1(K,A))$$

such that $\ker(\varphi )\subset J(K)/(B(K)+A(K))$. If $A(K)$ has rank 0, then $\ker(\varphi )=0$ (more generally, $\ker(\varphi )$ has order at most $n^r$ where $r$ is the rank of $A(K)$).

Proof. First we produce a map $\varphi :B(K)/n B(K) \rightarrow \Vis(H^1(K,A))$ by using that $B[n] \subset A$ hence a certain map factors through multiplication by $n$. Then we use the snake lemma and our hypothesis that $n$ does not divide the orders of certain torsion groups to bound the dimension of the kernel of $\varphi$.

The quotient $J/A$ is an abelian variety $C$ over $K$. The long exact sequence of Galois cohomology associated to the short exact sequence

$$0 \rightarrow A \rightarrow J \rightarrow C \rightarrow 0$$

begins

$$0\rightarrow A(K) \rightarrow J(K) \rightarrow C(K) \xrightarrow{\,\delta\,} H^1(K,A) \rightarrow \cdots.$$ (3.1)

Let $\psi$ be map $B\rightarrow C$ obtained by composing the inclusion $B\hookrightarrow J$ with the quotient map $J\rightarrow C$. Since $B[n] \subset A$, we see that $\psi$ factors through multiplication by $n$, so the following diagram commutes:

$$\xymatrix{ & B\ar[d]\ar[dr]^{\psi} \ar[r]^{n}& B\ar[d]\\ A\ar[r]&J\ar[r]&C.}$$

Using that $B[n](K)=\{0\}$, we obtain the following commutative diagram, all of whose rows and columns are exact:

$$\xymatrix{ & K_0\ar[d] & K_1\ar[d]& K_2\ar[d]\\ 0 \ar[r] & B(K) \... ...r[r] & J(K)/A(K)\ar[r]\ar[d] & C(K) \ar[r] & \delta(C(K)) \ar[r] & 0\\ & K_3, }$$ (3.2)

where $K_0$, $K_1$ and $K_2$ are the indicated kernels and $K_3$ is the indicated cokernel. Exactness of the top row expresses the fact that $B[n](K)=\{0\}$, and the bottom exact row arises from the exact sequence (3.1) above. The first vertical map $B(K)\rightarrow J(K)/A(K)$ is induced by the inclusion $B(K)\hookrightarrow J(K)$ composed with the quotient map $J(K)\rightarrow J(K)/A(K)$. The second vertical map $B(K)\rightarrow C(K)$ exists because the composition $B\hookrightarrow J\rightarrow C$ has kernel $B\cap A$, which contains $B[n]$, by assumption. The third vertical map exists because $\pi$ contains $nB(K)$ in its kernel, so that $\pi$ factors through $B(K)/nB(K)$.

The sequence (1.1) on page  implies that the image of $\varphi$ is contained in $\Vis_J(H^1(K,A))$. The snake lemma gives an exact sequence

$$K_0\rightarrow K_1 \rightarrow K_2 \rightarrow K_3.$$

Because $B\rightarrow C$ has finite kernel, $K_1\subset B(K)_{\tor}$. Since $B[n](K)=\left\{0\right\}$ and $K_2$ is an $n$-torsion group, the map $K_1\rightarrow K_2$ is the 0 map. Thus $K_2=\ker(\varphi )$ is isomorphic to a subgroup of $K_3=J(K)/(A(K)+B(K))$, as claimed.

Any torsion in the quotient $J(K)/B(K)$ is of order coprime to $n$ because $J(K)/B(K)$ is a subgroup of $(J/B)(K)$, and $\gcd(n,\char93 (J/B)(K)_{\tor})=1$, by assumption. Thus if $A(K)$ is a torsion group, $K_3 = (J(K)/B(K))/A(K)$ has no nontrivial torsion of order dividing $n$, so when $A(K)$ has rank zero, $\ker(\varphi )=0$.

Consider the map $\psi: A(K) \rightarrow J(K)/B(K)$. To show that $\ker(\phi)$ has order at most $n^r$, where $r$ is the rank of $A(K)$, it suffices to show that $\coker(\psi)[n]$ has order at most $n^r$. To prove the latter statement, by the structure theorem for finite abelian groups, it suffices to prove it for the case when $n$ is a power of a prime. Moreover, we may assume that $A(K)$ and $J(K)/B(K)$ have no prime-to-$n$ torsion. Then $J(K)/B(K)$ is in fact torsion-free, and so we may also assume $A(K)$ is torsion-free. With these assumptions, the statement we want to prove follows easily by elementary group-theoretic arguments (in particular, by considering of the Smith normal form of the matrix representing $\psi$). $\qedsymbol$