In this [lecture][1] of Serre on his open image theorem, around 6 minutes, Serre mentions the following theorem of Tate:

Let $A/k$ be an abelian variety over a number field and consider the Mordell-Weil group $A(k)$. It is a finitely generated group and we have the inclusion $A(k) \to \prod_v A(k_v)$ where $v$ ranges over the non archimedean valuations of $k.$

Consider the subgroups of $A(k)$ defined by finitely many congruence conditions under this map. Are these exactly the finite index groups? One direction is clear so really: Is every finite index subgroup of $A(k)$ determined by finitely many congruence conditions?

Serre mentions that Tate showed that this is equivalent to the following condition:

Let $\ell$ be a prime and consider $V_\ell(A) = \mathbb Q_\ell\otimes_{\mathbb Z_\ell}T_\ell(A)$ with the Galois action $G_k$. Then, if $H^1(G_k,V_\ell(A)) = 0$, then every finite index subgroup is cut out by congruence conditions.

How does one prove this? Answer in comments

I am a little confused because of the following argument:

By the standard Kummer pairing and taking an inverse limit, we have an inclusion:

$$0 \to A(k)\otimes\mathbb Z_\ell \to H^1(G_k,T_\ell).$$

To require $H^1(G_k,\mathbb V_\ell) = 0$ is to require $H^1(G_k,T_\ell)$ to be $\ell^{\infty}$ torsion but I don't think this is true for $A(k)\otimes\mathbb Z_\ell$?

Answer to above confusion: Tate/Cassels/Serre don't prove or need that the entire Galois cohomology is zero, they show that a particular Selmer group is $0$ and there is no contradiction. [1]: https://www.youtube.com/watch?v=0KqVP9We2LQ

In this lecture of Serre on his open image theorem, around 6 minutes, Serre mentions the following theorem of Tate:

Let $A/k$ be an abelian variety over a number field and consider the Mordell-Weil group $A(k)$. It is a finitely generated group and we have the inclusion $A(k) \to \prod_v A(k_v)$ where $v$ ranges over the non archimedean valuations of $k.$

Consider the subgroups of $A(k)$ defined by finitely many congruence conditions under this map. Are these exactly the finite index groups? One direction is clear so really: Is every finite index subgroup of $A(k)$ determined by finitely many congruence conditions?

Serre mentions that Tate showed that this is equivalent to the following condition:

Let $\ell$ be a prime and consider $V_\ell(A) = \mathbb Q_\ell\otimes_{\mathbb Z_\ell}T_\ell(A)$ with the Galois action $G_k$. Then, if $H^1(G_k,V_\ell(A)) = 0$, then every finite index subgroup is cut out by congruence conditions.

How does one prove this? Answer in comments

I am a little confused because of the following argument:

By the standard Kummer pairing and taking an inverse limit, we have an inclusion:

$$0 \to A(k)\otimes\mathbb Z_\ell \to H^1(G_k,T_\ell).$$

To require $H^1(G_k,\mathbb V_\ell) = 0$ is to require $H^1(G_k,T_\ell)$ to be $\ell^{\infty}$ torsion but I don't think this is true for $A(k)\otimes\mathbb Z_\ell$?

Answer to above confusion: Tate/Cassels/Serre don't prove or need that the entire Galois cohomology is zero, they show that a particular Selmer group is $0$ and there is no contradiction.

In this lecture of Serre on his open image theorem, around 6 minutes, Serre mentions the following theorem of Tate:

Let $A/k$ be an abelian variety over a number field and consider the Mordell-Weil group $A(k)$. It is a finitely generated group and we have the inclusion $A(k) \to \prod_v A(k_v)$ where $v$ ranges over the non archimedean valuations of $k.$

Consider the subgroups of $A(k)$ defined by finitely many congruence conditions under this map. Are these exactly the finite index groups? One direction is clear so really: Is every finite index subgroup of $A(k)$ determined by finitely many congruence conditions?

Serre mentions that Tate showed that this is equivalent to the following condition:

Let $\ell$ be a prime and consider $V_\ell(A) = \mathbb Q_\ell\otimes_{\mathbb Z_\ell}T_\ell(A)$ with the Galois action $G_k$. Then, if $H^1(G_k,V_\ell(A)) = 0$, then every finite index subgroup is cut out by congruence conditions.

How does one prove this? I am a little confused because of the following argument:

By the standard Kummer pairing and taking an inverse limit, we have an inclusion:

$$0 \to A(k)\otimes\mathbb Z_\ell \to H^1(G_k,T_\ell).$$

To require $H^1(G_k,\mathbb V_\ell) = 0$ is to require $H^1(G_k,T_\ell)$ to be $\ell^{\infty}$ torsion but I don't think this is true for $A(k)\otimes\mathbb Z_\ell$?

In this [lecture][1] of Serre on his open image theorem, around 6 minutes, Serre mentions the following theorem of Tate:

Let $A/k$ be an abelian variety over a number field and consider the Mordell-Weil group $A(k)$. It is a finitely generated group and we have the inclusion $A(k) \to \prod_v A(k_v)$ where $v$ ranges over the non archimedean valuations of $k.$

Consider the subgroups of $A(k)$ defined by finitely many congruence conditions under this map. Are these exactly the finite index groups? One direction is clear so really: Is every finite index subgroup of $A(k)$ determined by finitely many congruence conditions?

Serre mentions that Tate showed that this is equivalent to the following condition:

Let $\ell$ be a prime and consider $V_\ell(A) = \mathbb Q_\ell\otimes_{\mathbb Z_\ell}T_\ell(A)$ with the Galois action $G_k$. Then, if $H^1(G_k,V_\ell(A)) = 0$, then every finite index subgroup is cut out by congruence conditions.

How does one prove this? Answer in comments

I am a little confused because of the following argument:

By the standard Kummer pairing and taking an inverse limit, we have an inclusion:

$$0 \to A(k)\otimes\mathbb Z_\ell \to H^1(G_k,T_\ell).$$

To require $H^1(G_k,\mathbb V_\ell) = 0$ is to require $H^1(G_k,T_\ell)$ to be $\ell^{\infty}$ torsion but I don't think this is true for $A(k)\otimes\mathbb Z_\ell$?

Answer to above confusion: Tate/Cassels/Serre don't prove or need that the entire Galois cohomology is zero, they show that a particular Selmer group is $0$ and there is no contradiction. [1]: https://www.youtube.com/watch?v=0KqVP9We2LQ