Is $H^{1}_{Sel}\left(K,E_{p^{n}}\right)\rightarrow\prod_{q \nmid \infty} H^1\left(K_{q},E_{p^{n}}\right)$ an injection?

Question

If $K$ is a number field, $E$ an elliptic curve and $p$ a prime, does the Selmer group $$H^{1}_{\operatorname{Sel}}\left(K,E_{p^{n}}\right)$$ always inject into $$\prod_{q \text{ a nonarchimedean prime of }K}H^1 \left(K_{q},E_{p^{n}}\right)$$ (Excuse me if that question is stupid)

Answer 1

The answer is "no" in general.

By the definition of the Selmer group, you can replace the target of the map by the product of $E(K_q)/p^n E(K_q)$. Now $E(K)/p^n E(K)$ is a subgroup of the Selmer group. So if your map were always injective, this would imply that a global point in $E(K)$ is divisible by $p^n$ if and only if it is divisible by $p^n$ in all localisations at finite places. However there are known counter-examples for $K=\mathbb{Q}$ and $p^n$ with $p=2$ and $3$ and $n>1$. The first was found by Dvornicich and Zannier.

On the positive side, one can show that it is injective for $p>2$ and $n=1$ by a lemma of Tate. Also if $K=\mathbb{Q}$, $p>3$ and $Ш(E/\mathbb{Q})$ has no $p$-torsion, then it is injective for all $n$, because the local-global divisibility holds.

In the limit, however, the $p$-primary Selmer group $\varinjlim \operatorname{Sel}_{p^n}(E/\mathbb{Q})$ can only injects into $\prod_{v\nmid \infty} H^1(\mathbb{Q}_v, E[p^{\infty}])$ if the rank of $E(\mathbb{Q})$ is zero or one. This is because the image of this map is contained in $E(\mathbb{Q}_p)\otimes \mathbb{Q}_p/\mathbb{Z}_p$, which is of corank $1$.