Is there an elliptic curve analogue to the 4-term exact sequence defining the unit and class group of a number field?

Question

Let $K$ be a number field. One has the following exact sequence relating the unit group and ideal class group $\text{cl}(K)$: $$1\to \mathcal{O}_K^\times\to K^\times \to J_K\to \text{cl}(K)\to 1$$ where $J_K$ is the group of fractional ideals of $K$. This is essentially a restatement of the definition of unit group and class group respectively. However, there is an alternate way to construct $\text{cl}(K)$. One can consider the group of units of the algebraic integers $\mathcal{O}_{\bar{K}}^\times$, and this has a natural galois action. You can consider it's Tate-Shafaravich group $$\text{Sha}(K,\mathcal{O}_{\bar{K}}^\times)=\text{ker}\left(H^1(K,\mathcal{O}_{\bar{K}}^\times)\to \prod_{\mathfrak{p}}H^1(K_{\mathfrak{p}},\mathcal{O}_{\bar{K_{\mathfrak{p}}}}^\times)\right)$$ You can prove (see here) that there is a canonical isomorphism $\text{Sha}(K,\mathcal{O}_{\bar{K}}^\times)\simeq \text{cl}(K)$ Using the behavior of number fields as analogy, is it possible to create an analogous exact sequence for elliptic curves? I.e, one of the following form:

$$1\to E(K)\to X \to Y \to \text{Sha}(E/K)\to 1$$

for some "natural" objects $X,Y$.

Answer 1

As in the question $K$ is a number field and $E/K$ an elliptic curve.

Let me start by saying that I think the best analogues are the following two short exact sequences (the first two "$/n$" means modulo $n$-th powers): $$ \DeclareMathOperator{\Sel}{Sel} 0\to \mathcal{O}_K^{\times}/n \to \Bigl\{x\in K^{\times}/n \Bigm\vert \forall_v\, v(x)\equiv 0 \pmod{n}\Bigr\}\to \operatorname{Cl}(K)[n] \to 0 \\ 0\to E(K)/n E(K) \to \Sel_n(E/K) \to Ш(E/K)[n] \to 0 $$ where $n>1$ is any integer. This explains while the top middle set is called a Selmer group for $K$. One could view the following as a possibility to fill the two spaces in your sequence: $$ 0\to E(K)_{\text{tors}} \to E(K) \to E(K)\otimes \mathbb{Q} \to \operatorname{Sel}(E/K) \to Ш(E/K) \to 0 $$ where $\operatorname{Sel}(E/K)$ is the $\varinjlim_n$ of the Selmer groups $\operatorname{Sel}_n(E/K)$.

However there is another option. Let $S$ be a finite set of places containing all infinite places, all places dividing $n$ and all places of bad reduction for $E$. Write as usual $G_S(K)$ for the Galois group of the maximal extension of $K$ that is unramified outside $S$. The long exact sequence from global duality, in this case a theorem by Cassels, states $$ 0\to \Sel_n(E/K) \to H^1\bigl(G_S(K),E[n]\bigr) \to \bigoplus_{v\in S} \Bigl( E(K_v)/nE(K_v)\Bigr)^{\vee} \to \Sel_n(E/K)^{\vee} $$ where ${}^{\vee}$ is the Pontryagin dual. The local term is often written as $H^1_s\bigl(K_v, E[n]\bigr)$. Now assume that $Ш(E/K)$ is finite. Then the projective limit over $n$ gives $$ 0\to E(K)^{\ast} \to H^1\bigl(K, \varprojlim E[n] \bigr) \to \bigoplus_{\text{all } v} \Bigl( E(K_v)\otimes \mathbb{Q}/\mathbb{Z}\Bigr)^{\vee} \to \Sel(E/K)^{\vee} $$ where $E(K)^{\ast} = \varprojlim_n E(K)/nE(K)$ is the profinite completion of $E(K)$. Apart from the completion, this does not look like what you want because we have all of $\Sel(E/K)^{\vee}$ at the end and not just its finite torsion subgroup $Ш(E/K)^{\vee} \cong Ш(E/K)$. The good thing is that it is of the form, Mordell-Weil followed by global followed by sum of locals followed by something involving $Ш$.

An alternative is to switch local and global. Remark 6.14b after Theorem 6.13 in Milne's Arithmetic Duality Theorems gives the following exact sequence, still assuming that $Ш(E/K)$ is finite: $$ 0\to E(K)^{\ast}\to \prod_{\text{all }v} E(K_v)\to H^1\bigl(K, E\bigr)^{\vee} \to Ш(E/K)\to 0$$ where for the archimedean places $v$ one needs to replace $E(K_v)$ by its quotient with the connected component of $E(K_v)$. (I must admit I never fully worked through its proof, though.)

None of these are unique and, right now, I cannot see a good analogue of the initial sequence involving the group of fractional ideals. Maybe viewing $Ш$ as the Brauer group of the Néron model would help...

Edit: There is another four term sequence that was proposed by Zagier as an analogue. Let $S$ be the set of pairs $(C,P)$ where $C$ is a torsor representing an element in $Ш(E/K)$ and $P\in C(\bar{K})$. Two such are equivalent if there is a morphism $C\to C'$ of $E$-torsors sending $P$ to $P'$. Then there is the exact sequence $$ 0\to E(K) \to E(\bar{K}) \to S/\sim \to Ш(E/K) \to 0. $$