Bad prime of torsor and original elliptic curve ; Definition of Tate–Shafarevich group $Ш(E/K)$

Question

Let $E/K$ be an elliptic curve over number field $K$. Let $M_K$ be a set of all places of $K$.

My question is, Does there exist a finite set $S\subset M_K$ such that $\forall C$: $E/K$-torsor, $\forall v\in M_K\setminus S$, $C(K_v)\neq \emptyset$ ?

I had always thought it was true when $S=\{\text{Bad primes of }E/K, \text{ infinite places of }K\}$ until now because rational points on reduction mod $v$ lifts to rational points on $K_v$ thanks to Hensel lemma. But a good prime $v$ of $E/K$ is not always a good prime of its torsor $C/K$, so this argument does not work. Indeed, for positive integer $n$, $H^1(G_{K_v},E)[n]$ is not always $0$ since it is dual of $E(K_v)/nE(K_v)$ by Tate-duality.

Background :
I want to check whether we can characterize Tate–Shafarevich group $Ш(E/K) \stackrel{\text{def}}{=} \ker \left\{H^1(G_K, E) \to \prod_{v\in M_K} H^1(G_{K_v}, E) \right\}$

as following.

$Ш(E/K) \stackrel{\text{def}}{=} \ker \left\{H^1(G_K, E) \to \bigoplus_{v\in S} H^1(G_{K_v}, E) \right\}$ where $S$ is an appropriate finite set which is determined by $E/K$.

To establish this definition, the above proposition must hold; however, choosing such a finite set $S$ does not seem trivial for me as mentioned above.

Answer 1

I fear you wish for too much here.

If $Ш$ is finite, then we can represent each element by a torsor; each torsor has good reduction away from a finite set and the union of all bad places would then be a candidate for $S$.

However, if $C'$ is another torsor representing the same element as $C$, I cannot see any reason why they should both have good reduction outside a common fixed $S$. All we know is that they are isomorphic over $\bar{K}$. Yet, even in the trivial class of $Ш$ there are quadratic twists of $E$ which have bad reduction at any chosen place.

The analogous situation for class groups is thinking about elements $x$ in $K^{\times}$ modulo $m$-th powers. Such $x$ defines an element of the class group if its valuation is a multiple of $m$ at all finite places of $K$. For each $x$ this is true outside a finite set $S$ as the valuation is even $0$ there. If we fix an $S$ however in advance, we get a much larger group. A set of ideals representing all elements of the class group can be chosen in many ways; any choice gives a finite list of $S$ such that no place $v$ outside $S$ appears in any of these ideals; but you can't hope for an $S$ that works for all choices.

The point is that one needs to impose at least some condition outside $S$. I recommend, for any fixed $n$, to view $Ш[n]$ as the kernel of $$ H^1\bigl(G_S(K), E \bigr)[n] \to \bigoplus_{v \in S} H^1(K_v, E)[n]$$ where $S$ is the set containing all places where $E$ has bad reduction, all infinite places and all places dividing $n$. Here $G_S(K)$ is the Galois group of the maximal extension of $K$ unramified outside $S$. This already imposes on the torsor a condition at all places outside $S$ and each element of $Ш[n]$ has a torsor that satisfies this condition.

If $Ш$ is finite a finite $S$ will do without taking $n$-torsion. Proving that a finite fixed $S$ will do for all of $Ш$ seems to me as hard as showing that it is finite.