# Tate-Shafarevich groups over finitely generated fields

Let $G$ be an algebraic group over a number field $k$. One defines the Tate-Shafarevich set of $G$ to be $$Ш(k,G) = \ker\left(H^1(k,G) \to \prod_{v} H^1(k_v,G)\right),$$ where the product is over all places of $k$. Note that if $G$ is non-abelian, then this will only be a pointed set in general.

It is known that $Ш(k,G)$ is finite if $G$ is a linear algebraic group. It is conjectured that $Ш(k,G)$ is finite if $G$ is an abelian variety. This is known in some special cases, but is open in general.

Let now $G$ be an algebraic group over a finitely generated field extension $k$ of $\mathbb{Q}$.

Is there a natural analogue of $Ш(k,G)$ in this setting? Is it moreover known that $Ш(k,G)$ is finite when $G$ is linear algebraic?

Part of my motivation is the observation that results over number fields often generalise to finitely generated field extensions of $\mathbb{Q}$ (e.g. the Mordell-Weil theorem). So I would like to know if $Ш(k,G)$ makes sense in this more general setting.

I have a vague idea of how to proceed. Namely, to choose a model for $k$ (given as a proper flat scheme of finite type over $\mathbb{Z}$ with function field $k$, say), then take our notion of place to be a point of codimension one on this model. But I'm not really sure where to go from there, nor whether finiteness should hold when $G$ is linear algebraic.

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If $X/\mathbf{F}_q$ is a smooth projective scheme (a model of a function field in positive characteristic) and $\mathscr{A}/X$ an Abelian variety, then $H^1(X,\mathscr{A}) = \ker\Big(H^1(K,\mathscr{A}) \to \bigoplus_{x \in S}H^1(K_x^{nr},\mathscr{A})\Big)$ is an analogue of the Tate-Shafarevich group. Here, $S$ can be $X$, the set of closed points $|X|$ or the set of codimension-$1$ points $X^{(1)}$; $K_x^{nr}$ is the quotient field of the strict Henselisation of $\mathscr{O}_{X,x}$; in the case of $x \in X^{(1)}$, you can also use the completion.
What did you prove exactly in your thesis? The finiteness of this set? (Note that in the question $K$ is of characteristic zero.) –  Ariyan Javanpeykar Aug 15 '14 at 12:18
I just proved the equality $H^1(X,\mathscr{A}) = \ker(...)$. I am aware that the question is about characteristic $0$. –  Timo Keller Aug 15 '14 at 12:19