Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

I have proved this in my PhD thesis, which will be available soon.

share|improve this answer
1  
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 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 at 12:19
    
I see, thanks for the quick reply. Why does it not matter whether you take the sum over only the codim'n 1 points instead of all closed points? –  Ariyan Javanpeykar Aug 15 at 12:22
    
This is too difficult and would take too long to explain here. –  Timo Keller Aug 15 at 12:26
    
@Timo: Thanks for your answer. Whilst it is quite interesting, it does not really answer my question. So I have bumped my question in case someone else can answer it. –  Daniel Loughran Aug 18 at 9:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.