Question

Let $K = \mathbb{F}_q(C)$ be a global function field of an irreducible projective and smooth curve $C$ defined over a finite field of constants $\mathbb{F}_q$. Let $T$ be a $K$-torus. We choose one closed point $\infty$ to be the point at infinity, and consider the ring of ${\infty}$-integers of $K$, namely: $$ A_\infty = \{ a \in K : v_\frak{p}(a) \geq 0 \ \ \forall \frak{p} \neq \infty \}. $$ Then $T$ admits a global Néron-Raynaud model $\mathcal{T}$ defined over $\text{Spec}~A_\infty$, which is of finite type, obtained by glueing all local Néron-Raynaud models $\mathcal{T}_\frak{p}$ which are of finite type, defined each one over the corresponding local valuation ring $\mathcal{O}_p$. The glueing is along the generic fiber $T$. Denote by $\mathcal{T}_\frak{p}$ a connected reduction modulo $\frak{p}$. Let $\mathcal{T}^0$ denote the subscheme of $\mathcal{T}$ whose geometric fibers are $\mathcal{T}_\frak{p}^0$.

My question is can I express the finite index $[\mathcal{T}(A_\infty):\mathcal{T}^0(A_\infty)]$ as the product of $[\mathcal{T}_\frak{p}(\mathcal{O}_\frak{p}):\mathcal{T}_\frak{p}^0(\mathcal{O}_p)]$ (which may differ from 1 only if $\frak{p}$ is ramified on the minimal splitting field of $T$) running over all $\frak{p} \neq \infty$ ?

Answer 1

The direction to the following counterexample has been suggested to me by Prof. Liu, for whom I am grateful:

Suppose there are two distinct finite ramified primes $\frak{p}_1$ and $\frak{p}_2$ for which the corresponding geometric fibers have the same connected component. Then since $\mathcal{T}^0(A_\infty)$ is the intersection of all $\mathcal{T}^0_{\frak{p}}(\mathcal{O}_{\frak{p}})$, the resulting index maybe taken only once in the global scheme, while it may appear twice in the product of the local indices. So we have a divisiblity relation of course but not the equality necessarily.

Example: Let $K=\mathbb{F}_q(t)$ with odd characteristic and $L=K(\sqrt{\frak{p}_1 \cdot \frak{p}_2})$ where $\frak{p}_1$ and $\frak{p}_2$ are two distinct finite primes of degree $1$. Consider the corresponding norm torus: $$ T = Spec~K [x,y]/(x^2-{\frak{p}_1\frak{p}_2} y^2-1). $$ Then the reduction of both $\mathcal{T}^0_{\frak{p}_i}(\mathcal{O}_{\frak{p}_i})$ for $i=1,2$ is: $ \{ 1 + {\frak{p}_1 \frak{p}_2} y : y \in \mathbb{F}_q \}$. The $A_{\infty}$-scheme $\mathcal{T}$ obtained by the glueing process is not necessarily reductive as it may have (finitely many) fibers with a non-reductive reduction. An element in $T(K)$ having a proper pole at $\infty$ has a proper zero at some finite prime. But since: $$ \mathcal{T}(A_\infty) = T(K) \cap \prod_{\frak{p} \neq \infty} \mathcal{T}_{\frak{p}}(\mathcal{O}_{\frak{p}}), $$ this element has a proper pole at that prime, thus does not belong to $\mathcal{T}(A_\infty)$.
We may therefore conclude that $\mathcal{T}(A_\infty)$ consists only with elements which are regular everywhere, i.e. constants (as the curve is projective). In our case: $$ \mathcal{T}(A_{\infty}) = \mathcal{T}(\mathbb{F}_q) = \{x \in \mathbb{F}_q : x^2=1 \} = \{\pm 1\}. $$
So we get: $$ [\mathcal{T}(A_\infty):\mathcal{T}^0(A_\infty)] = 2 $$ while: $$ [\mathcal{T}_{\frak{p}_1}(\mathcal{O}_{\frak{p}_1}):\mathcal{T}_{\frak{p}_1}^0(\mathcal{O}_{\frak{p}_1})] \cdot [\mathcal{T}_{\frak{p}_2}(\mathcal{O}_{\frak{p}_2}):\mathcal{T}_{\frak{p}_2}^0(\mathcal{O}_{\frak{p}_2})] = 2 \cdot 2 =4. $$ Rony.