Group of connected components of the global Néron-Raynaud model of a torus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:28:49Z http://mathoverflow.net/feeds/question/102531 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102531/group-of-connected-components-of-the-global-neron-raynaud-model-of-a-torus Group of connected components of the global Néron-Raynaud model of a torus Rony Bitan 2012-07-18T11:29:37Z 2012-08-19T10:22:00Z <p>Let <code>$K = \mathbb{F}_q(C)$</code> be a global function field of an irreducible projective and smooth curve $C$ defined over a finite field of constants <code>$\mathbb{F}_q$</code>. 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: <code>$$A_\infty = \{ a \in K : v_\frak{p}(a) \geq 0 \ \ \forall \frak{p} \neq \infty \}.$$</code> 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 <code>$\mathcal{T}_\frak{p}$</code> 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 <code>$\mathcal{T}_\frak{p}$</code> a connected reduction modulo $\frak{p}$. Let $\mathcal{T}^0$ denote the subscheme of $\mathcal{T}$ whose geometric fibers are <code>$\mathcal{T}_\frak{p}^0$</code>. </p> <p>My question is can I express the finite index $[\mathcal{T}(A_\infty):\mathcal{T}^0(A_\infty)]$ as the product of <code>$[\mathcal{T}_\frak{p}(\mathcal{O}_\frak{p}):\mathcal{T}_\frak{p}^0(\mathcal{O}_p)]$</code> (which may differ from 1 only if $\frak{p}$ is ramified on the minimal splitting field of $T$) running over all $\frak{p} \neq \infty$ ?</p> http://mathoverflow.net/questions/102531/group-of-connected-components-of-the-global-neron-raynaud-model-of-a-torus/103100#103100 Answer by Rony Bitan for Group of connected components of the global Néron-Raynaud model of a torus Rony Bitan 2012-07-25T12:49:11Z 2012-08-05T10:05:32Z <p>The direction to the following counterexample has been suggested to me by Prof. Liu, for whom I am grateful: </p> <p>Suppose there are two distinct finite ramified primes <code>$\frak{p}_1$</code> and <code>$\frak{p}_2$</code> 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. </p> <p><strong>Example:</strong> Let <code>$K=\mathbb{F}_q(t)$</code> with odd characteristic and <code>$L=K(\sqrt{\frak{p}_1 \cdot \frak{p}_2})$</code> where <code>$\frak{p}_1$ and $\frak{p}_2$</code> are two distinct finite primes of degree $1$. Consider the corresponding norm torus: <code>$$T = Spec~K [x,y]/(x^2-{\frak{p}_1\frak{p}_2} y^2-1).$$</code> Then the reduction of both <code>$\mathcal{T}^0_{\frak{p}_i}(\mathcal{O}_{\frak{p}_i})$</code> for $i=1,2$ is: <code>$\{ 1 + {\frak{p}_1 \frak{p}_2} y : y \in \mathbb{F}_q \}$</code>. 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 <code>$T(K)$</code> having a proper pole at $\infty$ has a proper zero at some finite prime. But since: <code>$$\mathcal{T}(A_\infty) = T(K) \cap \prod_{\frak{p} \neq \infty} \mathcal{T}_{\frak{p}}(\mathcal{O}_{\frak{p}}),$$</code> this element has a proper pole at that prime, thus does not belong to <code>$\mathcal{T}(A_\infty)$</code>.<br> We may therefore conclude that <code>$\mathcal{T}(A_\infty)$</code> consists only with elements which are regular everywhere, i.e. constants (as the curve is projective). In our case: <code>$$\mathcal{T}(A_{\infty}) = \mathcal{T}(\mathbb{F}_q) = \{x \in \mathbb{F}_q : x^2=1 \} = \{\pm 1\}.$$</code><br> So we get: $$[\mathcal{T}(A_\infty):\mathcal{T}^0(A_\infty)] = 2$$ while: <code>$$[\mathcal{T}_{\frak{p}_1}(\mathcal{O}_{\frak{p}_1}):\mathcal{T}_{\frak{p}_1}^0(\mathcal{O}_{\frak{p}_1})] \cdot [\mathcal{T}_{\frak{p}_2}(\mathcal{O}_{\frak{p}_2}):\mathcal{T}_{\frak{p}_2}^0(\mathcal{O}_{\frak{p}_2})] = 2 \cdot 2 =4.$$</code> Rony.</p>