Let $T$ be an algebraic torus over a number field $K$. Following notations in Ono's The Arithmetic of Tori, http://www.jstor.org/discover/10.2307/1970307?sid=21105671135711&uid=3739888&uid=2&uid=3739256&uid=4

we define $T_A$ the adele points, $T_K$ the $K$-rational point of $T$ and $$T_{A,S_{\infty}}=\prod_{v \in S_{\infty}}T(K_v)\times \prod_{v \notin S_{\infty}}T_v^c $$ The class number of $T$ is defined as $$ h_T := [{T_A}:{T_{A,S_{\infty}}T_K}] $$ My question is how do we compute $h_T$ in pratice. If $T=\mathbb{G}_m$, $h_T$ is just the class number of $K$ and we compute it using Minkowski's bounds. But I don't know how to compute $h_T$ in general.

For an example, if $T$ is the torus $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$, what is $h_T$ ? Thank you very much.

  • $\begingroup$ You seem to have written the class number is a ratio of two groups. Presumably you mean the index. $\endgroup$ – Kimball Mar 19 '15 at 7:01
  • $\begingroup$ @Kimball Yes that's what i mean. Thank you. $\endgroup$ – raynor14 Mar 19 '15 at 15:08

I don't know how practical this is to compute in general, but Ono had a student, Shyr, who proved an analytic class number formula for $\mathbb Q$-tori in his thesis. This can at least be used to get an expression for "relative class numbers" (quotients of class numbers). Ono used this in this paper (and maybe Shyr in his thesis also) to compute the class number of a torus which is the kernel of a norm map of a cyclic extension of Q.

Shyr's formula was later generalized by Rony Bitan (2011, Journal of Number Theory).

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer ! I will look into these papers. It seems they are very relevant. $\endgroup$ – raynor14 Mar 19 '15 at 15:14

You can use relationships between different tori. For instance, your torus lives in a long exact sequence

$T\to \operatorname {Res} _{\mathbb Q(i) / \mathbb Q } \mathbb G_m \to \mathbb G_m $

This gives you a cohomology long exact sequence:

$ H^0( \mathbb Z, \operatorname {Res} _{\mathbb Q(i) / \mathbb Q } \mathbb G_m) \to H^0( \mathbb Z, \mathbb G_m) \to H^1(\mathbb Z, T) \to H^1 ( \mathbb Z, \operatorname {Res} _{\mathbb Q(i) / \mathbb Q }\mathbb G_m ) \to H^1(\mathbb Z, \mathbb G_m)$


$ H^0( \mathbb Z[i], \mathbb G_m) \to H^0( \mathbb Z, \mathbb G_m) \to H^1(\mathbb Z, T) \to H^1 ( \mathbb Z[i], \mathbb G_m ) \to H^1(\mathbb Z, \mathbb G_m)$

Evaluating the $H^0$s as unit groups and $H^1$s as class groups:

$ \mu_4 \to \mu_2 \to H^1(\mathbb Z, T) \to 0 \to 1$

The map $\mu_4 \to \mu_2$ is the norm map, which is trivial, hence $H^1(\mathbb Z, T) = \mu_2$, and the class number is $2$.

You can probably do this in general but you need a spectral sequence.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. Could you please elaborate a bit more about the exact sequence relates the class numbers and the unit groups ? $\endgroup$ – raynor14 Mar 23 '15 at 1:47
  • $\begingroup$ @raynor14 Done. $\endgroup$ – Will Sawin Mar 23 '15 at 2:51
  • $\begingroup$ are you using etale cohomology or Galois cohomology here ? $\endgroup$ – raynor14 Mar 23 '15 at 3:49
  • $\begingroup$ etale cohomology. $\endgroup$ – Will Sawin Mar 23 '15 at 3:55
  • 1
    $\begingroup$ I am confused. Then don't you need a ses of etale sheaves on $Spec(\mathbb{Z})$ ? The exact sequence we have is only an exact sequence on $Spec(\mathbb{Q})$. $\endgroup$ – raynor14 Mar 23 '15 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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