Let $k$ be a finite abelian extension of $\mathbb{Q}$. Class field theory states that $k$ corresponds to some open subgroup of finite index $U_k \subset \mathbb{A}_{\mathbb{Q}}^*/ \mathbb{Q}^*$ where $\mathbb{A}_{\mathbb{Q}}$ denotes the adeles of $\mathbb{Q}$. Specifically the subgroup $U_k$ is the image of the idelic norm map. Moreover every open subgroup of finite index arises this way for some unique abelian extension.

The subgroup $U_k$ should know everything about the abelian extension $k$. So I was wondering, does it know the class group?

Let $U \subset \mathbb{A}_{\mathbb{Q}}^*/ \mathbb{Q}^*$ be an open subgroup of finite index with corresponding abelian extension $k$. Is there some way to calculate the class group of $k$, just knowing $U$?

  • $\begingroup$ From $U$ you know some $N$ such that $V_N\subset U$ where $ V_N=\prod_{p^k\| N} (1+p^k\mathbb{Z}_p)\prod_{p\nmid N} \mathbb{Z}_p^\times $ then $k = \mathbb{Q}(\zeta_N)^H$ where $H = (U\cap \hat{\mathbb{Z}}^\times)/V_N,Gal(\mathbb{Q}(\zeta_N)/\mathbb{Q}) = \hat{\mathbb{Z}}^\times/V_N$ $\endgroup$
    – reuns
    Feb 10, 2019 at 23:27
  • 10
    $\begingroup$ The integer $N$ uniquely determines the field $k$ of $N$-th roots of unity, still it does not know its class group in any reasonable way. $\endgroup$ Feb 11, 2019 at 6:04


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