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Let $K$ be an finite abelian extension of $\mathbf{Q}$ conductor $p$, where $p$ is an odd prime. That is, $K \subset \mathbf{Q}(\mu_ p)$, the $p$-th cyclotomic field. Let $h_K$ be the class number of $K$.

If $[K:\mathbf{Q}]=2$, we know that $K=\mathbf{Q}(\sqrt{p})$ or $\mathbf{Q}(\sqrt{-p})$. Gauss's genus theory tells us that $h_K\equiv 1 \pmod 2$.

If $[K:\mathbf{Q}]=3$, a paper of G.Gras https://eudml.org/doc/151629 (Page 94 line 1-2) says that $h_K \equiv 1 \pmod 3$. Unfortunately, I did not find the proof of this result. I am very appreciated if someone gives references or ideas.

So my question is that

if $[K:\mathbf{Q}]=n$ is a prime number, then do we have that $h_K \equiv 1 \pmod n$?

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    $\begingroup$ google "Galois action on class groups" for references. $\endgroup$
    – Franz Lemmermeyer
    Commented Sep 1, 2016 at 16:44

1 Answer 1

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Maybe I am making a mistake here, but let me try:

Let $H$ be the Hilbert class field of $K$. Then $H\cap \mathbb{Q}(\zeta_p)=K$ as otherwise one prime in there should be totally ramified and unramified at the same time. This shows the triviality of the maximal quotient of the class group $C$ on which the Galois group $G=\operatorname{Gal}(K/\mathbb{Q})\cong \mathbb{Z}/n\mathbb{Z}$ acts trivially. Then the action of $G$ has no non-trivial fixed points on $C$, which means each orbit of $G$ on $C$, apart from the neutral element, is of size $n$, because $n$ is prime. Hence $\vert C\vert \equiv 1 \pmod{n}$.

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  • $\begingroup$ Can you explain in more detail how $H\cap \mathbb{Q}(\zeta_p)=K$ implies that $G$ has no non-trivial fixed point on $C$? Thanks in advance. $\endgroup$
    – GH from MO
    Commented Sep 1, 2016 at 17:37
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    $\begingroup$ @GHfromMO Look at the $\mathbb{Z}[G]$-map $g-1\colon C \to C$ for a genereator $g$ of $G$. The kernel and cokernel must have equal size. The latter corresponds to an extension $k/K$ within $H/K$. So it is unramified everywhere. The action of $G$ on the Galois group of $k/K$ is now trivial, which translate to saying that $k/\mathbb{Q}$ is abelian. As $k/\mathbb{Q}$ has conductor $p$, it is a subextension of $\mathbb{Q}(\zeta_p)$. $\endgroup$
    – Chris Wuthrich
    Commented Sep 2, 2016 at 9:02
  • $\begingroup$ Thank you for the additional details, which are nice and clean! $\endgroup$
    – GH from MO
    Commented Sep 2, 2016 at 13:05
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    $\begingroup$ @GHfromMO Or you can use the classic genus theory to show $C^G=0$. The genus theory states that if $L/F$ is a cyclic extension with Galois group $G$, then $$|C^G_L|=\frac{|C_F|\prod_{v}{e_v}}{|G|[\mathcal{O}^{*}_F:N(L^{*})\cap \mathcal{O}^{*}_F]}$$ where the product runs all places of $F$ and $e_v$ is the ramification index. For proof you can see Emerton's notes math.uchicago.edu/~emerton/number-theory/genus.pdf $\endgroup$
    – J.Li
    Commented Sep 3, 2016 at 3:12
  • $\begingroup$ @J.Li: Thank you very much, the notes are useful. $\endgroup$
    – GH from MO
    Commented Sep 3, 2016 at 12:33

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