9
$\begingroup$

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 $h_K \equiv 1 \pmod n$?

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

1 Answer 1

Reset to default
10
$\begingroup$

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 $\vert C\vert \equiv 1 \pmod{n}$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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
    Sep 1, 2016 at 17:37
  • 1
    $\begingroup$ @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
    Sep 2, 2016 at 9:02
  • $\begingroup$ Thank you for the additional details, which are nice and clean! $\endgroup$
    – GH from MO
    Sep 2, 2016 at 13:05
  • 1
    $\begingroup$ @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
    Sep 3, 2016 at 3:12
  • $\begingroup$ @J.Li: Thank you very much, the notes are useful. $\endgroup$
    – GH from MO
    Sep 3, 2016 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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