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I am having a look at the paper Explicit CM-theory for level 2-structures on abelian surfaces by Bröker, Gruenewald and Lauter, and there is an argument which I don't understand. The main reason being that the argument is not explicitly written down, but justified by the wild card "it is not hard to check".

Here is the situation (lemma 6.5 in the paper). We are given a cyclic, CM-field $K$ of degree 4 (i.e., a cyclic extension $K/\mathbb Q$ of degree 4, such that $K$ is a totally imaginary quadratic extension of a totally real quadratic number field $K_0$). Let $\sigma$ be a generator of $\mathrm{Gal}(K/\mathbb Q)$, and define the map $$m : \mathrm{Cl}(\mathcal O_K) \longrightarrow \mathrm{Cl}(\mathcal O_K) : [\mathfrak p] \longmapsto [\mathfrak p \mathfrak p^\sigma],$$ for degree 1 prime ideals $\mathfrak p$ (any ideal class can be represented by such a $[\mathfrak p]$). They claim that the image of $m$ contains all the squares of $\mathrm{Cl}(\mathcal O_K)$. This is described as "not hard to check", but I have to admit that I cannot find any simple argument.

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  • $\begingroup$ I am not sure if this statement is true. Let us assume that the class numbers of $K$ and $K_0$ are odd. Then any $c_0\in\mathrm{Cl}(\mathcal O_{K_0})$ can be written as $c_0=N_{K/K_0}c$ for some $c\in\mathrm{Cl}(\mathcal O_K)$, which in turn is of the form $c=[\mathfrak p \mathfrak p^\sigma]$. Hence $c_0=N_{K/K_0}c=cc^{\sigma^2}=[\mathfrak p \mathfrak p^\sigma \mathfrak p^{\sigma^2} \mathfrak p^{\sigma^3}]=1$, which implies that $\mathrm{Cl}(\mathcal O_{K_0})$ is trivial. $\endgroup$
    – GH from MO
    Aug 28, 2014 at 20:39
  • $\begingroup$ In a group of odd order, every element is a square. $\endgroup$
    – GH from MO
    Aug 28, 2014 at 20:48
  • $\begingroup$ Right. This claim is very suspicious, there is either something wrong, or I am misinterpreting the paper. $\endgroup$
    – Calodeon
    Aug 28, 2014 at 20:52
  • $\begingroup$ If I were you, I would simply email the authors. This is why they have their email addresses listed at the end of the paper. Naturally, if you learn anything please share with us! $\endgroup$
    – GH from MO
    Aug 28, 2014 at 20:53
  • $\begingroup$ No answer from the authors so far, but sure, I will. $\endgroup$
    – Calodeon
    Sep 2, 2014 at 15:16

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