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On page 149 of Algebraic Number Theory by Serge Lang, I'm trying to understand why the inclusion $$k^{\ast}N_k^K(J_K) \cap J_c \subseteq \psi^{-1}(P_c \mathfrak N(c))$$ is true. I've been trying for like 3 hours. Can anyone please explain?

EDIT: What all that means: $K/k$ is an abelian extension of number fields. $c = (m(v))$ is an admissible cycle of $k$ divisible by all ramified primes, $J_c$is the subgroup of the ideles of $k$: $$\prod\limits_{v \mid c} W_c(v) \prod\limits_{v \not\mid c}' k_v^{\ast}$$ where $W_c(v) = 1 + \mathfrak p_v^{m(v)}$ for $v$ finite, and $W_c(v)$ is $(0, \infty)$ when $v$ is real. Since $c$ is admissible, that means that $W_c(v)$ is contained in the group of local norms for all $v \mid c$.

$\phi$ is a homomorphism from $J_c$ to the group of fractional ideals which are relatively prime to $c$, given by $x \mapsto \prod\limits_{v \nmid c, v < \infty} \mathfrak p_v^{\nu_v(x_v)}$. Also $P_c$ is the group of principal ideals $x \mathcal O_k$, where $x \in J_c \cap k^{\ast}$ (in more standard notation, $x \equiv 1 \mod^{\ast} c$). Finally $\mathfrak N(c)$ is the group of norms of fractional ideals of $K$ which are relatively prime to $c$.

Bless you
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    What if we don't have the book? – Eoin May 25 '15 at 03:21
  • I'll edit the question then to show you what all those letters mean. Thank you for your time – Bless you May 25 '15 at 03:21
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    Is it possible to upvote the title to this question? – user134824 May 25 '15 at 04:59
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    The title is very funny. On the other hand, it seems to me that most people find that things actually get worse once you leave Lang's library. Even in this subject I think Cassels-Frölich and Neukirch are brutal in their own ways. – Hoot Jun 02 '15 at 22:42

1 Answers1

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$\newcommand{\fraka}{\mathfrak{a}} \newcommand{\frakA}{\mathfrak{A}} \newcommand{\frakc}{\mathfrak{c}} \newcommand{\frakN}{\mathfrak{N}} \newcommand{\frakb}{\mathfrak{b}}$ Let $\alpha \in J_\mathfrak{c}$ be an idèle that's relatively prime to $\mathfrak{c}$, such that $\alpha = aN_{K/k}(\beta)$ for some $\beta\in J_K$ and $a\in k^*$. We want to show $\psi(\alpha)\in P_\frakc \frakN_\frakc$.

We have $\psi(a)\in P$, and $\psi(N_{K/k}(\beta))\in \frakN$, so $\psi(\alpha) \in P\frakN$. Furthermore, since $\alpha\in J_\frakc$, we also have $\psi(\alpha) \in I(\frakc)$. It's then enough to show that $I(\frakc)\cap P\frakN \subset P_\frakc \frakN_\frakc$.

So let $\frakb\in \frakN$ and $(a)\in P$ be such that $(a)\frak{b}$ is relatively prime to $\frakc$ (e.g. $\frak{b} = \psi(\beta)$ and $(a)=\psi(a)$ in the above.)

Now by the approximation theorem, there exists some $\gamma\in K^\times$ such that $N_{K/k}(\gamma)\frakb$ is prime to $\frakc$. For instance for all $v|\frakc$, $v<\infty$, if $w_1,\cdots,w_r$ are the places over $v$, we can pick $\gamma$ such that $w_1(\gamma)=-v(\frakb)$ and $w_i(\gamma)=0$ for $i>1$. Now let $a' = aN_{K/k}(\gamma^{-1})$ and $\frakb' = N_{K/k}(\gamma)\frakb$, so that $a'\frakb'=a\frakb$. Then $\frakb'$ and $a'\frakb'$ are relatively prime to $\frakc$, so $(a')$ is also relatively prime to $\frakc$. In other words $(a')\in P_\frakc$, $\frakb'\in \frakN_\frakc$ and so $a\frakb=a'\frakb'\in P_\frakc\frakN_\frakc$.

Zavosh
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