# Ideal theoretic proof of the first inequality of global class field theory

In the old days(namely in the 1920s), global class field theory was stated and proved without using $\mathfrak p$-adic fields. I am interested in their methods, but unfortunately they were written in German and they are difficult to decipher at least for me. In particular, I would like to prove the first inequality(see below) of global class field theory without using $\mathfrak p$-adic fields.

We use the terminology of Milne's online course note: Class Field Theory(2013). Let $k$ be an algebraic number field(we always assume $[k : \mathbb Q] \lt \infty$). Let $K/k$ be a finite cyclic extension of degree $n$. Let $\mathfrak m$ be a modulus(Milne p.144) of $k$ such that $\mathfrak p | \mathfrak m$ if and only if $\mathfrak p$ is ramified(for the definition of ramification of a real infinite prime, see Milne p.4) in $K/k$. We denote by $I_k(\mathfrak m)$ the group of nonzero fractional ideals of $k$ that are relatively prime to $\mathfrak m_0$, where $\mathfrak m_0$ is the finite part of $\mathfrak m$. We denote by $I_K(\mathfrak m)$ the group of nonzero fractional ideals of $K$ that are relatively prime to $\mathfrak m_0 O_K$, where $O_K$ is the ring of integers in $K$. We denote $i(k_{\mathfrak m, 1})$ (Milne p.145) by $R(\mathfrak m)$. Hence $I_k(\mathfrak m)/R(\mathfrak m)$ is the ray class group modulo $\mathfrak m$ (Milne p.145).

The theorem of the first inequality of global class field theory states that there exists such $\mathfrak m$ such that $(I_k(\mathfrak m) : R(\mathfrak m)N_{K/k}(I_K(\mathfrak m))) \ge n$.

Reading a proof of this theorem in Takagi's Algebraic Number Theory(in Japanese), I found that the aforementioned problem is reduced to proving the following theorem without using $\mathfrak p$-adic analysis.

Theorem Let $K/k$ be as above. Let $\mathfrak p$ be a prime ideal of $k$ which is ramified in $K/k$. We denote by $O_{k, \mathfrak p}$ the localization of $O_k$ at $\mathfrak p$. Then there exists an integer $r_0 \ge 0$ such that $(k^{\times}(\mathfrak p) : U_{\mathfrak p}^{(i)} N_{K/k}(K^{\times}(\mathfrak p O_K)) = e$ for all $i \ge r_0$, where $k^{\times}(\mathfrak p)$ is the set of nonzero elements of $k$ that are relatively prime to $\mathfrak p$, $K^{\times}(\mathfrak p O_K)$ is the set of nonzero elements of $K$ that are relatively prime to $\mathfrak p O_K$, $U_{\mathfrak p}^{(i)} = \{\alpha \in O_{k, \mathfrak p}\ |\ \alpha \equiv 1$ (mod $\mathfrak p^i O_{k, \mathfrak p})\}$, and $e$ is the ramification index of $\mathfrak p$.

My question: how do we prove this without using $\mathfrak p$-adic field?

Remark For the reader's convenience, I will write a sketch of the usual proof of the theorem using $\mathfrak p$-adic analysis. Let $Z$ be the decomposition group of $\mathfrak p$ with respect to $K/k$, $K^Z$ be the corresponding subfield of $K/k$. By replacing $K$ by $K^Z$, we may assume that there exists only one prime ideal $\mathfrak P$ of $K$ lying over $\mathfrak p$. Namely $\mathfrak p = \mathfrak P^e$. By taking $\mathfrak p$-adic completion, we may assume that $k$ is the $\mathfrak p$-adic field and that $K$ is the $\mathfrak P$-adic field. Hence it sufices to prove that $(U_k : U_k^{(i)} N_{K/k}(U_K) = e$ for all $i \ge r_0$, where $U_k$ and $U_K$ are the groups of units in $k$ and $K$ respectively and $U_k^{(i)} = \{\alpha \in O_k\ |\ \alpha \equiv 1$ (mod $\mathfrak p^i )\}$. It is well known that $U_k^{(i)} \subset (k^{\times})^n$ for sufficiently large $i$. Hence $U_k^{(i)} \subset N_{K/k}(U_K)$ for sufficiently large $i$. Hence it sufices to prove that $(U_k : N_{K/k}(U_K)) = e$. But this is well known(see for example Lang's Algebraic Number Theory).

• Take a look at the appendix in Classical Invitation to Algebraic Numbers and Class Field Theory by Cohn. It accurately represents the early state of things. – Rene Schipperus Dec 31 '15 at 0:07
• @ReneSchipperus Thanks. But it uses the $\mathfrak p$-adic method to prove the theorem. – Makoto Kato Dec 31 '15 at 1:12
• ok so dont read it... – Rene Schipperus Dec 31 '15 at 1:56
• @ReneSchipperus Don't get me wrong. I thanked you for the info. – Makoto Kato Jan 1 '16 at 22:50
• @A.P. One reason is that ideals are more concrete and elementary than $\mathfrak p$-adic numbers. For example, ideals are more suitable in concrete calculations. Another reason is that knowing various proofs of a theorem is usually useful for better understanding of the theorem. Finally I'm curious just like you. – Makoto Kato Jan 5 '16 at 0:50