# Proving the index formula of the norm residue group without $\mathfrak{p}$-adic completion

Let $K$ be an algebraic number field. Let $A$ be the ring of integers in $K$. Let $\mathfrak{p}$ be a prime ideal of $K$. Let $A_\mathfrak{p}$ be the localization of $A$ at $\mathfrak{p}$. Let $A_\mathfrak{p}^\times$ be the group of invertible elements of $A_\mathfrak{p}$. Elements of $A_\mathfrak{p}^\times$ are called $\mathfrak{p}$-units.

Let $I$ be a non-zero ideal of $A$. An element $\alpha$ of $A$ is called an $I$-unit if $\alpha$ is a $\mathfrak{p}$-unit for every prime divisor $\mathfrak{p}$ of $I$.

Let $L$ be a finite etension of $K$. Let $B$ be the ring of integers in $L$. Let $\mathfrak{p}$ be a prime ideal of $K$. Let $\lambda > 0$ be an integer. Let $\alpha \in A_\mathfrak{p}^\times$. If there exists a $\mathfrak{p}B$-unit $\Gamma \in L$ such that $\frac{\alpha}{N_{L/K}(\Gamma)} \equiv 1$ (mod $\mathfrak{p}^\lambda A_\mathfrak{p})$, $\alpha$ is called a norm residue of $\mathfrak{p}^\lambda$ with respect to $L/K$. Such $\alpha$ form a subgroup of $A_\mathfrak{p}^\times$. This group is called the norm residue group of $\mathfrak{p}^\lambda$ with respect to $L/K$. We denote this group by $\mathfrak{N}(\mathfrak{p}^\lambda)$.

The following proposition is crucial in the classical proof of the global class field theory. The usual proof uses $\mathfrak{p}$-adic completion.

Proposition Let $L$ be a finite cyclic extension of $K$. Let $\mathfrak{p}$ be a prime ideal of $K$. Let $e$ be the ramification index of $\mathfrak{p}$ in $L$. Then there exists an integer $\lambda_0 > 0$ such that $[A_\mathfrak{p}^\times \colon \mathfrak{N}(\mathfrak{p}^\lambda)] = e$ for all integer $\lambda \ge \lambda_0$.

My question How do we prove this without using the $\mathfrak{p}$-adic completion?

Remark I think Hasse's "Bericht" proves this without using the $\mathfrak{p}$-adic completion. However, my German is not good.

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 I asked this question in MathOverflow. mathoverflow.net/questions/107137/… – Makoto Kato Sep 14 '12 at 1:27