# Why is the order of $A/m$ is $p^f$?

Let $K/\mathbb{Q}_p$ be a finite Galois extension, $v:\mathbb{Q}_p^\times \rightarrow \mathbb{Z}$ the p-adic valuation and $N:K^\times \rightarrow \mathbb{Q}_p^\times$ the norm given by $N(x) = \prod_{\sigma \in G} \sigma(x)$.

The composition of homomorphisms $v \circ N : K^\times \rightarrow \mathbb{Z}$ is nonzero with some image $f\mathbb{Z}$. We define $w: K^\times \longrightarrow \mathbb{Z}$ by $w := \frac{1}{f}v \circ N$. Then $w$ is a discrete valuation on $K$. Moreover, $w$ is the unique discrete valuation on $K$ which extends $v$.

Now, let $A=\{ x \in K\ |\ w(x) \ge 0 \}$ be the valuation ring of $K$ and let $m = \{ x \in K \ | \ w(x) > 0 \}$ be the maximal ideal of $A$.

I want show that the order of residue field $A/m$ is $p^f$, but I can only show that $A/m$ is a finite field.

Any hints on how to prove this?

This is an exercise on page xxix in Boston's book "The Proof of Fermat's last theorem", available in http://www.math.wisc.edu/~boston/869.pdf .

Thanks.

• This is usually proved in books on ANT. Did you look into J. Milne's notes? – Dietrich Burde Oct 13 '17 at 20:30
• In J. Milne's notes I have found only that A/m is finite. – Adler Marques Oct 13 '17 at 20:44

The key point is the formula (in loose words) : degree = ramification index $\times$ inertia index. This is shown in any ANT textbook, but generally for Dedekind domains. Here, for discrete valuation rings, the situation is much simpler since we deal only with a single prime ideal which is moreover principal. I outline the proof for your convenience.
Sticking to your notations, the inertia index is the $f$ which appears in your definition of the valuation $w$, and the ramification index is $e=w(p)$, so that $p=u\pi^{e}$, where $u$ is a unit and $\pi$ a uniformizer of $K$ . Taking norms and valuations in $\mathbf Q_p$ we get immediately that $n=ef$, where $n=[K:\mathbf Q_p]$. Let $p^{f'}$ be the order of the residual field $k=A/\mathfrak m$. It remains to show that $f'=f$. The vector space $A/pA$ over $\mathbf F_p$ (=the residual field of $\mathbf Q_p$) has the sequence of quotient spaces $A/\mathfrak m, \mathfrak m /{\mathfrak m}^2,..., {\mathfrak m}^{e-1}/{\mathfrak m}^e$, and moreover all the quotients ${\mathfrak m}^{n-1}/{\mathfrak m}^n$ are isomorhic to $k$ via multiplication by $\pi ^{n-1}$, so that $dim_k A/pA = ef'$. On the other hand, $A$ is a free $\mathbf Z_p$-module of rank $n$, so that $n=ef'$. This shows $f'=f$ as desired.
NB : (a) The Galois hypothesis is not needed; (b) The proof above carries over to any finite extension of $p$-adic local fields; (c) The usual formula for an extension of Dedekind rings (which involves a sum over all prime ideals $\mathfrak Q$ above a given prime ideal $\mathfrak p$) consists in using localization w.r.t. $\mathfrak p$ to reduce to the case of a single prime ideal.