Inertia degree of primes in p-adic extensions

Question

I'm reading through some number theory and ran across a theorem where the proofs referenced were incomprehensible to me, and I was hoping there might be a simpler proof than slogging through another $\approx$100 pages of background.

The statement is that if you have a finite extension $K$ of the $p$-adic rationals $\mathbb{Q}_p$, and $\pi$ is a uniformizer of $\mathcal{O}_K$, then

$$[\mathcal{O}_K/\pi\mathcal{O}_K : \mathbb{Z}_p/p\mathbb{Z}_p] = v_p(\vert N(\pi)\vert)$$

In the non-$p$-adic case, where $K'$ is some extension of $\mathbb{Q}$ and where $\pi \vert p$, showing that $[\mathcal{O}_{K'}/\pi\mathcal{O}_{K'}:\mathbb{Z}/p\mathbb{Z}] = v_p(\vert N(\pi)\vert)$ is more straightforward to me: Since $\mathcal{O}_{K'}$ is a $\mathbb{Z}$-module, I can imagine it as $\mathbb{Z}^n$, and then the quotient by $\pi$ is like a lattice, with $\pi$ as matrix giving a basis of that lattice, and the volume of the fundamental parallelepiped is the determinant of that matrix, which is $N(\pi)$. The number of integer points in the fundamental parallelepiped will be equal to its volume, showing that $\vert \mathcal{O}_{K'}/\pi\mathcal{O}_{K'}\vert = \vert N(\pi)\vert$. Since $\vert\mathbb{Z}/p\mathbb{Z}\vert =p$, that gives the result.

But this doesn't seem to extend to the $p$-adic case. $\mathbb{Z}_p^n$ doesn't look like a lattice to me, and even if it were, I'm not sure how to relate the volume to the number of points in it.

Maybe I could argue that $\mathbb{Z}$ is a subring of $\mathbb{Z}_p$, and that $\mathbb{Z}_p/p\mathbb{Z}_p$ can take representatives of each coset from $\mathbb{Z}$, so I can treat $\mathcal{O}_K/\pi\mathcal{O}_K$ as a $\mathbb{Z}$-module. But then I can't seem to treat $\pi$ as a linear operator without causing problems with the determinant.

Or maybe, since $\mathbb{Z}_p/p\mathbb{Z}_p \cong \mathbb{Z}/p\mathbb{Z}$, I could show that $\mathcal{O}_K/\pi\mathcal{O}_K\cong \mathcal{O}_{K'}/\pi\mathcal{O}_{K'}$. Certainly I can say that every linear operator over $\mathbb{Z}_p/p\mathbb{Z}_p$ is a linear operator over $\mathbb{Z}/p\mathbb{Z}$, but I'm not sure that every element of $\mathcal{O}_K$ will form a linear operator that is equal to one from $\mathcal{O}_{K'}$. In fact this is almost certainly false, so the quotient by $\pi$ must be crucial, but I don't know how to incorporate this fact.

Is there a relatively simple proof of this, or any way to complete my attempts?

Answer 1

The problem is that you haven't really told us what you think is too much theory. I'll give a proof below with some annotations, hopefully this is enough.

Consider the function $\phi:K^\times \to \mathbb{R}$ given by

$$\phi(x):=\left|N(x)\right|_p^{\frac{1}{n}}$$

where $N:L^\times\to \mathbb{Q}_p^\times$ is the usual field norm, $n=[K:\mathbb{Q}_p]$, and $|\cdot|_p$ is the usual $p$-adic absolute value.

Since $\pi$ is a uniformizer of $K$ we know that $p=\pi^e u$ for some unit $u$ of $K$ $\color{red}{(1)}$. So then, we have that

$$\phi(\pi)^e\overset{{\color{red}{(2)}}}{=}\phi(\pi^e)\overset{{\color{red}{(3)}}}{=}\phi(\pi^e u)=\phi(p)=|p^n|_p^{\frac{1}{n}}=p^{-1}$$

From this, we see that

$$|N(\pi)|_p^{\frac{1}{n}}=p^{\frac{-1}{e}}$$

which implies that

$$|N(\pi)|=p^{\frac{-n}{e}}\overset{\color{red}{(4)}}{=}p^{-f}$$

But,

$$|N(\pi)|=p^{-v_p(N(\pi))}$$

so we deduce that

$$v_p(N(\pi))=f$$

where $f=[\mathcal{O}_K/\pi\mathcal{O}_K:\mathbb{F}_p]$.

$\color{red}{(1)}$ This is, essentially, a definition. Namely, we know that $p=\pi^e u$ for some $e$, and so I'm just naming this power--it's e.

$\color{red}{(2)}$ I am using the fact here that $\phi$ is multiplicative. This is clear since $N$ is multiplicative (since it's just a determinant!) and $|\cdot|_p$ is multiplicative.

$\color{red}{(3)}$ I've snuck in a sneaky $u$ here. The reason this is ok is that $\phi(u)=1$. Indeed, note that $N$ sends elements of $\mathcal{O}_K$ into elements of $\mathbb{Z}_p$. The reason is that $N(x)$ can be thought of as the constant term of the minimal polynomial for $x$, but if $x$ is $\mathcal{O}_K$ these minimal polynomial coefficients are in $\mathbb{Z}_p$. So, since $u$ is a unit it has an inverse $v$ in $\mathcal{O}_K$ and since $1=N(uv)=N(u)N(v)$ you know that $N(u)$ is not just in $\mathbb{Z}_p$ but $\mathbb{Z}_p^\times$ and thus $|N(u)|=1$ so $\phi(u)=1$.

$\color{red}{(4)}$ This is the only truly hard part I'm using. Namely, I am using the fac that $n=ef$. The proof of this is pretty simple though. Namely, note that $\mathcal{O}_K$ is a torsion free $\mathbb{Z}_p$ module and since $\mathbb{Z}_p$ is a PID this implies (by the structure theory of modules over PIDs) that $\mathcal{O}_K=\mathbb{Z}_p^r$ for some $r$. But, note that

$$\mathbb{Q}_p^r\cong \mathbb{Z}_p^r\otimes_{\mathbb{Z}_p}\mathbb{Q}_p\cong \mathcal{O}_K\otimes_{\mathbb{Z}_p}\mathbb{Q}_p=K$$

as $\mathbb{Q}_p$-modules. But, since $K=\mathbb{Q}_p^n$ where $n=[K:\mathbb{Q}_p]$ we know that $n=r$. So, we see that $\mathcal{O}_K$ is a free $\mathbb{Z}_p$-module of rank $n=[K:\mathbb{Q}_p]$. Thus, we see that

$$\mathcal{O}_K/p\mathcal{O}_K\cong \mathcal{O}_K\otimes_{\mathbb{Z}_p}\mathbb{Z}_p/p\mathbb{Z}_p$$

is a free module over $\mathbb{Z}_p/p\mathbb{Z}_p\cong \mathbb{F}_p$ of rank $n$. But, we can count the dimension of the $\mathbb{F}_p$-module $\mathcal{O}_K/p\mathcal{O}_K$ in another way. Namely, writing $p=\pi^e u$ for $u$ a unit it's easy to see using the PIDness that $\mathcal{O}_K/p\mathcal{O}_K$ has a filtration

$$\mathcal{O}_K/p\mathcal{O}_K\supseteq \pi\mathcal{O}_K/p\mathcal{O}_K\supseteq\cdots p^{e-1}\mathcal{O}_K/p\mathcal{O}_K\supseteq p^e\mathcal{O}_K/p\mathcal{O}_K=0$$

and that each $i$ we have an isomorphism of $\mathbb{F}_p$-modules

$$(\pi^i\mathcal{O}_K/p\mathcal{O}_K)/(\pi^{i+1}\mathcal{O}_K/p\mathcal{O}_K)\cong \mathcal{O}_K/\pi\mathcal{O}_K$$

So, since this filtration has length $e$ we see that

$$n=\dim_{\mathbb{F}_p} \mathcal{O}_K/p\mathcal{O}_K=e\dim_{\mathbb{F}_p}\mathcal{O}_K/\pi\mathcal{O}_K=ef$$

as desired.

Answer 2

Here is a solution that is similar in many ways to Alex's solution, but the (second) way I explain $n = ef$ will be quite different.

The definition of $e$ is that $p = \pi^e u$ for some $u \in \mathcal O_K^\times$. Taking the norm from $K$ down to $\mathbf Q_p$ of both sides, $$ p^n = {\rm N}(\pi)^e{\rm N}(u), $$ where ${\rm N} = {\rm N}_{K/\mathbf Q_p}$. Because the $p$-adic absolute value on $K$ is defined by the rule $$ |x| = |{\rm N}(x)|_p^{1/n} $$ for all $x$ in $K$ (the function $|\cdot|$ is the unique extension of $|\cdot|_p$ as an absolute value from $\mathbf Q_p$ to $K$).

Let's recall why the norm maps units to units: ${\rm N}(\mathcal O_K^\times) \subset \mathbf Z_p^\times$. From $u \in \mathcal O_K$ we have $|{\rm N}(u)|_p = |u|^n = 1$. Thus $$ p^n = {\rm N}(\pi)^e{\rm N}(u) \Longrightarrow |p^n| = |{\rm N}(\pi)|^e \Longrightarrow \left(\frac{1}{p}\right)^n = |{\rm N}(\pi)|^e $$ since $|p| = 1/p$. From the fundamental identity $ef = n$, which is how $e$, $f$, and $n$ are related for $p$-adic fields, $(1/p)^{ef} = |{\rm N}(\pi)|^e$. Take $e$th roots of both sides and we're left with $|{\rm N}(\pi)| = (1/p)^f$.

Why does $n = ef$? Surely you saw this already, since it's hard to imagine how you could be reading about extensions of $p$-adic fields without having this fact being proved somewhere in the treatment. An argument for $n = ef$ that is different from the approach Alex describes (or at least it feels different) is to build a $\mathbf Q_p$-basis of $K$ of size $ef$, and since that dimension is $n$, we'd get $ef = n$. This is proved in Gouvea's book "$p$-adic Numbers" and the idea is this: $\mathcal O_K/(\pi)$ is a vector space over $\mathbf F_p = \mathbf Z_p/(p)$ of dimension $f$ (that's the definition of $f$), so pick an $\mathbf F_p$-basis of $\mathcal O_K/(\pi)$ and lift up the terms in the basis of $\mathcal O_K$, say to $\omega_1, \ldots, \omega_f$. Gouvea shows that the set $\{\omega_i\pi^j\}$ for $1 \leq i \leq f$ and $0 \leq j \leq e-1$ is a $\mathbf Q_p$-basis of $K$. (He really shows first that it is a $\mathbf Z_p$-basis of $\mathcal O_K$.) So that gives a $\mathbf Q_p$-basis of $K$ of size $ef$.

An argument for $n = ef$ that in some ways feels more different from what Alex wrote can be based on measure theory. (You did express an interest in a volume-theoretic approach since that is how you think about some norm results in number fields.) Since $K$ as an additive group is locally compact and Hausdorff, it has a Haar measure $\mu$. Normalize this Haar measure to give the compact open subset $\mathcal O_K$ measure 1. (Haar measure is always nonzero on open sets and finite on compact sets, so it has a positive finite value on a compact open subset like $\mathcal O_K$.) The size of the residue field $\mathcal O_K/(\pi)$ is $p^f$. Let $\alpha_1, \ldots, \alpha_{p^f}$ be coset representatives for that residue field: $$ \mathcal O_K = \bigcup_{1 \leq i \leq p^f} (\alpha_i + \pi\mathcal O_K), $$ where the union is disjoint. Because Haar measure on the additive group $K$ is invariant under additive translations, and that union above is disjoint, $$ 1 = \mu(\mathcal O_K) = \sum_{i=1}^{p^f} \mu(\alpha_i + \pi\mathcal O_K) = p^f\mu(\pi\mathcal O_K). $$ Therefore $\mu(\pi\mathcal O_K) = 1/p^f$.

Now we're going to calculate the Haar measure of $\pi\mathcal O_K$ in a second way. The function $\mu_\pi(A) = \mu(\pi A)$ for all Borel sets $A$ is a Haar measure on $K$: it is invariant under left translation since multiplication distributes over addition, and it satisfies the regularity conditions for Haar measure since multiplication by a nonzero number is a homeomorphism of $K$. Haar measure is determined up to scaling by a positive constant, so there is some $c > 0$ such that $\mu_\pi = c \mu$: $\mu(\pi A) = c\mu(A)$ for all Borel sets $A$. Taking $A = \mathcal O_K$, $\mu(\pi\mathcal O_K) = c\mu(\mathcal O_K) = c$. Thus $c = \mu(\pi\mathcal O_K) = 1/p^f$. That means $\mu(\pi A) = (1/p^f)\mu(A)$ for all Borel sets $A$.

Let's compute the measure of $p\mathcal O_K$ in two ways. Since $p = \pi^e u$, $p\mathcal O_K = \pi^e\mathcal O_K$, so $$ \mu(p\mathcal O_K) = \mu(\pi^e\mathcal O_K). $$ We saw above that the effect on Haar measure of multiplying by $\pi$ is to scale Haar measure by $1/p^f$. Therefore
$$ \mu(p\mathcal O_K) = \mu(\pi^e\mathcal O_K) = \left(\frac{1}{p^f}\right)^e\mu(\mathcal O_K) = \frac{1}{p^{ef}} $$ since $\mathcal O_K$ has Haar measure $1$.

Since $\mathcal O_K$ is a free $\mathbf Z_p$-module of rank $n$, $[\mathcal O_K:p\mathcal O_K] = p^n$. Therefore $\mathcal O_K$ is a disjoint union of $p^n$ cosets of $p\mathcal O_K$, and $\mu(\alpha + p\mathcal O_K) = \mu(p\mathcal O_K)$ for all $\alpha$, so $$ \mu(\mathcal O_K) = p^n \mu(p\mathcal O_K). $$ Since $\mathcal O_K$ has measure $1$, $\mu(p\mathcal O_K) = 1/p^n$. Comparing this value with what we found in the previous paragraph, $1/p^n = (1/p)^{ef}$, so $n = ef$.

Since this argument dealt a lot with cosets of $p\mathcal O_K$ and $\pi\mathcal O_K$ in $\mathcal O_K$, as Alex's argument does, his approach to $n = ef$ and the approach here with measure theory clearly have some underlying similarities even if the surface details are not the same.

It was not essential to normalize the Haar measure of $\mathcal O_K$ to be $1$: whatever value it has must be finite and positive since $\mathcal O_K$ is compact and open, and if we did not normalize the Haar measure by setting $\mu(\mathcal O_K) = 1$ then we'd be dragging a scaling factor $\mu(\mathcal O_K)$ around in many calculations, so at the end we'd have $(1/p^n)\mu(\mathcal O_K) = (1/p^{ef})\mu(\mathcal O_K)$, and then cancelling the unspecified finite positive number $\mu(\mathcal O_K)$ from both sides we'd get $1/p^n = 1/p^{ef}$, so $n = ef$.