Finite Residue Field Case

Let $K$ be a field with a non-archimedean valuation $v=\left\vert \cdot \right\vert$. Recall that the set of $a\in K$ with $\left\vert a\right\vert\leq 1$ forms a ring $\O$, the ring of integers for $v$. The set of $u\in K$ with $\left\vert u\right\vert=1$ are a group $U$ under multiplication, the group of units for $v$. Finally, the set of $a\in K$ with $\left\vert a\right\vert<1$ is a maximal ideal $\mathfrak{p}$, so the quotient ring $\O /\mathfrak{p}$ is a field. In this section we consider the case when $\O /\mathfrak{p}$ is a finite field of order a prime power $q$. For example, $K$ could be $\mathbf{Q}$ and $\left\vert \cdot \right\vert{}$ could be a $p$-adic valuation, or $K$ could be a number field and $\left\vert \cdot \right\vert{}$ could be the valuation corresponding to a maximal ideal of the ring of integers. Among other things, we will discuss in more depth the topological and measure-theoretic nature of the completion of $K$ at $v$.

Suppose further for the rest of this section that $\left\vert \cdot \right\vert{}$ is discrete. Then by Lemma 15.2.8, the ideal $\mathfrak{p}$ is a principal ideal $(\pi)$, say, and every $a\in K$ is of the form $a=\pi^n\varepsilon$, where $n\in\mathbf{Z}$ and $\varepsilon \in U$ is a unit. We call

$$n = \ord (a) = \ord _\pi(a) = \ord _\mathfrak{p}(a) = \ord _v(a)$$

the ord of $a$ at $v$. (Some authors, including me (!) also call this integer the of $a$ with respect to $v$.) If $\mathfrak{p}=(\pi')$, then $\pi/\pi'$ is a unit, and conversely, so $\ord (a)$ is independent of the choice of $\pi$.

Let $\O _v$ and $\mathfrak{p}_v$ be defined with respect to the completion $K_v$ of $K$ at $v$.

Lemma 17.1.1   There is a natural isomorphism

$$\varphi :\O _v/\mathfrak{p}_v \to \O /\mathfrak{p},$$

and $\mathfrak{p}_v = (\pi)$ as an $\O _v$-ideal.

Proof. We may view $\O _v$ as the set of equivalence classes of Cauchy sequences $(a_n)$ in $K$ such that $a_n\in \O$ for $n$ sufficiently large. For any $\varepsilon$, given such a sequence $(a_n)$, there is $N$ such that for $n,m\geq N$, we have $\left\vert a_n-a_m\right\vert<\varepsilon$. In particular, we can choose $N$ such that $n,m\geq N$ implies that $a_n\equiv a_m\pmod{\mathfrak{p}}$. Let $\varphi ((a_n)) = a_N\pmod{\mathfrak{p}}$, which is well-defined. The map $\varphi$ is surjective because the constant sequences are in $\O _v$. Its kernel is the set of Cauchy sequences whose elements are eventually all in $\mathfrak{p}$, which is exactly $\mathfrak{p}_v$. This proves the first part of the lemma. The second part is true because any element of $\mathfrak{p}_v$ is a sequence all of whose terms are eventually in $\mathfrak{p}$, hence all a multiple of $\pi$ (we can set to 0 a finite number of terms of the sequence without changing the equivalence class of the sequence). $\qedsymbol$

Assume for the rest of this section that $K$ is complete with respect to $\left\vert \cdot \right\vert$.

Lemma 17.1.2   Then ring $\O$ is precisely the set of infinite sums

$$a = \sum_{j=0}^{\infty} a_j \cdot \pi^j$$ (17.1)

where the $a_j$ run independently through some set $\mathcal{R}$ of representatives of $\O$ in $\O /\mathfrak{p}$.

By (17.1.1) is meant the limit of the Cauchy sequence $\sum_{j=0}^n a_j\cdot \pi^j$ as $j\to\infty$.

Proof. There is a uniquely defined $a_0\in \mathcal{R}$ such that $\left\vert a-a_0\right\vert<1$. Then $a' = \pi^{-1}\cdot (a-a_0) \in \O$. Now define $a_1\in \mathcal{R}$ by $\left\vert a'-a_1\right\vert<1$. And so on. $\qedsymbol$

Example 17.1.3   Suppose $K=\mathbf{Q}$ and $\left\vert \cdot \right\vert=\left\vert \cdot \right\vert _p$ is the $p$-adic valuation, for some prime $p$. We can take $\mathcal{R}=\{0,1,\ldots, p-1\}$. The lemma asserts that

$$\O =\mathbf{Z}_p = \left\{ \sum_{j=0}^{\infty} a_n p^n : 0\leq a_n\leq p-1\right\}.$$

Notice that $\O$ is uncountable since there are $p$ choices for each $p$-adic ``digit''. We can do arithmetic with elements of $\mathbf{Z}_p$, which can be thought of ``backwards'' as numbers in base $p$. For example, with $p=3$ we have

$$(1+2\cdot 3 + 3^2 + \cdots ) + (2 + 2\cdot 3 + 3^2 + \cdots )$$

$$= 3+4\cdot 3 + 2\cdot 3^2 + \cdots$$

$$= 0 + 2\cdot 3 + 3\cdot 3 + 2\cdot 3^2 + \cdots$$

$$= 0 + 2\cdot 3 + 0\cdot 3^2 + \cdots$$

Basic arithmetic with the $p$-adics in is really weird (even weirder than it was a year ago... There are presumably efficiency advantages to using the formalization, and it's supposed to be better for working with extension fields. But I can't get it to do even the calculation below in a way that is clear.) In PARI (gp) the $p$-adics work as expected:

   ? a = 1 + 2*3 + 3^2 + O(3^3);
   ? b = 2 + 2*3 + 3^2 + O(3^3); 
   ? a+b
   %3 = 2*3 + O(3^3)
   ? sqrt(1+2*3+O(3^20))  
   %5 = 1 + 3 + 3^2 + 2*3^4 + 2*3^7 + 3^8 + 3^9 + 2*3^10 + 2*3^12 
          + 2*3^13 + 2*3^14 + 3^15 + 2*3^17 + 3^18 + 2*3^19 + O(3^20)
   ? 1/sqrt(1+2*3+O(3^20))   
   %6 = 1 + 2*3 + 2*3^2 + 2*3^7 + 2*3^10 + 2*3^11 + 2*3^12 + 2*3^13 
          + 2*3^14 + 3^15 + 2*3^16 + 2*3^17 + 3^18 + 3^19 + O(3^20)

Theorem 17.1.4   Under the conditions of the preceding lemma, $\O$ is compact with respect to the $\left\vert \cdot \right\vert{}$-topology.

Proof. Let $V_\lambda$, for $\lambda$ running through some index set $\Lambda$, be some family of open sets that cover $\O$. We must show that there is a finite subcover. We suppose not.

Let $\mathcal{R}$ be a set of representatives for $\O /\mathfrak{p}$. Then $\O$ is the union of the finite number of cosets $a+\pi\O$, for $a\in \mathcal{R}$. Hence for at lest one $a_0\in \mathcal{R}$ the set $a_0+\pi \O$ is not covered by finitely many of the $V_\lambda$. Then similarly there is an $a_1\in \mathcal{R}$ such that $a_0 + a_1\pi + \pi^2\O$ is not finitely covered. And so on. Let

$$a = a_0 + a_1\pi + a_2 \pi^2 + \cdots \in \O .$$

Then $a\in V_{\lambda_0}$ for some $\lambda_0\in\Lambda$. Since $V_{\lambda_0}$ is an open set, $a+\pi^J\cdot \O\subset V_{\lambda_0}$ for some $J$ (since those are exactly the open balls that form a basis for the topology). This is a contradiction because we constructed $a$ so that none of the sets $a+\pi^n\cdot \O$, for each $n$, are not covered by any finite subset of the $V_{\lambda}$. $\qedsymbol$

Definition 17.1.5 (Locally compact)   A topological space $X$ is at a point $x$ if there is some compact subset $C$ of $X$ that contains a neighborhood of $x$. The space $X$ is locally compact if it is locally compact at each point in $X$.

Corollary 17.1.6   The complete local field $K$ is locally compact.

Proof. If $x\in K$, then $x \in C=x+\O$, and $C$ is a compact subset of $K$ by Theorem 17.1.4. Also $C$ contains the neighborhood $x+\pi\O = B(x,1)$ of $x$. Thus $K$ is locally compact at $x$. $\qedsymbol$

Remark 17.1.7   The converse is also true. If $K$ is locally compact with respect to a non-archimedean valuation $\left\vert \cdot \right\vert{}$, then

1. $K$ is complete,
2. the residue field is finite, and
3. the valuation is discrete.

For there is a compact neighbourhood $C$ of 0. Let $\pi$ be any nonzero with $\left\vert\pi\right\vert<1$. Then $\pi^n\cdot \O\subset C$ for sufficiently large $n$, so $\pi^n\cdot \O$ is compact, being closed. Hence $\O$ is compact. Since $\left\vert \cdot \right\vert$ is a metric, $\O$ is sequentially compact, i.e., every fundamental sequence in $\O$ has a limit, which implies (1). Let $a_\lambda$ (for $\lambda\in\Lambda$) be a set of representatives in $\O$ of $\O /\mathfrak{p}$. Then $\O _{\lambda} = \{z : \left\vert z-a_{\lambda}\right\vert<1\}$ is an open covering of $\O$. Thus (2) holds since $\O$ is compact. Finally, $\mathfrak{p}$ is compact, being a closed subset of $\O$. Let $S_n$ be the set of $a\in K$ with $\left\vert a\right\vert<1-1/n.$ Then $S_n$ (for $1\leq n < \infty$) is an open covering of $\mathfrak{p}$, so $\mathfrak{p}=S_n$ for some $n$, i.e., (3) is true.

If we allow $\left\vert \cdot \right\vert{}$ to be archimedean the only further possibilities are $k=\mathbf{R}$ and $k=\mathbf{C}$ with $\left\vert \cdot \right\vert{}$ equivalent to the usual absolute value.

We denote by $K^+$ the commutative topological group whose points are the elements of $K$, whose group law is addition and whose topology is that induced by $\left\vert \cdot \right\vert$. General theory tells us that there is an invariant Haar measure defined on $K^+$ and that this measure is unique up to a multiplicative constant.

Definition 17.1.8 (Haar Measure)   A on a locally compact topological group $G$ is a translation invariant measure such that every open set can be covered by open sets with finite measure.

Lemma 17.1.9   Haar measure of any compact subset $C$ of $G$ is finite.

Proof. The whole group $G$ is open, so there is a covering $U_\alpha$ of $G$ by open sets each of which has finite measure. Since $C$ is compact, there is a finite subset of the $U_\alpha$ that covers $C$. The measure of $C$ is at most the sum of the measures of these finitely many $U_\alpha$, hence finite. $\qedsymbol$

Remark 17.1.10   Usually one defined Haar measure to be a translation invariant measure such that the measure of compact sets is finite. Because of local compactness, this definition is equivalent to Definition 17.1.8. We take this alternative viewpoint because Haar measure is constructed naturally on the topological groups we will consider by defining the measure on each member of a basis of open sets for the topology.

We now deduce what any such measure $\mu$ on $G=K^+$ must be. Since $\O$ is compact (Theorem 17.1.4), the measure of $\O$ is finite. Since $\mu$ is translation invariant,

$$\mu_n = \mu(a + \pi^n \O )$$

is independent of $a$. Further,

$$a + \pi^n\O = \bigcup_{1\leq j\leq q} a + \pi^n a_j + \pi^{n+1}\O ,$$   (disjoint union)

where $a_j$ (for $1\leq j \leq q$) is a set of representatives of $\O /\mathfrak{p}$. Hence

$$\mu_n = q\cdot \mu_{n+1}.$$

If we normalize $\mu$ by putting

$$\mu(\O ) = 1$$

we have $\mu_0 = 1$, hence $\mu_1 = q$, and in general

$$\mu_n = q^{-n}.$$

Conversely, without the theory of Haar measure, we could define $\mu$ to be the necessarily unique measure on $K^+$ such that $\mu(\O )=1$ that is translation invariant. This would have to be the $\mu$ we just found above.

Everything so far in this section has depended not on the valuation $\left\vert \cdot \right\vert$ but only on its equivalence class. The above considerations now single out one valuation in the equivalence class as particularly important.

Definition 17.1.11 (Normalized valuation)   Let $K$ be a field equipped with a discrete valuation $\left\vert \cdot \right\vert$ and residue class field with $q<\infty$ elements. We say that $\left\vert \cdot \right\vert$ is if

$$\left\vert\pi\right\vert = \frac{1}{q},$$

where $\mathfrak{p}=(\pi)$ is the maximal ideal of $\O$.

Example 17.1.12   The normalized valuation on the $p$-adic numbers $\mathbf{Q}_p$ is $\left\vert u\cdot p^n\right\vert = p^{-n}$, where $u$ is a rational number whose numerator and denominator are coprime to $p$.

Next suppose $K=\mathbf{Q}_p(\sqrt{p})$. Then the $p$-adic valuation on $\mathbf{Q}_p$ extends uniquely to one on $K$ such that $\left\vert\sqrt{p}\right\vert^2 = \left\vert p\right\vert = 1/p$. Since $\pi=\sqrt{p}$ for $K$, this valuation is not normalized. (Note that the ord of $\pi=\sqrt{p}$ is $1/2$.) The normalized valuation is $v=\left\vert \cdot \right\vert' = \left\vert \cdot \right\vert^2$. Note that $\left\vert \cdot \right\vert'{p} = 1/p^2$, or $\ord _v(p)=2$ instead of $1$.

Finally suppose that $K=\mathbf{Q}_p(\sqrt{q})$ where $x^2-q$ has not root mod $p$. Then the residue class field degree is $2$, and the normalized valuation must satisfy $\left\vert\sqrt{q}\right\vert = 1/p^2$.

The following proposition makes clear why this is the best choice of normalization.

Theorem 17.1.13   Suppose further that $K$ is complete with respect to the normalized valuation $\left\vert \cdot \right\vert{}$. Then

$$\mu(a + b\O ) = \left\vert b\right\vert,$$

where $\mu$ is the Haar measure on $K^+$ normalized so that $\mu(\O )=1$.

Proof. Since $\mu$ is translation invariant, $\mu(a+b\O ) = \mu(b\O )$. Write $b=u\cdot \pi^n$, where $u$ is a unit. Then since $u\cdot \O =\O$, we have

$$\mu(b\O ) = \mu(u\cdot \pi^n\cdot \O ) = \mu(\pi^n \cdot u\cdot\... ...i^n\cdot \O ) = q^{-n} = \left\vert\pi^n\right\vert = \left\vert b\right\vert.$$

Here we have $\mu(\pi^n\cdot \O ) = q^{-n}$ by the discussion before Definition 17.1.11. $\qedsymbol$

We can express the result of the theorem in a more suggestive way. Let $b\in K$ with $b\neq 0$, and let $\mu$ be a Haar measure on $K^+$ (not necessarily normalized as in the theorem). Then we can define a new Haar measure $\mu_b$ on $K^+$ by putting $\mu_b(E) = \mu(bE)$ for $E\subset K^+$. But Haar measure is unique up to a multiplicative constant and so $\mu_b(E) = \mu(bE) = c\cdot \mu(E)$ for all measurable sets $E$, where the factor $c$ depends only on $b$. Putting $E=\O$, shows that the theorem implies that $c$ is just $\left\vert b\right\vert$, when $\left\vert \cdot \right\vert$ is the normalized valuation.

Remark 17.1.14   The theory of locally compact topological groups leads to the consideration of the dual (character) group of $K^+$. It turns out that it is isomorphic to $K^+$. We do not need this fact for class field theory, so do not prove it here. For a proof and applications see Tate's thesis or Lang's Algebraic Numbers, and for generalizations see Weil's Adeles and Algebraic Groups and Godement's Bourbaki seminars 171 and 176. The determination of the character group of $K^*$ is local class field theory.

The set of nonzero elements of $K$ is a group $K^*$ under multiplication. Multiplication and inverses are continuous with respect to the topology induced on $K^*$ as a subset of $K$, so $K^*$ is a topological group with this topology. We have

$$U_1 \subset U \subset K^*$$

where $U$ is the group of units of $\O\subset K$ and $U_1$ is the group of $1$-units, i.e., those units $\varepsilon \in U$ with $\left\vert\varepsilon -1\right\vert<1$, so

$$U_1 = 1 + \pi\O .$$

The set $U$ is the open ball about 0 of radius $1$, so is open, and because the metric is nonarchimedean $U$ is also closed. Likewise, $U_1$ is both open and closed.

The quotient $K^*/U = \{ \pi^n \cdot U : n \in \mathbf{Z}\}$ is isomorphic to the additive group $\mathbf{Z}^+$ of integers with the discrete topology, where the map is

$$\pi^n\cdot U \mapsto n$$    for $$n\in\mathbf{Z}.$$

The quotient $U/U_1$ is isomorphic to the multiplicative group $\mathbf{F}^*$ of the nonzero elements of the residue class field, where the finite gorup $\mathbf{F}^*$ has the discrete topology. Note that $\mathbf{F}^*$ is cyclic of order $q-1$, and Hensel's lemma implies that $K^*$ contains a primitive $(q-1)$th root of unity $\zeta$. Thus $K^*$ has the following structure:

$$K^* = \{ \pi^n\zeta^m\varepsilon : n\in \mathbf{Z}, m\in\mathbf{Z... ...cong \mathbf{Z} \times \mathbf{Z}/(q-1)\mathbf{Z} \times U_1.$$

(How to apply Hensel's lemma: Let $f(x) = x^{q-1}-1$ and let $a\in \O$ be such that $a\mod\mathfrak{p}$ generates $K^*$. Then $\left\vert f(a)\right\vert<1$ and $\left\vert f'(a)\right\vert=1$. By Hensel's lemma there is a $\zeta\in K$ such that $f(\zeta)=0$ and $\zeta\equiv a\pmod{\mathfrak{p}}$.)

Since $U$ is compact and the cosets of $U$ cover $K$, we see that $K^*$ is locally compact.

Lemma 17.1.15   The additive Haar measure $\mu$ on $K^+$, when restricted to $U_1$ gives a measure on $U_1$ that is also invariant under multiplication, so gives a Haar measure on $U_1$.

Proof. It suffices to show that

$$\mu(1+\pi^n\O ) = \mu(u\cdot(1+\pi^n\O )),$$

for any $u\in U_1$ and $n>0$. Write $u=1+a_1\pi + a_2\pi^2 +\cdots$. We have

$$u\cdot (1+\pi^n\O ) = (1+a_1\pi + a_2\pi^2 +\cdots)\cdot (1+\pi^n\O )$$

$$= 1+a_1\pi + a_2\pi^2 + \cdots + \pi^n\O$$

$$= a_1\pi + a_2\pi^2 + \cdots + ( 1+\pi^n\O ),$$

which is an additive translate of $1+\pi^n\O$, hence has the same measure. $\qedsymbol$

Thus $\mu$ gives a Haar measure on $K^*$ by translating $U_1$ around to cover $K^*$.

Lemma 17.1.16   The topological spaces $K^+$ and $K^*$ are totally disconnected (the only connected sets are points).

Proof. The proof is the same as that of Proposition 16.2.13. The point is that the non-archimedean triangle inequality forces the complement an open disc to be open, hence any set with at least two distinct elements ``falls apart'' into a disjoint union of two disjoint open subsets. $\qedsymbol$

Remark 17.1.17   Note that $K^*$ and $K^+$ are locally isomorphic if $K$ has characteristic 0. We have the exponential map

$$a \mapsto \exp(a) = \sum_{n=0}^{\infty} \frac{a^n}{n!}$$

defined for all sufficiently small $a$ with its inverse

$$\log(a) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(a-1)^n}{n},$$

which is defined for all $a$ sufficiently close to $1$.