What is meant by 'the completion of Z'?

Question

In the first chapter of Algebraic Number Theory (lecture notes collected by Cassels-Fröhlich), page 28 has the following paragraph:

"We suppose now that $k$ is a finite field of characteristic $p$ with $q=p^m$ elements. Denote by $\bar{\mathbb{Z}}$ the completion of $\mathbb{Z}$ with respect to the topology defined by the subgroups $n\mathbb{Z}$ ($n>0$). Then $\Gamma(\bar{k}^s/k)$ is an isomorphic copy of $\bar{\mathbb{Z}}$ under the map $$v \mapsto w_q^v$$ where $$\alpha w_q = \alpha^q."$$

$\Gamma(\bar{k}^s/k)$ refers to the Galois group of the maximal seperable extension of $k$ over $k$, and given an element $\sigma$ of the Galois group Fröhlich writes $x\sigma := \sigma(x)$.

Firstly, I'm not positive what it means when it says the topology defined by those subgroups; is it saying the topology generated by taking those sets as a basis? Even if so, I don't see what the completion here would be (i.e. how it could be described), and if the elements aren't integers, I don't see how one 'exponentiates' the automorphism $w_q$.

Answer 1

Yes, the ideals $n\mathbf Z$ are taken as a basis of neighborhoods of $0$ (it is sufficient to describe a neighborhood basis of $0$ in a topological group, by translation). A Cauchy sequence in this topology is a sequence $\{a_k\}$ of integers such that, modulo any integer $n>0$, $a_k-a_l$ is eventually congruent to $0$.

A more abstract definition of the completion (not using Cauchy sequences) is as the inverse limit $\widehat{\mathbf Z} = \varprojlim \mathbf Z/n\mathbf Z$. The group $\widehat{\mathbf Z}$ is an example of a profinite group (an inverse limit of finite groups). It is an uncountable topological group, which is compact and totally disconnected (somewhat like a Cantor set). By the Chinese Remainder Theorem, it is actually isomorphic to the direct product of all $\mathbf Z_p$'s, the additive groups of the $p$-adic integers, over all primes $p$.

In the category of profinite abelian groups, $\widehat{\mathbf Z}$ plays the role of $\mathbf Z$, being the profinite abelian group freely generated by a single element (sometimes called a "topological generator"). This generator is the image of $1$ under the canonical map from $\mathbf Z$ to its completion (or, if you prefer, the constant Cauchy sequence $\equiv 1$). Under your setup, the Frobenius automorphism is the topological generator.

It is a deep and important fact that the absolute Galois group of a finite field is canonically isomorphic with $\widehat{\mathbf Z}$ (as anon explains). This should be very surprising if you are used to Galois theory over a field like $\mathbf Q$. Indeed, the absolute Galois group of $\mathbf Q$ is an incredibly complicated object, very far from being abelian, let alone of having a simple explicit description!

Answer 2

This is $\widehat{\Bbb Z}$, the profinite integers. It is the inverse limit $\varprojlim\Bbb Z/n\Bbb Z$. Probably the best way to think about it is as a subgroup of $\prod_{n\in\Bbb N}\Bbb Z/n\Bbb Z$ where the coordinates are congruent to each other wherever possible under the appropriate projection maps (i.e. reducing mod $n$ to mod $d\mid n$).

For the topology, take as nbhd basis of $0$ all kernels of coordinate projections $\widehat{\Bbb Z}\to\Bbb Z/n\Bbb Z$.

The best way to think about $\bar{k}^s$ is as the "union" of all finite fields of characteristic $p$. This makes sense intuitively because, recall, $d\mid n\implies {\Bbb F}_{q^d}\hookrightarrow{\Bbb F}_{q^n}$ and thus any two fields of characteristic $p$ say $\Bbb F_{q^a}$ and $\Bbb F_{q^b}$ lie inside $\Bbb F_{q^n}$ for any multiple $n$ of both $a$ and $b$.

Now let's see about how elements of $\widehat{\Bbb Z}$ act on $\bar{k}^s$ while fixing $k=\Bbb F_q$ pointwise. Let $x\in \bar{k}^s$, so in particular it lies in some finite field, $x\in{\Bbb F}_q(x)=\Bbb F_{q^r}$. Then $G(\Bbb F_{q^r}/\Bbb F_q)\cong C_r$ is cyclic generated canonically by the Frobenius morphism $F:\alpha\mapsto\alpha^q$. If $g\in\widehat{\Bbb Z}$, take the $\Bbb Z/r\Bbb Z$ coordinate of $g$, call it $g(r)$, and let it act on $x$ by Frobenius, which means that $gx=F^{g(r)}x=x^{q^{\large g(r)}}$.

You can check that $(1,1,1,\cdots)\in\widehat{\Bbb Z}$ is a topological generator corresponding to $x\mapsto x^q$ on $\bar{k}^s$.