The absoulte Galois group of \(\mathbb{C}(\!(t)\!)\) is procyclic

1 Introduction

Let \(k\) be an algebraically closed field of characteristic zero. For example one may take \(k = \mathbb{C}\) if desired. We shall expose the proof of the following theorem.

Theorem. The Galois group of the Laurent field \(K = k(\!(t)\!)\) is the profinite completion \(\widehat{\mathbb{Z}}\) of the group \(\mathbb{Z}\) of integers.

In algebraic geometry, the spectrum of the ring \(\mathcal{O}_K = k[\![t]\!]\) of power series with coefficients in \(k\) is regarded as an algebraic analogue of the unit disk \(\Delta = \{z \in \mathbb{C} : |z| < 1\}\), and \(\mathrm{Spec}(K)\) regarded as an analogue of the purectured disk \(\Delta^* = \Delta \setminus \{0\}\) accordingly. So the theorem is an algebraic analogue of the fact that \(\pi_{1}(\Delta^*) = \mathbb{Z}\).

For the reader familiar with the theory of local fields, the idea of the proof can be summarized as follows: by comparing with the higher unit groups, one shows that all the higher ramifications vanish. This plus some extra argument suffices to conclude the proof of the theorem. In these notes, we give a direct proof (Proposition 3/1) that \(G_1 = \{1\}\), avoiding the idea of higher ramifications.

2 Unit groups

The study of the absolute Galois group \(G_K\) is accomplished by the study of its finite quotients. Let \(L/K\) be a finite extension of \(K\). Denote by \(\mathcal{O}_L\) the ring of integral elements over \(K\). Then, by the Hensel’s lemma, the order valuation of \(K\) extends uniquely to a valuation on \(L\), and \(\mathcal{O}_L\) is a complete discrete valuation ring as well. Denote by \(\varpi\) a uniformizer of \(\mathcal{O}_L\). Let \(U_{L}^{(0)}\) be the multiplicative group of the invertible elements in \(\mathcal{O}_L\), and define

\begin{equation*} \label{eq:definition-unit-group} U^{(1)}_L = 1 + \varpi \cdot \mathcal{O}_L \subset U_{L}^{(0)}. \end{equation*}

Clearly

\begin{equation*}\tag{\(\dagger\)} \label{eq:unit-group-subquotients} U^{(0)}_L/U^{(1)}_L \cong k^{\times}, \end{equation*}

We shall eventually realize \(G\) as a subgroup of \(U^{(0)}_L/U^{(1)}_L\), so the following elementary lemma becomes relevant.

Lemma. Let \(F\) be a field. Then any finite subgroup \(G\) of \(F^\times\) is cyclic.

Proof. Let \(n\) be the order of \(G\). Then any \(a \in G\) satisfies the condition \(a^n = 1\). If we use \(\mu_{m}\) to denote the group of \(m\)th roots of \(1\), then \(G \subset F^{\times}\) is a subgroup of \(\mu_n(F) \subset \mu_n(F^{\mathrm{a}})\). Thus it suffices to prove the group \(\mu_n\) of \(n\)th roots of \(1\) in \(F^{\mathrm{a}}\) is a cyclic group (since any subgroup of a cyclic group is again cyclic). No matter how large \(F\) may be, the elements in \(\mu_{n}(F^{\mathrm{a}})\) are defined over the algebraic closure of the prime field. The result for \(\mathbb{Q}^{\mathrm{a}}\) follows from the fact that the group of \(n\)th roots of \(1\) in \(\mathbb{C}\) is cyclic; by reduction mod \(p\), the result for \(\mu_{n}(\mathbb{F}_{p}^{\mathrm{a}})\) also follows.

3 Triviality of ramification groups

Retain the notations in §2. We shall denote \(\mathrm{Gal}(L/K)\) by \(G\) and assume from now on that \(L/K\) is Galois. The group \(G\) acts on \(\mathcal{O}_L\) because \(\mathcal{O}_L\) is, by definition, the integral closure of \(\mathcal{O}_K\) in \(L\). Therefore \(G\) preserves the unique maximal ideal \(\varpi{}\cdot \mathcal{O}_L\) of \(\mathcal{O}_L\) and induces an automorphism of the residue field \(\mathcal{O}_{L}/\varpi\). The residue field \(\mathcal{O}_{L}/\varpi\) is necessarily a finite extension of \(k\); but as \(k\) is algebraically closed, we must have \(\mathcal{O}_{L}/\varpi = k\). Therefore \(K\)-automorphism of \(L\) are all equal to the identity up to order zero.

We next study the actions of \(G\) on the ring \(\mathcal{O}_L\) up to first order, i.e., we shall look at the natural map

\begin{equation*}\tag{\(\ast\)} \label{eq:representation-1st-order} \rho: G \to \mathrm{Aut}_{k}(\mathcal{O}_{L}/\varpi^{2}). \end{equation*}

Proposition 1. The map \(\rho\) defined in \eqref{eq:representation-1st-order} is injective.

Before proving Proposition 1 let’s first recall the following simple fact regarding the structure of \(\mathcal{O}_{L}\).

Lemma 2. There is an isomorphism \(\mathcal{O}_{L} \cong k[\![\varpi]\!]\) of rings.

Proof. By the universal property of the formal power series, the assignment \(x \mapsto \varpi\) defines a homomorphism \(\phi\) from \(k[\![x]\!]\) to \(\mathcal{O}_{L}\). Since \(k\) is a subfield of \(\mathcal{O}_{L}\) which maps isomorphically onto the residue field, for each element \(f \in \mathcal{O}_{L}\), we can write \(f\) uniquely as an infinite sum

\begin{equation*} f = \sum_{n=0}^{\infty} a_{n}\varpi^{n}. \end{equation*}

Here, \(a_0\) is the the residue class of \(f\), \(a_1\) is the residue class of \((f-a_{0})/\varpi\), and so on. This proves that \(\phi\) is bijective.

Proof of Proposition 1. Using the identification from Lemma 2 we can write \(f \in \mathcal{O}_{L}\) into a power series

\begin{equation*} f = \sum_{n=0}^\infty a_n \varpi^{n}. \end{equation*}

Also, it makes sense to write \(a_0 = f(0)\). Since the characteristic of \(k\) is zero, the Taylor expansion holds true, hence

\begin{equation*} a_{n} = \frac{f^{(n)}(0)}{n!}. \end{equation*}

Now let \(\sigma\) be any automorphism of \(L\) such that \(\sigma\) induces the identify on the \(k\)-vector space \(\mathcal{O}/\varpi^2\). Then we know that for any \(f \in \mathcal{O}_{L}\), \(\sigma(f)'(0) = f'(0)\). Applying this to the all the derivatives of \(f\), we see \(\sigma(f)^{(n)}(0) = f^{(n)}(0)\) for all \(n\). Thus we must have \(\sigma(f) = f\), thanks to the Taylor formula.

Corollary 3. \(G\) is cyclic.

Proof. Consider the map \(G \to U^{(0)}_L/U^{(1)}_L\) defined by the map

\begin{equation*} \sigma \mapsto \sigma(\varpi)/\varpi \text{ mod } U^{(1)}. \end{equation*}

Clearly this is a homomorphism of groups. If \(\sigma\) is in the kernel, then this means that \(\sigma(\varpi) = \varpi\) mod \(\varpi^{2}\), i.e., \(\sigma\) lies in the kernel of \(\rho\) \eqref{eq:representation-1st-order}. By Proposition 1, \(\sigma=\mathrm{Id}_{L}\). Now we can apply \eqref{eq:unit-group-subquotients} and the lemma in §2.

4 The absolute Galois group of \(K\)

Now we are in the position to prove the Theorem. In the view of the results of §3, we already know all finite quotients of \(G_{K}\) are cyclic groups. Moreover, for any \(n \in \mathbb{Z}\), we can find an explicit example of cyclic Galois extension of \(K\) with group \(\mathbb{Z}/n{}\mathbb{Z}\), namely the extension \(L_{n} = K(t^{1/n})\). Therefore, the only thing that we need to establish is the following lemma.

Lemma. If \(L/K\) finite Galois, then \(L \subset L_{n}\) for some large \(n\).

Proof. Let’s say that \(L/K\) is Galois of degree \(d\). Then by §2, Lemma, \(L/K\) is cyclic. Consider the composite \(M = L_{n}L\). Then, again, \(M/K\) is also cyclic. Take \(n = dp\) with gcd\((d,p) = 1\). Then \(\mathrm{Hom}(\mathbb{Z}/n, \mathbb{Z}/d) = 1\). It follows from the snake lemma that \(\mathrm{Gal}(M/L_n) \to \mathrm{Gal}(M/K)\) factors through \(\mathrm{Gal}(M/L)\). This shows than \(L\) is contained in \(L_n\) for some suitable \(n\).