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Properties of Galois Groups
Structure of infinite Galois groups with the Krull topology
This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.
We have previously defined Galois extensions and Galois groups. The central result of Galois theory is that for a field extension \(\mathbb{L}/\mathbb{K}\), there exists an inclusion-reversing bijection between the lattice of closed subgroups of the Galois group \(\Gal(\mathbb{L}/\mathbb{K})\) and the lattice of subextensions of \(\mathbb{L}/\mathbb{K}\). Many treatments discuss this result only when \(\Gal(\mathbb{L}/\mathbb{K})\) is finite, but we shall also cover the case where \(\Gal(\mathbb{L}/\mathbb{K})\) is infinite; for this we must endow \(\Gal(\mathbb{L}/\mathbb{K})\) with an appropriate topology.
Topology of the Galois Group
Let \(\mathbb{L}/\mathbb{K}\) be a Galois extension, and let \(\Gal(\mathbb{L}/\mathbb{K})\) be its Galois group. Since the Galois group is a collection of functions from the set \(\mathbb{L}\) to itself, if we equip the set \(\Fun(\mathbb{L},\mathbb{L})=\mathbb{L}^\mathbb{L}\) of all functions from \(\mathbb{L}\) to \(\mathbb{L}\) with a topology, we can induce a topology on \(\Gal(\mathbb{L}/\mathbb{K})\) as a subset. ([Topology] §Subspaces, ⁋Definition 1)
To this end, we endow \(\mathbb{L}\) with the discrete topology. ([Topology] §Open Sets, ⁋Example 2) Then, by the discussion following [Topology] §Product Spaces, ⁋Definition 1, the sets of the form \(\pr_x^{-1}(U)\) for the projections \(\pr_x:\mathbb{L}^\mathbb{L}\rightarrow\mathbb{L}\) constitute a subbase for \(\mathbb{L}^\mathbb{L}\), and since \(\mathbb{L}\) is discrete we may restrict to singletons \(U\) and still obtain a subbase. That is, a subbase for this set is given by the collection of sets of the form
\[U_{x,y}=\left\{\sigma\mid\sigma(x)=y \right\}\]so regarding \(\Gal(\mathbb{L}/\mathbb{K})\) as a subspace, for any \(\sigma\in\Gal(\mathbb{L}/\mathbb{K})\) we know that the collection of sets of the form
\[U_{x_1,\ldots,x_n}=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \text{$\tau(x_i)=\sigma(x_i)$ for all $i$}\right\}\]constitutes a local base at \(\sigma\). ([Topology] §Bases of Topological Spaces, ⁋Definition 4)
On the other hand, the functions satisfying the above condition are precisely those that agree with \(\sigma\) when restricted to the finite subextension \(\mathbb{M}=\mathbb{K}(x_1,\ldots,x_n )\) of \(\mathbb{L}\), and conversely any finite subextension \(\mathbb{M}/\mathbb{K}\) defines an element of the local base at \(\sigma\) in this manner. Thus, letting \(\Ext_{\fin}(\mathbb{L}/\mathbb{K})\) denote the collection of finite subextensions of \(\mathbb{L}/\mathbb{K}\), and for any \(\mathbb{M}/\mathbb{K}\in \Ext_{\fin}(\mathbb{L}/\mathbb{K})\) and any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\) defining the subset \(U_\mathbb{M}(\sigma)\) of \(\Gal(\mathbb{L}/\mathbb{K})\) by
\[U_\mathbb{M}(\sigma)=\left\{\tau\in \Gal(\mathbb{L}/\mathbb{K})\mid \sigma\vert_\mathbb{M}=\tau\vert_\mathbb{M}\right\}\]this set becomes an element of the local base at \(\sigma\), and the collection \((U_\mathbb{M}(\sigma))_{\mathbb{M}\in\Ext_{\fin}(\mathbb{L}/\mathbb{K})}\) is exactly the local base at \(\sigma\). The topology on \(\Gal(\mathbb{L}/\mathbb{K})\) obtained in this way is called the Krull topology.
Example 1 In particular, consider the case where \(\mathbb{L}/\mathbb{K}\) is a finite degree Galois extension. Then from the discussion following [Galois Extension] §Galois Extension, ⁋Definition 12 we know that \(\Gal(\mathbb{L}/\mathbb{K})\) is a finite set. On the other hand, since \(\mathbb{L}/\mathbb{K}\) is of finite degree, \(\mathbb{L}/\mathbb{K}\) itself is already an element of \(\Ext_{\fin}(\mathbb{L}/\mathbb{K})\), and thus for any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\)
\[U_\mathbb{L}(\sigma)=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \sigma\vert_\mathbb{L}=\tau\vert_\mathbb{L}\right\}=\left\{\sigma\right\}\]is an element of the local base at \(\sigma\) described above. Hence every singleton \(\left\{\sigma\right\}\) is an open set, and in this case \(\Gal(\mathbb{L}/\mathbb{K})\) carries the discrete topology.
Meanwhile, the topological space \(\Gal(\mathbb{L}/\mathbb{K})\) defined above is originally a group under composition of \(\mathbb{K}\)-automorphisms, and it is not difficult to show that composition of functions is compatible with this topology.
Proposition 2 \(\Gal(\mathbb{L}/\mathbb{K})\) defined above is a topological group.
Proof
We must show that the two maps
\[\Gal(\mathbb{L}/\mathbb{K})\times\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}/\mathbb{K});\quad (\sigma,\sigma')\mapsto \sigma\sigma',\qquad \Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}/\mathbb{K});\quad \sigma\mapsto \sigma^{-1}\]are continuous. First, considering an arbitrary element \(U_\mathbb{M}(\sigma\sigma')\) of the local base at \(\sigma\sigma'\), by definition
\[U_\mathbb{M}(\sigma\sigma')=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\sigma\sigma'\vert_\mathbb{M}\right\}\]Since \(\sigma'\) is a \(\mathbb{K}\)-automorphism of \(\mathbb{L}\), \(\sigma'(\mathbb{M})\) is also a finite subextension of \(\mathbb{L}\), and if \(\tau\in U_{\sigma'(\mathbb{M})}(\sigma)\) and \(\tau'\in U_\mathbb{M}(\sigma')\), then for any \(x\in \mathbb{M}\) we have \(\tau'(x)=\sigma'(x)\in\sigma'(\mathbb{M})\), so \(\tau\tau'(x)=\sigma\sigma'(x)\). Thus the open set \(U_{\sigma'(\mathbb{M})}(\sigma)\times U_\mathbb{M}(\sigma')\) in \(\Gal(\mathbb{L}/\mathbb{K})\times\Gal(\mathbb{L}/\mathbb{K})\) is contained in the preimage of the above set, and hence the multiplication map is continuous.
Similarly, the local base element \(U_\mathbb{M}(\sigma^{-1})\) at \(\sigma^{-1}\) is given by
\[U_\mathbb{M}(\sigma^{-1})=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\sigma^{-1}\vert_\mathbb{M}\right\}\]and since \(\sigma^{-1}(\mathbb{M})\) is also a finite subextension, we may consider \(U_{\sigma^{-1}(\mathbb{M})}(\sigma)\). For any \(x\in \mathbb{M}\), since \(\sigma^{-1}(x)\in\sigma^{-1}(\mathbb{M})\), if \(\tau\in U_{\sigma^{-1}(\mathbb{M})}(\sigma)\) then \(\tau(\sigma^{-1}(x))=\sigma(\sigma^{-1}(x))=x\), that is \(\tau^{-1}(x)=\sigma^{-1}(x)\), and thus this set is contained in the preimage of the above set.
In particular, the local base at any \(\sigma\) is obtained by translating the local base at the identity \(\id_\mathbb{L}\) via the left translation map. That is, for any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\)
\[U_\mathbb{M}(\sigma)=\sigma U_\mathbb{M}(\id_\mathbb{L})\]holds. From this we see that it suffices to consider only the sets
\[U_\mathbb{M}(\id_\mathbb{L})=\left\{\tau\in \Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\id_\mathbb{M}\right\}\]instead of the above. Then by definition, as a set
\[U_\mathbb{M}(\id_\mathbb{L})=\Gal(\mathbb{L}/\mathbb{M})\]Here, the third condition of Galois Extensions in [Galois Extension] §Galois Extension holds with \(\mathbb{M}\) in place of \(\mathbb{K}\), so \(\mathbb{L}/\mathbb{M}\) is also a Galois extension, and the inclusion of the group on the right into \(\Gal(\mathbb{L}/\mathbb{K})\) is simply obtained by viewing an \(\mathbb{M}\)-automorphism as a \(\mathbb{K}\)-automorphism. Moreover, the topology on \(\Gal(\mathbb{L}/\mathbb{M})\) coincides with the subspace topology inherited from \(\Gal(\mathbb{L}/\mathbb{K})\). Then by the first condition of the same theorem, since \(\mathbb{L}^{\Gal(\mathbb{L}/\mathbb{M})}=\mathbb{M}\),
\[U_\mathbb{M}(\id_\mathbb{L})\subseteq U_\mathbb{N}(\id_\mathbb{L})\iff \mathbb{M}\supseteq \mathbb{N}\]holds. The right-to-left direction follows immediately from the definition, and the left-to-right direction follows from \(\mathbb{N}=\mathbb{L}^{\Gal(\mathbb{L}/\mathbb{N})}\subseteq\mathbb{L}^{\Gal(\mathbb{L}/\mathbb{M})}=\mathbb{M}\).
Now consider the collection \(\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})\) of finite degree Galois subextensions. By Galois Extensions in [Galois Extension] §Galois Extension, this is a cofinal subset of \(\Ext_{\fin}(\mathbb{L}/\mathbb{K})\). Hence \((U_\mathbb{M}(\id_\mathbb{L}))_{\mathbb{M}\in\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})}\) is also a local base at \(\id_\mathbb{L}\). Then for any \(\mathbb{M}\in \Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})\), considering the restriction homomorphism \(\rho:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{M}/\mathbb{K})\) examined in Galois Extensions of [Galois Extension] §Galois Extension, since any finite degree subextension of \(\mathbb{M}\) is also a finite degree subextension of \(\mathbb{L}\), this restriction homomorphism is continuous with respect to the topology defined above. In this situation, \(\rho\) is a continuous map from \(\Gal(\mathbb{L}/\mathbb{K})\) to the finite discrete space \(\Gal(\mathbb{M}/\mathbb{K})\) (Example 1), so \(\ker\rho\) is a closed subgroup of \(\Gal(\mathbb{L}/\mathbb{K})\). However, by definition
\[\sigma\in\ker\rho\iff \sigma\vert_\mathbb{M}=\id\vert_\mathbb{M}\iff\sigma\in U_\mathbb{M}(\id_\mathbb{L})\]so each \(U_\mathbb{M}(\id_\mathbb{L})\) is clopen. On the other hand, any clopen set can always be written as a union of connected components, and therefore any nonempty intersection of clopen sets must contain a connected component. Yet the following holds.
Proposition 3 In the above situation,
\[\{\id_\mathbb{L}\}=\bigcap_{\mathbb{M}\in \Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})}U_\mathbb{M}(\id_\mathbb{L})\]holds.
Proof
Let \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\) be given. If \(\sigma\neq\id_\mathbb{L}\), then there exists \(x\in \mathbb{L}\) such that \(\sigma(x)\neq x\). Taking \(\mathbb{M}=\mathbb{K}(x)\), we have \(\sigma\not\in U_\mathbb{M}(\id_\mathbb{L})\). Now, as observed earlier, since \(\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})\) is a cofinal subset of \(\Ext_{\fin}(\mathbb{L}/\mathbb{K})\), we obtain the desired result.
Therefore, by this proposition, the connected component containing \(\id_\mathbb{L}\) is \(\left\{\id_\mathbb{L}\right\}\). On the other hand, by Proposition 2, left translation by any \(\sigma\) is a homeomorphism, so the connected component containing any point is also a singleton, and hence \(\Gal(\mathbb{L}/\mathbb{K})\) is a totally disconnected space. ([Topology] §Connected Spaces, ⁋Definition 7) Moreover, the following holds.
Proposition 4 \(\Gal(\mathbb{L}/\mathbb{K})\) is compact.
Proof
First, for each \(x\in \mathbb{L}\), since \(\mathbb{L}/\mathbb{K}\) is an algebraic extension, \(x\) is algebraic, and hence there are only finitely many elements conjugate to \(x\). ([Galois Extension] §Galois Extension, ⁋Proposition 3) In other words, considering
\[\Gal(\mathbb{L}/\mathbb{K})\hookrightarrow \prod_{x\in \mathbb{L}}\mathbb{L}\overset{\pr_x}{\longrightarrow}\mathbb{L};\qquad \sigma\mapsto \sigma(x)\]the image of this map is a finite set. Therefore \(\Gal(\mathbb{L}/\mathbb{K})\) is a subset of a product of finite sets, and since finite sets are compact, this product is also compact. ([Topology] §Compactness and Paracompactness, ⁋Theorem 2 (Tychonoff)) Thus proving the given proposition amounts to showing that \(\Gal(\mathbb{L}/\mathbb{K})\) is closed in \(\mathbb{L}^\mathbb{L}\).
Suppose a function \(u\) belongs to the closure of \(\Gal(\mathbb{L}/\mathbb{K})\) in \(\mathbb{L}^\mathbb{L}\). First, a field homomorphism \(u:\mathbb{L}\rightarrow\mathbb{L}\) fixing \(\mathbb{K}\) is always an element of \(\Gal(\mathbb{L}/\mathbb{K})\), because \(u\) is injective and for any \(x\in \mathbb{L}\), since the finite set of roots of the minimal polynomial of \(x\) in \(\mathbb{L}\) is mapped to itself, \(u\) is surjective on this set, and hence \(x\) belongs to the image of \(u\). Therefore, if \(u\) is not an element of \(\Gal(\mathbb{L}/\mathbb{K})\), then either \(u\) is not a field homomorphism or \(u\) does not fix \(\mathbb{K}\). Adopting the first assumption, suppose there exist \(x,y\in\mathbb{L}\) such that \(u(x+y)\neq u(x)+u(y)\). Then the set
\[\left\{f\in \mathbb{L}^\mathbb{L}\mid f(x)=u(x),f(y)=u(y),f(x+y)=u(x+y)\right\}\]is a basic open set in \(\mathbb{L}^\mathbb{L}\) and hence is open, and moreover it contains \(u\). That is, this set is an open neighborhood of \(u\). However, by assumption
\[f(x+y)=u(x+y)\neq u(x)+u(y)=f(x)+f(y)\]so these \(f\) also fail to be field homomorphisms. Thus the above open neighborhood does not meet \(\Gal(\mathbb{L}/\mathbb{K})\), which contradicts the assumption that \(u\) belongs to the closure of \(\Gal(\mathbb{L}/\mathbb{K})\). By similar reasoning all other cases can also be ruled out, and from this we conclude that \(\Gal(\mathbb{L}/\mathbb{K})\) is closed in \(\mathbb{L}^\mathbb{L}\).
Now let \(\mathbb{L}/\mathbb{K}\) be a Galois extension, and let \(\mathbb{L}_i/\mathbb{K}\) be Galois subextensions of this extension satisfying \(\mathbb{L}=\bigcup_{i\in I}\mathbb{L}_i\), and suppose that for any \(i,j\in I\) there exists \(k\in I\) such that \(\mathbb{L}_i\cup\mathbb{L}_j\subseteq \mathbb{L}_k\). Then we endow this with the partial order
\[i\leq j \iff \mathbb{L}_i\subseteq \mathbb{L}_j\]and under this partial order we can define the following restriction maps
\[\rho_{ij}:\Gal(\mathbb{L}_j/\mathbb{K}) \rightarrow \Gal(\mathbb{L}_i/\mathbb{K})\qquad \text{whenever $i\leq j$}\]These are continuous homomorphisms, and therefore their inverse limit
\[\varprojlim_{i\in I}\Gal(\mathbb{L}_i/\mathbb{K})=\left\{(\sigma_i)\in\prod_{i\in I}\Gal(\mathbb{L}_i/\mathbb{K})\mid\text{$\rho_{ij}(\sigma_j)=\sigma_i$ whenever $i\leq j$}\right\}\]and the canonical morphisms \(\rho_i:\varprojlim \Gal(\mathbb{L}_i/\mathbb{K})\rightarrow\Gal(\mathbb{L}_i/\mathbb{K})\) exist. ([Category Theory] §Limits, ⁋Example 5)
On the other hand, considering the restriction maps
\[\lambda_i:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}_i/\mathbb{K})\]these satisfy \(\lambda_i=\rho_{ij}\circ\lambda_j\), so there exists an induced continuous homomorphism \(\lambda:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\varprojlim\Gal(\mathbb{L}_i/\mathbb{K})\).
Proposition 5 The \(\lambda\) defined above is an isomorphism of topological groups.
Proof
Each \(\Gal(\mathbb{L}_i/\mathbb{K})\) is a subspace of the Hausdorff space \(\mathbb{L}_i^{\mathbb{L}_i}\) and hence is Hausdorff, and since products and subspaces of Hausdorff spaces are again Hausdorff, their inverse limit \(\varprojlim \Gal(\mathbb{L}_i/\mathbb{K})\) is also Hausdorff. On the other hand, since \(\Gal(\mathbb{L}/\mathbb{K})\) is compact by Proposition 4, by [Topology] §Compact Spaces, ⁋Proposition 9 it suffices to show that \(\lambda\) is bijective.
First, if \(\lambda(\sigma)\) is the identity, then \(\sigma\vert_{\mathbb{L}_i}=\id_{\mathbb{L}_i}\) for all \(i\), and since \(\mathbb{L}=\bigcup_i\mathbb{L}_i\), we have \(\sigma=\id_\mathbb{L}\). That is, \(\lambda\) is injective. Now let \((\sigma_i)\in\varprojlim\Gal(\mathbb{L}_i/\mathbb{K})\) be given, and for \(x\in \mathbb{L}_i\) define \(\sigma(x)=\sigma_i(x)\). If \(x\) belongs to both \(\mathbb{L}_i\) and \(\mathbb{L}_j\), then taking \(k\) with \(\mathbb{L}_i\cup\mathbb{L}_j\subseteq \mathbb{L}_k\) we have \(\sigma_i(x)=\rho_{ik}(\sigma_k)(x)=\sigma_k(x)\) and by the same reasoning \(\sigma_j(x)=\sigma_k(x)\), so \(\sigma\) is well-defined, and any two elements of \(\mathbb{L}\) also belong to some common \(\mathbb{L}_k\), so \(\sigma\) is a field homomorphism fixing \(\mathbb{K}\). On the other hand, since the \(\rho_{ij}\) are homomorphisms, \((\sigma_i^{-1})\) is also an element of \(\varprojlim\Gal(\mathbb{L}_i/\mathbb{K})\), and the function obtained in the same way is the inverse of \(\sigma\). That is, \(\sigma\in\Gal(\mathbb{L}/\mathbb{K})\) and \(\lambda(\sigma)=(\sigma_i)\), so \(\lambda\) is surjective.
In particular, the family of finite degree Galois subextensions \(\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})\) satisfies the conditions of this proposition. The compositum of any two elements of this family is again a finite degree Galois subextension by Galois Extensions in [Galois Extension] §Galois Extension, and any element \(x\) of \(\mathbb{L}\) belongs to an element of \(\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})\) containing \(\mathbb{K}(x)\). That is, the Galois group of any Galois extension is an inverse limit of finite groups, namely a profinite group.
Galois Cohomology
The Galois group is not merely a group, but a group acting on \(\mathbb{L}\), and in particular on the multiplicative group \(\mathbb{L}^\times\). The standard tool for extracting the arithmetic information encoded in this action is Galois cohomology, and to close we examine the classical theorem that stands at its origin: Hilbert’s Theorem 90. In this section \(\mathbb{L}/\mathbb{K}\) is a finite degree Galois extension and \(G=\Gal(\mathbb{L}/\mathbb{K})\).
Definition 6 A function \(\varphi:G \rightarrow \mathbb{L}^\times\) is called a 1-cocycle if for any \(\sigma,\tau\in G\)
\[\varphi(\sigma\tau)=\varphi(\sigma)\cdot\sigma\bigl(\varphi(\tau)\bigr)\]holds. In particular, a 1-cocycle of the form \(\varphi(\sigma)=\sigma(c)/c\) for some \(c\in\mathbb{L}^\times\) is called a 1-coboundary.
First, we verify that a 1-coboundary is indeed a 1-cocycle:
\[\varphi(\sigma)\cdot\sigma(\varphi(\tau))=\frac{\sigma(c)}{c}\cdot\sigma\left(\frac{\tau(c)}{c}\right)=\frac{\sigma(c)}{c}\cdot\frac{\sigma\tau(c)}{\sigma(c)}=\frac{\sigma\tau(c)}{c}=\varphi(\sigma\tau)\]Also, since \(\mathbb{L}^\times\) is abelian, the 1-cocycles form an abelian group under pointwise multiplication, and since \(c\mapsto(\sigma\mapsto\sigma(c)/c)\) is a group homomorphism, the 1-coboundaries form a subgroup. Therefore we can form the quotient group, and we denote it by \(H^1(G,\mathbb{L}^\times)\). Hilbert’s Theorem 90 states that this group carries no information at all.
Theorem 7 (Hilbert 90) For a finite degree Galois extension \(\mathbb{L}/\mathbb{K}\), every 1-cocycle \(\varphi:G \rightarrow \mathbb{L}^\times\) is a 1-coboundary. That is, \(H^1(G,\mathbb{L}^\times)\) is trivial.
Proof
The elements of \(G\) are distinct homomorphisms from \(\mathbb{L}\) to \(\mathbb{L}\), so they are linearly independent over \(\mathbb{L}\) by [Étale Algebras] §Étale Algebras, ⁋Corollary 3. Since the values of \(\varphi\) are all nonzero, the linear combination
\[\sum_{\tau\in G}\varphi(\tau)\tau\]is not the zero map, and therefore for some \(x\in\mathbb{L}\)
\[b=\sum_{\tau\in G}\varphi(\tau)\tau(x)\neq0\]Now for any \(\sigma\in G\), rewriting the cocycle condition as \(\sigma(\varphi(\tau))=\varphi(\sigma)^{-1}\varphi(\sigma\tau)\) and computing,
\[\sigma(b)=\sum_{\tau\in G}\sigma(\varphi(\tau))\sigma\tau(x)=\varphi(\sigma)^{-1}\sum_{\tau\in G}\varphi(\sigma\tau)\sigma\tau(x)=\varphi(\sigma)^{-1}b\]The last equality holds because as \(\tau\) ranges over all of \(G\), so does \(\sigma\tau\). Therefore, setting \(c=b^{-1}\),
\[\varphi(\sigma)=\frac{b}{\sigma(b)}=\frac{\sigma(c)}{c}\]so \(\varphi\) is a 1-coboundary.
The classical form of Hilbert 90 concerns cyclic extensions. Let \(G=\langle\sigma\rangle\) be a cyclic group of order \(n\), and define the norm of \(x\in\mathbb{L}\) by
\[N_{\mathbb{L}/\mathbb{K}}(x)=\prod_{i=0}^{n-1}\sigma^i(x)\]Applying \(\sigma\) merely permutes the factors, so \(N_{\mathbb{L}/\mathbb{K}}(x)\) is \(G\)-invariant, and since \(\mathbb{L}/\mathbb{K}\) is Galois, by Galois Extensions in [Galois Extension] §Galois Extension we have \(N_{\mathbb{L}/\mathbb{K}}(x)\in\mathbb{K}\).
Corollary 8 Let \(\mathbb{L}/\mathbb{K}\) be a finite degree Galois extension and let \(G=\Gal(\mathbb{L}/\mathbb{K})=\langle\sigma\rangle\) be cyclic. Then for \(x\in\mathbb{L}^\times\) the following are equivalent.
- \(N_{\mathbb{L}/\mathbb{K}}(x)=1\).
- There exists \(y\in\mathbb{L}^\times\) such that \(x=\sigma(y)/y\).
Proof
First, assuming the second condition,
\[N_{\mathbb{L}/\mathbb{K}}\bigl(\sigma(y)/y\bigr)=\prod_{i=0}^{n-1}\frac{\sigma^{i+1}(y)}{\sigma^i(y)}=\frac{\sigma^n(y)}{y}=1\]The middle equality is telescoping and the last equality follows from \(\sigma^n=\id_\mathbb{L}\).
Conversely, assume \(N_{\mathbb{L}/\mathbb{K}}(x)=1\). Define \(\varphi:G \rightarrow \mathbb{L}^\times\) by
\[\varphi(\sigma^i)=\prod_{k=0}^{i-1}\sigma^k(x)\qquad(0\leq i\leq n-1)\]where for \(i=0\) the empty product gives \(\varphi(\id)=1\). Let us verify that this is a 1-cocycle. For \(0\leq a,b\leq n-1\),
\[\varphi(\sigma^a)\cdot\sigma^a\bigl(\varphi(\sigma^b)\bigr)=\prod_{k=0}^{a-1}\sigma^k(x)\cdot\prod_{k=0}^{b-1}\sigma^{a+k}(x)=\prod_{k=0}^{a+b-1}\sigma^k(x)\]If \(a+b\leq n-1\), then by definition this is \(\varphi(\sigma^{a+b})=\varphi(\sigma^a\sigma^b)\). If \(a+b\geq n\), then since \(\sigma^k=\sigma^{k-n}\) for \(k\geq n\),
\[\prod_{k=0}^{a+b-1}\sigma^k(x)=\prod_{k=0}^{n-1}\sigma^k(x)\cdot\prod_{k=n}^{a+b-1}\sigma^k(x)=N_{\mathbb{L}/\mathbb{K}}(x)\cdot\prod_{k=0}^{a+b-n-1}\sigma^k(x)=\varphi(\sigma^{a+b-n})\]and since \(\sigma^a\sigma^b=\sigma^{a+b-n}\), the cocycle condition again holds. The assumption \(N_{\mathbb{L}/\mathbb{K}}(x)=1\) was used in the last equality.
Now by Theorem 7 (Hilbert 90), \(\varphi\) is a 1-coboundary. That is, for some \(c\in\mathbb{L}^\times\) we have \(\varphi(\sigma^i)=\sigma^i(c)/c\), and in particular for \(i=1\)
\[x=\varphi(\sigma)=\frac{\sigma(c)}{c}\]so we may take \(y=c\).
References
[Bou] N. Bourbaki. Algebra II: Chapters 4–7. Springer, 2003.
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