Properties of Galois Groups

We have previously defined Galois extensions and Galois groups. The central result of Galois theory is that for a Galois 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 $\mathbb{L}/\mathbb{K}$. In many cases, this result is discussed only when the Galois group $\Gal(\mathbb{L}/\mathbb{K})$ is finite, but we shall also treat 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 the Galois group of this extension. Since the Galois group is in any case a collection of functions from the set $\mathbb{L}$ to $\mathbb{L}$, if we endow the collection $\Fun(\mathbb{L},\mathbb{L})=\mathbb{L}^\mathbb{L}$ of functions from $\mathbb{L}$ to $\mathbb{L}$ with a topology, we can endow $\Gal(\mathbb{L}/\mathbb{K})$ with a topology as a subset of this set. ([Topology] §Subspaces, ⁋Definition 1)

To this end, let us endow $\mathbb{L}$ with the discrete topology. ([Topology] §Open Sets, ⁋Example 2) Then, by the discussion following [Topology] §Product Spaces, ⁋Definition 1, sets of the form $\pr_x^{-1}(U)$ for the projections $\pr_x:\mathbb{L}^\mathbb{L}\rightarrow\mathbb{L}$ form a subbase of $\mathbb{L}^\mathbb{L}$, and since $\mathbb{L}$ is discrete, restricting $U$ to singletons here still yields a subbase. That is, since we know that a subbase of this set is the collection of sets that can be written in the form

\[U_{x,y}=\left\{\sigma\mid\sigma(x)=y \right\}\]

viewing $\Gal(\mathbb{L}/\mathbb{K})$ as a subspace of this, we know that for any $\sigma\in\Gal(\mathbb{L}/\mathbb{K})$, the collection of sets of the form

\[U_{x_1,\ldots,x_n}(\sigma)=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \text{$\tau(x_i)=\sigma(x_i)$ for all $i$}\right\}\]

is a local base at $\sigma$. ([Topology] §Bases of a Topological Space, ⁋Definition 4)

On the other hand, the functions satisfying the above condition are 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. That is, let $\Ext_{\fin}(\mathbb{L}/\mathbb{K})$ be the collection of finite subextensions of the extension $\mathbb{L}/\mathbb{K}$; for any $\mathbb{M}/\mathbb{K}\in \Ext_{\fin}(\mathbb{L}/\mathbb{K})$ and any $\sigma\in \Gal(\mathbb{L}/\mathbb{K})$, if we define the subset $U_\mathbb{M}(\sigma)$ of $\Gal(\mathbb{L}/\mathbb{K})$ by the formula

\[U_\mathbb{M}(\sigma)=\left\{\tau\in \Gal(\mathbb{L}/\mathbb{K})\mid \sigma\vert_\mathbb{M}=\tau\vert_\mathbb{M}\right\}\]

then this set becomes an element of the local base at $\sigma$, and the collection $(U_\mathbb{M}(\sigma))_{\mathbb{M}\in\Ext_{\fin}(\mathbb{L}/\mathbb{K})}$ of these sets is precisely 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 Extensions, ⁋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. That is, since the singleton $\left\{\sigma\right\}$ is an open set, in this case $\Gal(\mathbb{L}/\mathbb{K})$ becomes a set equipped with 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 the composition of functions is compatible with this topological structure.

Proposition 2 $\Gal(\mathbb{L}/\mathbb{K})$ defined above is a topological group.

Proof

That is, 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, for $\sigma\sigma'$, if we consider an arbitrary element $U_\mathbb{M}(\sigma\sigma')$ of the local base, by definition

\[U_\mathbb{M}(\sigma\sigma')=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\sigma\sigma'\vert_\mathbb{M}\right\}\]

holds. Here, 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 $x\in \mathbb{M}$, since $\tau'(x)=\sigma'(x)\in\sigma'(\mathbb{M})$, we have $\tau\tau'(x)=\sigma\sigma'(x)$. That is, 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 therefore the multiplication map is continuous.

Similarly, for $\sigma^{-1}$, the local base element $U_\mathbb{M}(\sigma^{-1})$ is given by the formula

\[U_\mathbb{M}(\sigma^{-1})=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\sigma^{-1}\vert_\mathbb{M}\right\}\]

and since $\sigma^{-1}(\mathbb{M})$ is also a finite subextension, we can 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 therefore this set is contained in the preimage of the above set.

In particular, the local base at any $\sigma$ is given by translating the local base at the identity $\id_\mathbb{L}$ along the left translation map. That is, for any $\sigma\in \Gal(\mathbb{L}/\mathbb{K})$, the equation

\[U_\mathbb{M}(\sigma)=\sigma U_\mathbb{M}(\id_\mathbb{L})\]

holds. From this, we see that it suffices to consider only the set

\[U_\mathbb{M}(\id_\mathbb{L})=\left\{\tau\in \Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\id_\mathbb{M}\right\}\]

instead of the sets above. Then by definition,

\[U_\mathbb{M}(\id_\mathbb{L})=\Gal(\mathbb{L}/\mathbb{M})\]

as sets. Here, since the third condition of §Galois Extensions, ⁋Theorem 8 holds even when replacing $\mathbb{K}$ with $\mathbb{M}$, $\mathbb{L}/\mathbb{M}$ is also a Galois extension, and the inclusion from 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 topological structure of $\Gal(\mathbb{L}/\mathbb{M})$ is the same as the subspace topology that $U_\mathbb{M}(\id_\mathbb{L})$ inherits 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, considering the collection $\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})$ of finite degree Galois subextensions, by §Galois Extensions, ⁋Proposition 11 we know that this is a cofinal subset of $\Ext_{\fin}(\mathbb{L}/\mathbb{K})$. That is, $(U_\mathbb{M}(\id_\mathbb{L}))_{\mathbb{M}\in\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})}$ 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, ⁋Proposition 13, 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, since $\rho$ is a continuous map from $\Gal(\mathbb{L}/\mathbb{K})$ to the finite discrete space $\Gal(\mathbb{M}/\mathbb{K})$ (Example 1), $\ker\rho$ is a closed subgroup of $\Gal(\mathbb{L}/\mathbb{K})$. Now, 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. However, the following holds.

Proposition 3 In the above situation, the following equation

\[\{\id_\mathbb{L}\}=\bigcap_{\mathbb{M}\in \Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})}U_\mathbb{M}(\id_\mathbb{L})\]

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 $\sigma(x)\neq x$. Then, 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 the result of 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 from this we see that $\Gal(\mathbb{L}/\mathbb{K})$ 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 Extensions, ⁋Proposition 3) In other words, if we consider

\[\Gal(\mathbb{L}/\mathbb{K})\hookrightarrow \prod_{x\in \mathbb{L}}\mathbb{L}\overset{\pr_x}{\longrightarrow}\mathbb{L};\qquad \sigma\mapsto \sigma(x)\]

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 Convergence of Filters, ⁋Theorem 19 (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 $x\in \mathbb{L}$, maps the finite set of roots in $\mathbb{L}$ of the minimal polynomial of $x$ to itself and is thus bijective on it, so $x$ 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 for instance that 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. That is, the above open neighborhood does not intersect $\Gal(\mathbb{L}/\mathbb{K})$, 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 can prove that $\Gal(\mathbb{L}/\mathbb{K})$ is closed in $\mathbb{L}^\mathbb{L}$.

Now let $\mathbb{L}/\mathbb{K}$ be a Galois extension, let Galois subextensions $\mathbb{L}_i/\mathbb{K}$ of this extension satisfy $\mathbb{L}=\bigcup_{i\in I}\mathbb{L}_i$, 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

\[\rho_{ij}:\Gal(\mathbb{L}_j/\mathbb{K}) \rightarrow \Gal(\mathbb{L}_i/\mathbb{K})\qquad \text{whenever $i\leq j$}\]

Then 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 canonical morphisms $\rho_i:\varprojlim \Gal(\mathbb{L}_i/\mathbb{K})\rightarrow\Gal(\mathbb{L}_i/\mathbb{K})$ exist. ([Category Theory] §Limits, ⁋Example 5 (Inverse limit))

On the other hand, considering the restriction maps

\[\lambda_i:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}_i/\mathbb{K})\]

they 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 defines an isomorphism between 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 for any $i$ we have $\sigma\vert_{\mathbb{L}_i}=\id_{\mathbb{L}_i}$, 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 with $\mathbb{L}_i\cup\mathbb{L}_j\subseteq \mathbb{L}_k$ for some $k$, 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 together to a single $\mathbb{L}_k$, 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. This is because the compositum of two elements of this family is again a finite degree Galois subextension by §Galois Extensions, ⁋Proposition 10, and in $\mathbb{L}$, any element $x$ belongs to an element containing $\mathbb{K}(x)$ in $\Ext_{\fin,\gal}(\mathbb{L}/\mathbb{K})$. 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 contained in this action is Galois cohomology, and to conclude this post, we examine Hilbert’s Theorem 90, the classical theorem at its starting point. 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 $\sigma,\tau\in G$ the equation

\[\varphi(\sigma\tau)=\varphi(\sigma)\cdot\sigma\bigl(\varphi(\tau)\bigr)\]

holds. In particular, for some $c\in\mathbb{L}^\times$, a 1-cocycle of the form $\varphi(\sigma)=\sigma(c)/c$ 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 of it. Therefore we can consider the quotient group, which we denote 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 by §Étale Algebras, ⁋Corollary 3 (Dedekind) they are linearly independent in the $\mathbb{L}$-vector space. Since the values of $\varphi$ are all non-$0$, the linear combination

\[\sum_{\tau\in G}\varphi(\tau)\tau\]

is not the zero map, and therefore for some $x\in\mathbb{L}$ we have

\[b=\sum_{\tau\in G}\varphi(\tau)\tau(x)\neq0\]

Now for any $\sigma\in G$, writing the cocycle condition as $\sigma(\varphi(\tau))=\varphi(\sigma)^{-1}\varphi(\sigma\tau)$, we compute

\[\sigma(b)=\sum_{\tau\in G}\sigma(\varphi(\tau))\sigma\tau(x)=\varphi(\sigma)^{-1}\sum_{\tau\in G}\varphi(\sigma\tau)\sigma\tau(x)=\varphi(\sigma)^{-1}b\]

The last equality holds because as $\tau$ ranges over all of $G$, $\sigma\tau$ also ranges over all of $G$. Therefore, setting $c=b^{-1}$, we have

\[\varphi(\sigma)=\frac{b}{\sigma(b)}=\frac{\sigma(c)}{c}\]

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, ⁋Theorem 8 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, we have

\[N_{\mathbb{L}/\mathbb{K}}\bigl(\sigma(y)/y\bigr)=\prod_{i=0}^{n-1}\frac{\sigma^{i+1}(y)}{\sigma^i(y)}=\frac{\sigma^n(y)}{y}=1\]

The middle equality is telescoping and the last equality is because $\sigma^n=\id_\mathbb{L}$.

Conversely, assume $N_{\mathbb{L}/\mathbb{K}}(x)=1$. Define the function $\varphi:G \rightarrow \mathbb{L}^\times$ by

\[\varphi(\sigma^i)=\prod_{k=0}^{i-1}\sigma^k(x)\qquad(0\leq i\leq n-1)\]

Here, when $i=0$, the empty product gives $\varphi(\id)=1$. Let us verify that this is a 1-cocycle. For $0\leq a,b\leq n-1$, we have

\[\varphi(\sigma^a)\cdot\sigma^a\bigl(\varphi(\sigma^b)\bigr)=\prod_{k=0}^{a-1}\sigma^k(x)\cdot\prod_{k=0}^{b-1}\sigma^{a+k}(x)=\prod_{k=0}^{a+b-1}\sigma^k(x)\]

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 $\sigma^k=\sigma^{k-n}$ ($k\geq n$),

\[\prod_{k=0}^{a+b-1}\sigma^k(x)=\prod_{k=0}^{n-1}\sigma^k(x)\cdot\prod_{k=n}^{a+b-1}\sigma^k(x)=N_{\mathbb{L}/\mathbb{K}}(x)\cdot\prod_{k=0}^{a+b-n-1}\sigma^k(x)=\varphi(\sigma^{a+b-n})\]

and since $\sigma^a\sigma^b=\sigma^{a+b-n}$, the cocycle condition holds as well. 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 $i=1$,

K

Properties of Galois Groups

Topology of the Galois Group

Galois Cohomology

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