§ 10. Galois extensions

Throughout this paragraph K denotes a field.

1. Definition of Galois extensions

Theorem 1

— Let N be an algebraic extension of K and $\Gamma$ the group of K-automorphisms of N. Then the following assertions are equivalent: a) Every element of N invariant under $\Gamma$ belongs to the image of K in N. b) N is a separable quasi-Galois extension of K. c) For each $x \in N$ the minimal polynomial of x over K splits in $N[X]$ into a product of distinct polynomials of degree 1. The equivalence of b) and c) follows from the Cor. of Prop. 6 (V, p. 40) and the definition of quasi-Galois extension (V, p. 53, Def. 2). Let us identify K with its canonical image in N. $a) \Rightarrow c)$ : Suppose that K is the field of invariants of $\Gamma$ . Let $x \in N$ , with minimal polynomial $f$ over K and let A be the set of all roots of f in N. Put

g(X) = \prod_{y \in A} (X - y).

Every automorphism

a \in \Gamma

induces a permutation of A, and hence leaves invariant the coefficients of the polynomial

g \in N[X]

. We thus have

g \in K[X]

and since

g(x) = 0

, the polynomial g is a multiple of

f

K[X]

(V, p. 16, Th. 1). Further,

f

and g are monic and g divides

f

(IV, p. 16, Prop. 5); thus we have

f = g

, which means that the minimal polynomial

f

x

over

K

is a product in

N[X]

of distinct polynomials of degree 1.

c) ⇒ a): let x be an element of N not belonging to K. Denote by Ω an algebraic closure of K containing N as subextension (V, p. 23, Th. 2). Let f be the minimal polynomial of x over K, which is of degree ≥ 2 by hypothesis and let A be the set of roots of $f(X)$ in N. If condition c) holds, we have $f(X) = \prod_{y \in A} (X - y)$ and so (V, p. 53, Cor. 1) A is the set of conjugates of x in Ω. Since f has degree ≥ 2, there exists in A an element $y \neq x$ , hence a K-automorphism u of Ω such that $u(x) = y$ . Now under the hypothesis c), the extension N of K is quasi-Galois, hence $u(N) = N$ (V, p. 54, Cor. 1); it follows that u induces a K-automorphism a of N such that $\sigma(x) = y \neq x$ , so K is the field of invariants of Γ.

Definition 1

— An extension N of K is said to be Galois if it is algebraic and satisfies the equivalent conditions a), b), c) of Th. 1.

Let N be a field, Γ a group of automorphisms of N and $N_0$ the field of invariants of Γ. When N is algebraic over $N_0$ it is a Galois extension of $N_0$ . This need not always be so: for example, suppose that K is infinite and take N to be the field of rational fractions K(X); for each $a \in K$ let $\sigma_a$ be the automorphism of K(X) which maps $f(X)$ to $f(X + a)$ . The set of all $a$ , is a group of automorphisms of K(X) whose field of invariants is easily seen to be K; however K(X) is not algebraic over K.

Let Ω be an algebraic closure of K, let A denote a set of elements of Ω separable over K and B the set of conjugates over K of elements of A. Then B consists of elements that are algebraic and separable over K. Therefore (V, p. 39, Prop. 6 and p. 56, Prop. 5) the field K(B) is a separable quasi-Galois extension of K; in other words, the quasi-Galois extension generated by A (V, p. 55) is a Galois extension of K; we shall also say that it is the Galois extension of K generated by the subset A of O.

In particular, the splitting field in Ω of a family of separable polynomials over K, a separable closure of K, are Galois extensions of K.

Proposition 1

— Let N be an extension of K and $(N_i)_{i \in I}$ , a non-empty family of subextensions of N. We put $E = \cap_{i \in I} N_i$ and $F = K \left( \bigcup_{i \in I} N_i \right)$ . If all the extensions N, are Galois over K then the same is true of E and F.

In the first place E is algebraic and separable over K (V, p. 36, Prop. 1) and the same holds for F (V, p. 41, Prop. 8). Moreover, E and F are quasi-Galois over K by Prop. 4 (V, p. 55).

Proposition 2

— Let N be a Galois extension of K and E a subextension of N, of finite degree over K. There exists a subextension F of N containing E, Galois and of finite degree over K.

Since N is quasi-Galois over K, Cor. 1 of V, p. 56 proves the existence of a quasi-Galois subextension F of N containing E and of finite degree over K. Since N is separable over K, the same holds for F (V, p. 36, Prop. 1), hence F is Galois over K.

Prop. 2 implies the following result: let R be an algebraic closure of K and E₁, ..., Eₙ separable algebraic extensions of finite degree over K, contained in R. Then there exists a Galois extension N of K, of finite degree, contained in R and containing E₁, ..., Eₙ.

2. The Galois group

Definition 2

Let N be a Galois extension of the field K. The group of all K-automorphisms of N will be called the Galois group of N over K and denoted by Gal(N/K).

Let N be a finite Galois extension of K. Then N is a finite separable and quasi-Galois extension of K. Therefore (V, p. 32, Prop. 4 and V, p. 54, Prop. 3), the order of Gal(N/K) is equal to [N : K]. We shall prove later that if N is a Galois extension of K such that Gal(N/K) is finite, then N is of finite degree over K (V, p. 66, Th. 3).

Let Ω be an algebraically closed extension of K and let A be the set of roots in Ω of a separable polynomial f ∈ K[X]. Then the field N = K(A) is a Galois extension of K. It is clear that every K-automorphism of N leaves A stable, and since A generates N over K, the mapping a ↦ σ|A is an isomorphism of Gal(N/K) onto a subgroup Γ of the symmetric group S_A of the set A, which will be called the Galois group of the polynomial f. From Remark 3 (V, p. 55) it follows that if x and y belong to A, the following properties are equivalent:

a) x and y are conjugate over K, b) x and y belong to the same orbit under Γ, c) x and y are roots of the same irreducible factor of f.

In particular f is irreducible if and only if A is non-empty and Γ operates transitively on A.

Example 1

Suppose that the characteristic of K is different from 2 and let N be a quadratic extension of K. If x ∈ N − K, then we have N = K(x) and the minimal polynomial of x over K is of the form f(X) = X² − aX + b, with a, b ∈ K. We thus have f(X) = (X − x)(X − y) where y = a − x, so y is conjugate to x; since f(X) is separable, the extension N is Galois. The group Gal(N/K) has two elements which induce the two permutations of the set {x, y}.

Example 2

\* Let f = X³ + X² − 2X − 1 ∈ Q[X]. The polynomial f is irreducible, for if not, then it would have a root x ∈ Q; on writing x = a/b with a, b ∈ Z, a and b relatively prime, we should find a(a² + ab − 2b²) = b³ and a³ = b(b² + 2ab − a²); but this implies that a divides b and b divides a, hence x = ±1, which is impossible. Put ξ = e^{2\pi i / 7} ∈ C; then the polynomial f has the roots α = ξ + ξ⁻¹, β = ξ² + ξ⁻², γ = ξ³ + ξ⁻³. We have β = a² − 2 and γ = a³ − 3α, hence the extension Q(α) is Galois over Q. The Galois group of Q(α) over Q is cyclic of order 3 and is generated by an element σ such that σ(α) = β, σ(β) = γ, σ(γ) = α. \

Example 3

Suppose that $K = \mathbf{Q}$ and take $f = X^3 - 2$ . Using the decomposition of integers into products of prime factors (I, p. 51), we easily see that 2 is not the cube of an element of $\mathbf{Q}$ . Hence the polynomial $f$ is irreducible, for if not, it would have a root in $\mathbf{Q}$ . Let $A = \{ x_1, x_2, x_3 \}$ be the set of roots of $f$ in $\Omega$ and $\Gamma$ the Galois group of $f$ . It acts transitively on $A$ ; its order is therefore divisible by three. On the other hand, the quotient $j = \frac{x_2}{x_1}$ is different from 1 and we have $j^3 = 1$ . Hence $j$ satisfies the relation $j^2 + j + 1 = 0$ ; now the polynomial $T^2 + T + 1 = \left( T + \frac{1}{2} \right)^2 + \frac{3}{4}$ has no roots in $\mathbf{Q}$ , which shows (Example 1) that $[\mathbf{Q}(j) : \mathbf{Q}] = 2$ . Thus $[\mathbf{N} : \mathbf{Q}]$ is divisible by 2, and it follows that the order of $\Gamma$ is divisible by 6. Since $\Gamma$ is contained in the group $\mathfrak{S}_A$ of order 6, we have $\Gamma = \mathfrak{S}_A$ .

Example 4

Suppose that $K$ is of characteristic $p \neq 0$ and let $K(T)$ be the field of rational fractions and $U = T^p - T$ . Put $E = K[U]$ and $F = K[T]$ , then the polynomial $f(X) = X^p - X - U$ of $E[X]$ has the roots $T, T+1, ..., T+p-1$ in $F$ . Let $\sigma$ be the $K$ -automorphism of $F$ such that $\sigma(T) = T+1$ . We have $\sigma^i(T) = T+i$ and $\sigma(U) = U$ . The group $G = \{ 1, \sigma, ..., \sigma^{p-1} \}$ is cyclic of order $p$ , and its field of invariants contains $E$ ; since $[F : E] \leq p$ , Dedekind's theorem (V, p. 27, Cor. 2) implies that $E$ is the field of invariants of $G$ and $[F : E] = p$ . The polynomial $f$ is thus irreducible in $E[X]$ ; the extension $F$ of $E$ is Galois, its Galois group $G$ is cyclic of order $p$ , and the group $\Gamma$ is the group of cyclic permutations of $T, T+1, ..., T+p-1$ .

For a generalization of this example see V, p. 93, Example 2.

Example 5

Let $F = K(X_1, ..., X_n)$ be the field of rational fractions in $n$ indeterminates $X_1, ..., X_n$ with coefficients in $K$ . Put

s_k = \sum_{1 \leq i_1 < ... < i_k \leq n} X_{i_1} ... X_{i_k}

for

1 \leq k \leq n

and

E = K(s_1, ..., s_n)

; we denote by

f(T)

the polynomial

T^n - s_1 T^{n-1} + \cdots + (-1)^n s_n .

We have

f(T) = \prod_{i=1}^n (T - X_i)

, so that

F

is a splitting field of the separable polynomial

f(T) \in E[T]

. Further, for every permutation

a \in \mathfrak{S}_n

there exists a unique

K

-automorphism

h_a

F

such that

h_a(X_i) = X_{a(i)}

for

1 \leq i \leq n

; we have

h_a(s_k) = s_k

for

1 \leq k \leq n

, hence

h_a

is an

E

-automorphism of

F

. In other words,

F

is a Galois extension of

E

and the restriction to the set of roots

\{ X_1, ..., X_n \}

f(T)

defines an isomorphism of

\mathrm{Gal}(F/E)

onto the group

\mathfrak{S}_n

. In particular,

E

consists of rational fractions

f

such that

f(X_{\sigma(1)}, ..., X_{\sigma(n)}) = f(X_1, ..., X_n)

for every

\sigma \in \mathfrak{S}_n

(cf. IV, p. 67, Cor.).

Example 6

Suppose that f is monic of degree > 0 and K is of characteristic ≠ 2. Define on A a total ordering, denoted by ≤, and put δ(f) = $\prod_{\alpha < \beta} (\beta - a)$ , $(\alpha, \beta) \in A \times A$ , and for each $\sigma \in \mathfrak{S}_A$ put $\delta_\sigma(f) = \prod_{\alpha < \beta} (\sigma(\beta) - \sigma(\alpha))$ . We have $\delta_\sigma(f) = \varepsilon(u) \delta(f)$ , where $\varepsilon(u)$ is the signature of $\sigma$ (I, p. 64) and $\delta(f) \neq 0$ . For every $\tau \in \mathrm{Gal}(N/K)$ we have $\tau(\delta(f)) = \delta_{\tau|_A}(f)$ . Hence $\Gamma$ is contained in the alternating group $\mathfrak{A}_A$ if and only if $\delta(f) \in K$ . Moreover $\delta(f)^2 = \prod_{\alpha < \beta} (\beta - a)' = d(f)$ is the discriminant of the polynomial f (IV, p. 81). Hence $\Gamma \subset \mathfrak{A}_A$ if and only if $d(f)$ is the square of an element of K. Thus in Example 2 we have $d(f) = 49 = 7^2$ and in Example 3, $d(f) = -108$ (IV, p. 85).

Let N be a Galois extension of K and let L be a subextension of N which is Galois over K. Every K-automorphism $\sigma$ of N induces a K-automorphism $\sigma_L$ of L (V, p. 55, Remark 1). Therefore the mapping $\sigma \mapsto \sigma_L$ is a homomorphism of $\mathrm{Gal}(N/K)$ into $\mathrm{Gal}(L/K)$ , called the restriction homomorphism.

Proposition 3

— The restriction homomorphism of $\mathrm{Gal}(N/K)$ into $\mathrm{Gal}(L/K)$ is surjective.

More generally, consider two subextensions L and L' of N, and a K-isomorphism $u$ of L onto L'. Choose an algebraic closure $\Omega$ of K containing N as subextension (V, p. 23, Th. 2). There exists a K-automorphism $v$ of $\Omega$ which coincides with $u$ on L (V, p. 52, Prop. 1), and since N is a quasi-Galois extension of K, $v$ induces a K-automorphism $\sigma$ of N (V, p. 55, Remark 1). In other words, the element $\sigma$ of $\mathrm{Gal}(N/K)$ coincides with $u$ on L.

3. Topology of the Galois group

Let N be a Galois extension of K and $\Gamma$ the Galois group of N over K. We equip N with the discrete topology, the set $N^N$ of all mappings of N into itself with the product topology of the discrete topology of the factors ("topology of simple convergence in N") and the group $\Gamma$ with the topology induced from $N^N$ .

Let A be the set of all subextensions of N of finite degree over K. For $\sigma \in \Gamma$ and $E \in A$ we shall write $U_E(\sigma)$ for the set of elements $\tau$ of $\Gamma$ which have the same restriction as $\sigma$ to E. If $E = K(x_1, ..., x_n)$ , the set $U_E(\sigma)$ consists of the elements $\tau \in \Gamma$ such that $\tau(x_i) = \sigma(x_i), ..., \tau(x_n) = \sigma(x_n)$ . It follows that the family $(U_F(\sigma))_{F \in \mathscr{F}}$ is a base of the filter of neighbourhoods of $\sigma$ in $\Gamma$ .

When N is of finite degree over K, we have $N \in A$ and $U_N(\sigma) = \{\sigma\}$ , hence the topology of $\mathrm{Gal}(N/K)$ is discrete; we recall (V, p. 58), that the group $\mathrm{Gal}(N/K)$ is finite in this case.

This description of the topology of $\mathrm{Gal}(N/K)$ shows that the restriction homomorphism of $\mathrm{Gal}(N/K)$ onto $\mathrm{Gal}(L/K)$ is continuous for every subextension L of N which is Galois over N.

Let $A$ be a subset of $\Gamma$ . To say that $A$ is open means that for each $\sigma \in A$ there exists $E$ in $A$ such that the set $U_E(\sigma)$ is contained in $A$ . The closure $\overline{A}$ of $A$ consists of all $\sigma \in \Gamma$ such that for any $E \in A$ there exists $\tau \in A$ having the same restriction to $E$ as $\sigma$ ; the field of invariants of $\overline{A}$ is the same as that of $A$ .

Let $\varepsilon$ be the neutral element of $\Gamma$ and let $A'$ be the set of subextensions of $N$ which are Galois and of finite degree over $K$ . By Prop. 2 (V, p. 57) the set $A'$ is cofinal in $A$ and the family $(U_E(\varepsilon))_{E \in A'}$ is thus a base of the filter of neighbourhoods of $\varepsilon$ in $\Gamma$ . Further, for $E \in A'$ the set $U_E(E)$ is the kernel of the restriction homomorphism of $\mathrm{Gal}(N/K) = \Gamma$ into $\mathrm{Gal}(E/K)$ . Since $\mathrm{Gal}(E/K)$ is finite, it follows that $U_{\Gamma}(\varepsilon)$ is an open and closed subgroup, normal and of finite index in $\Gamma$ .

Clearly we have $U_{\Gamma}(\sigma) = \sigma U_{\Gamma}(\varepsilon) = U_{\Gamma}(\varepsilon) \sigma$ for $\sigma \in \Gamma$ and $E \in A'$ . Since $U_{\Gamma}(\varepsilon)$ is a normal subgroup of $\Gamma$ for all $E \in A'$ and the family $(U_E(\varepsilon))_{E \in A'}$ is a base of neighbourhoods at $\varepsilon$ , the topology of $\Gamma$ is compatible with the group structure (Gen. Top., III, p. 223). In other words, the mapping $(\sigma, \tau) \mapsto \sigma \tau^{-1}$ of $\Gamma \times \Gamma$ into $\Gamma$ is continuous.

Proposition 4

— Let $N$ be a Galois extension of $K$ . Then the Galois group $\Gamma = \mathrm{Gal}(N/K)$ is compact and totally disconnected.

Every element $\sigma$ of $\Gamma$ has a base of neighbourhoods consisting of the open and closed sets $U_E(\sigma)$ , hence $\Gamma$ is totally disconnected (Gen. Top., I, p. 111). We have $\{\sigma\} = \bigcap_{E \in \Lambda} U_E(\sigma)$ , hence $\Gamma$ is separated. For each $x \in N$ the set of conjugates $\sigma(x)$ of $x$ , as $\sigma$ ranges over $\Gamma$ , is finite because $x$ is algebraic over $K$ (V, p. Cor. 1); hence all the projections of $\Gamma$ on the factor spaces of $N^N$ are finite sets, and this shows $\Gamma$ to be relatively compact in $N^N$ (Gen. Top., I, p. 88). It remains to show that $\Gamma$ is closed in $N^N$ . Now if $u$ is in the closure of $\Gamma$ in $N^N$ , then for each pair of points $(x, y)$ of $N$ there exists $a \in \Gamma$ with $u(x) = \sigma(x), \ u(y) = \sigma(y), \ u(x+y) = \sigma(x+y), \ u(xy) = \sigma(xy)$ , whence $u(x+y) = u(x) + u(y)$ and $u(xy) = u(x)u(y)$ . By the same reasoning we have $u(x) = x$ for all $x \in K$ , hence $u$ is a $K$ -homomorphism of $N$ into $N$ ; since $N$ is algebraic over $K$ , $u$ is a $K$ -automorphism of $N$ (V, p. 52, Prop. 1), hence $u \in T$ .

Let $N$ be a Galois extension of $K$ and $(N_i)_{i \in I}$ an increasing directed family of subextensions of $N$ . Suppose that $N_i$ is Galois over $K$ for all $i \in I$ and that $N = \bigcup_{i \in I} N_i$ . For each $i \in I$ denote by $\Gamma_i$ the Galois group of $N_i$ over $K$ ; for $i \leq j$ in $I$ we have $N_i \subset N_j$ and the restriction homomorphism $\varphi_{ij}$ of $\Gamma_j$ into $\Gamma_i$ is defined. It is continuous and so the family $(\Gamma_i, \varphi_{ij})$ is an inverse system of topological groups. Further, for each $i \in I$ denote by $\lambda_i$ the restriction homomorphism of $\mathrm{Gal}(N/K)$ into $\mathrm{Gal}(N_i/K) = \Gamma_i$ ; it is continuous and we have $\lambda_i = \varphi_{ij} \circ \lambda_j$ for $i \leq j$ , hence the family $(\lambda_i)_{i \in I}$ defines a continuous homomorphism $\Lambda$ of $\mathrm{Gal}(N/K)$ into $\varprojlim \Gamma_i$ .

Proposition 5

— The homomorphism $\lambda$ of $\mathrm{Gal}(N/K)$ into $\varprojlim \mathrm{Gal}(N_i/K)$ is an isomorphism of topological groups.

Since $\mathrm{Gal}(N/K)$ is compact, $\lambda$ continuous and the group $\varprojlim \mathrm{Gal}(N_i/K)$ separated, it suffices to show that $h$ is bijective (Gen. Top., I, p. 87, Cor. 2). Let $u = (u_i)_{i \in I}$ be an element of $\varprojlim \mathrm{Gal}(N_i/K)$ ; for each $i \in I$ , $u_i$ is a K-automorphism of $N_i$ and $u_i$ is the restriction of $u_j$ to $N_i$ for $i \leq j$ . Since $N = \bigcup_{i \in I} N_i$ , there exists a unique element $\sigma$ of $\mathrm{Gal}(N/K)$ which coincides with $u_i$ on $N_i$ for all $i \in I$ . Thus $\sigma$ is the unique element of $\mathrm{Gal}(N/K)$ such that $\lambda(u) = u$ , hence $\lambda$ is bijective.

This applies in particular when we take for the family $(N_i)$ the family of all finite Galois subextensions of $N$ ; then each group $\mathrm{Gal}(N_i/K)$ is discrete and finite. The topological group $\mathrm{Gal}(N/K)$ is thus isomorphic to a directed inverse limit of finite groups, equipped with the discrete topology; this is sometimes called a profinite topological group.

4. Galois descent

In this No. we denote by $N$ a field, $\Gamma$ a group of automorphisms of $N$ , $e$ the neutral element of $\Gamma$ and $K$ the field of invariants of $\Gamma$ .

Let $V$ be a vector space over $N$ . We recall (II, p. 317) that a K-structure on $V$ is a vector sub-K-space $V_0$ of $V$ such that the K-linear mapping $\varphi : N \otimes_K V_0 \to V$ which maps $h \otimes x$ to $\lambda x$ is bijective. Let $V_0$ be such a K-structure; for each $\sigma \in \Gamma$ we put $u_\sigma = \varphi \circ (\sigma \otimes \mathrm{Id}_{V_0}) \circ \varphi^{-1}$ ; then we have $u_\sigma \left( \sum_{i \in I} \lambda_i e_i \right) = \sum_{i \in I} \sigma(\lambda_i) e_i$ for every family of elements $\lambda_i$ of $N$ and $e_i$ of $V_0$ whence we obtain the relations

\begin{align*} u_\sigma(x + y) &= u_\sigma(x) + u_\sigma(y) \\ u_\sigma(\lambda x) &= \sigma(\lambda) u_\sigma(x) \\ u_\sigma \circ u_\tau &= u_{\sigma \tau} \\ u_e &= \mathrm{Id}_V \end{align*}

for

\sigma, \tau

\Gamma

x, y

V

and

\lambda

N

Proposition 6

— a) Let $V$ be a vector space over $N$ with a $K$ -*structure*. For a vector $x \in V$ to be rational over $K$ it is necessary and sufficient that $u_\sigma(x) = x$ for all $\sigma \in \Gamma$ . For a vector sub-N-space $W$ of $V$ to be rational over $K$ it is necessary and sufficient that $u_\sigma(W) \subset W$ for all $\sigma \in \Gamma$ .

b) Let $V_1$ and $V_2$ be two vector spaces over $N$ , each with a $K$ -*structure*. For a linear mapping $f$ of $V_1$ into $V_2$ to be rational over $K$ it is necessary and sufficient that $f(u_\sigma(x)) = u_\sigma(f(x))$ for all $\sigma \in \Gamma$ and all $x \in V_1$ .

It is clear that $K$ is the set of $x \in N$ such that $\sigma(xy) = x \sigma(y)$ for all $\sigma \in \Gamma$ and all $y \in N$ . The proposition therefore follows from Th. 1 (II, p. 324).

Corollary

— Let $V_0$ be a vector space over $K$ and let $W$ be a vector sub-N-space of $N \otimes_K V_0$ . Suppose that $W$ is stable under the mappings $\sigma \otimes \mathrm{Id}_{V_0}$ for all $\sigma \in \Gamma$ . Let $W_0$ be the set of $x \in V_0$ such that $1 \otimes x \in W$ ; then $W_0$ is the unique vector sub-K-space of $V_0$ such that $W = N \otimes_K W_0$ .

It suffices to remark that the set of elements of the form $1 \otimes x \ (x \in V_0)$ is a K-structure on $N \otimes_K V_0$ for which we have $u_\sigma = \sigma \otimes \mathrm{Id}_{V_0}$ for $\sigma \in \Gamma$ .

Proposition 7

— Let $V$ be a vector space over $N$ , $(u_\sigma)_\sigma$ , r a family of mappings of $V$ into itself satisfying (1) to (4) and $V_0$ the set of $x \in V$ such that $u_\sigma(x) = x$ for all $\sigma \in \Gamma$ .

a) $V_0$ is a vector sub-K-space of $V$ and the K-linear mapping $\varphi$ of $N \otimes_K V_0$ into $V$ which maps $\lambda \otimes x$ to $\lambda x$ is injective.

b) If $\Gamma$ is finite, then $\varphi$ is bijective and $V_0$ is a K-structure on $V$ .

It is clear that $V_0$ is a vector sub-K-space of $V$ .

The formula $u_\sigma \circ \varphi = \varphi \circ (\sigma \otimes \mathrm{Id}_{V_0})$ shows that the kernel $W$ of $\varphi$ is stable under the mappings $\sigma \otimes \mathrm{Id}_{V_0}$ ; by the Cor. to Prop. 6 there exists therefore a subspace $W_0$ of $V_0$ such that $W = N \otimes_K W_0$ . If $x$ belongs to $W_0$ we then have $x = \varphi(1 \otimes x) = 0$ , hence $W_0 = 0$ and so $W = 0$ . This proves a).

Suppose now that $\Gamma$ is finite; we have to show that $\varphi$ is surjective, or equivalently that $V_0$ generates the vector N-space $V$ . Thus let $f$ be an N-linear form on $V$ whose restriction to $V_0$ is zero. Let $x \in V$ ; for every $\lambda \in N$ the element $y_\lambda = \sum_{\sigma \in \Gamma} u_\sigma(\lambda x)$ of $V$ clearly belongs to $V_0$ , whence $f(y_\lambda) = 0$ , that is,

\sum_{\sigma \in \Gamma} f(u_\sigma(x)) \sigma(A) = 0.

By Dedekind's theorem (V, p. 27, Cor. 2) we thus have

f(u_\sigma(x)) = 0

for each

\sigma \in \Gamma

; in particular, taking

\sigma = \varepsilon

we find

f(x) = 0

, which means that

f = 0

. This proves b).

Let $M$ be a vector space over $N$ ; for each $\sigma \in \Gamma$ let $M^\sigma$ be the vector space over $N$ with the same underlying additive group as $M$ , with the external law $(A, x) \mapsto \sigma(\lambda) x$ . Write $V = \prod_{\sigma \in \Gamma} M^\sigma$ ; the underlying additive group of $V$ is that of all mappings of $\Gamma$ into $M$ , with the external law defined by

(\lambda . h)(\sigma) = \sigma(\lambda) h(\sigma) \quad (\lambda \in N, h \in V, \sigma \in \Gamma).

(The product

\sigma(\lambda) h(\sigma)

is calculated in the vector space

M

.) Further, we define on

N \otimes_K M

a vector space structure over

N

by the formula

\lambda \left( \sum_i \mu_i \otimes x_i \right) = \sum_i \lambda \mu_i \otimes x_i

Finally we denote by

\psi

the K-linear mapping of

N \otimes_K M

into

V

characterized by the relation

\psi(\lambda \otimes x)(\sigma) = \sigma(\lambda) . x

for

A \in N, x \in M

and

\sigma \in \Gamma

. It is clear that

\psi

is N-linear

Proposition 8

— The N-linear mapping $\psi$ of $N \otimes_K M$ into $V = \prod_{\sigma \in \Gamma} M^\sigma$ is injective, and it is bijective if $\Gamma$ *is finite.

For every $a \in \Gamma$ we define a mapping $u_\sigma$ of $V$ into $V$ by

(u_\sigma h)(\tau) = h(\tau \sigma)

for $h \in V$ and $\tau \in \Gamma$ . The verification of (1)-(4) is immediate. Denote by $V$ , the set of $h \in V$ such that $u_\sigma(h) = h$ for all $\sigma \in \Gamma$ . For each $x \in M$ let $\theta(x)$ be the constant mapping of $\Gamma$ into $M$ with value $x$ ; then $\theta$ is a K-isomorphism of $M$ onto $V$ . If we define the homomorphism $\varphi : N \otimes_K V, \to V$ as above, we have $\psi = \varphi \circ (\mathrm{Id}_N \otimes \theta)$ and now Prop. 8 follows from Prop. 7.

Corollary

— Let $\psi$ be the K-linear mapping of $N \otimes_K N$ into the product vector space $N^\Gamma$ such that $\psi(x \otimes y)\,(a) = \sigma(x)\,y$ for $x,\,y$ in $N$ and $\sigma \in \Gamma$ . Then $\psi$ is injective and it is bijective when $\Gamma$ *is finite.

This is the particular case $M = N$ of Prop. 8.

Remark 1

Let F be an extension of K and $N$ a subextension of F, and let $\Gamma$ be a finite group of automorphisms of $N$ , of which K is the field of invariants. Prop. 8 implies the existence of an isomorphism of K-algebras $0 : N \otimes_F F \to F^\Gamma$ characterized by $\theta(x \otimes y)\,(a) = \sigma(x)\,y$ for $x \in N,\,y \in F$ and $\sigma \in \Gamma$ .

Remark 2

The notation K, N and $\Gamma$ has the same meaning as before. For each integer $n \geqslant 1$ , let $A_n$ be the tensor product of n K-algebras identical with $N$ ; let $B_n$ be the set of mappings of $\Gamma^{n-1}$ into $N$ . By induction on $n$ we can deduce from the Cor. of Prop. 8 the existence of an isomorphism $\varphi_n : A_n \to B_n$ , mapping $x_1 \otimes \ldots \otimes x_n$ to the function $(\sigma_1, \ldots, a, ) \quad \sigma_1(x_1) \cdot \sigma_n(x_{n-1})\,x_n$ .

5. Galois cohomology

Let $N$ be a field, $\Gamma$ a finite group of automorphisms of $N$ and K the field of invariants of $\Gamma$ . For every integer $n \geqslant 1$ we denote by $\mathbf{GL}(n, N)$ the group of square matrices of order $n$ with coefficients in $N$ and non-zero determinant (II; p. 349). We let the group $\Gamma$ operate on the group $\mathbf{GL}(n, N)$ by the rule $\sigma(A) = (\sigma(a_{ij}))$ for $A = (a_{ij})$ .

Proposition 9

— Let $(U_\sigma)_{\sigma \in \Gamma}$ be a family of elements of $\mathbf{GL}(n, N)$ . For $A$ to exist in $GL(n, N)$ such that $U_\sigma = A^{-1}\,a(A)$ for all $\sigma \in \Gamma$ it is necessary and sufficient that $U_{\sigma \tau} = U_\sigma \cdot \sigma(U_\tau)$ for $\sigma,\,\tau$ in $\Gamma$ .

The condition is necessary: if $U_\sigma = A^{-1}\,a(A)$ , then we have

U_\sigma \cdot \sigma(U_\tau) = A^{-1}\,\sigma(A)\,\sigma(A^{-1}\,\tau(A)) = A^{-1}\sigma\tau(A) = U_{\sigma \tau}.

The condition is sufficient: we identify the elements of $N^n$ with matrices of $n$ rows and one column with coefficients in $N$ . We let the groups $\Gamma$ act on $N^n$ by

\sigma(x) = (\sigma(x_i))_{1 \leq i \leq n} \quad \text{for} \quad x = (x_i)_{1 \leq i \leq n}.

For each

a \in \Gamma

we denote by

u_\sigma

the mapping

x \mapsto U_\sigma \cdot \sigma(x)

N^n

into itself. The verification of Formulae (1) to (3) of V, p. 62 is immediate. Moreover we have

u_\varepsilon \circ u_\varepsilon = u_\varepsilon

and since

u_\varepsilon

is bijective, we have

u_\varepsilon = \mathrm{Id}_{N^n}

. Let

V_0

be the set of vectors

x \in N^n

such that

u_\sigma(x) = x

for all

\sigma \in \Gamma

. By Prop. 7 (V, p. 63),

V_0

is a K-structure on

N^n

; in particular there exist in

V_0

vectors

b_1, \ldots, b_n

forming a basis of

N^n

over

N

. The matrix

B

with columns

b_1, \ldots, b_n

is therefore invertible and the relation

u_\sigma(b_i) = b_i

for

1 \leq i \leq n

is equivalent to

U_\sigma \cdot \sigma(B) = B

. Writing

A = B^{-1}

, we obtain

U_\sigma = A^{-1} \sigma(A)

for all

a \in \Gamma

Corollary 1

Let $(c_\sigma)_{\sigma \in \Gamma}$ be a family of non-zero elements of $N$ . For $a \neq 0$ to exist in $N$ such that $c_\sigma = \sigma(a) \cdot a^{-1}$ for all $\sigma \in \Gamma$ it is necessary and sufficient that $c_{\sigma \tau} = c_\sigma \cdot \sigma(c_\tau)$ for $\sigma, \tau$ in $\Gamma$ .

Corollary 2

Let $(a_\sigma)_{\sigma \in \Gamma}$ be a family of elements of $N$ . For $b$ to exist in $N$ such that $a_\sigma = \sigma(b) - b$ for all $\sigma \in \Gamma$ it is necessary and sufficient that $a_{\sigma \tau} = a_\sigma + \sigma(a_\tau)$ for $\sigma, \tau$ in $\Gamma$ .

We have $\sigma \tau(b) - b = [\sigma(b) - b] + \sigma[\tau(b) - b]$ for all $b$ in $N$ and $\sigma, \tau$ in $\Gamma$ , whence the necessity.

Conversely suppose that $a_{\sigma \tau} = a_\sigma + \sigma(a_\tau)$ for any $\sigma$ and $\tau$ in $\Gamma$ . Put $U_\sigma = \begin{pmatrix} 1 & a_\sigma \\ 0 & 1 \end{pmatrix}$ for $\sigma \in \Gamma$ ; then we have $U_{\sigma \tau} = U_\sigma \cdot \sigma(U_\tau)$ for $\sigma, \tau$ in $\Gamma$ ; by Prop. 9, there exists thus a matrix $A = \begin{pmatrix} x & y \\ z & t \end{pmatrix}$ with non-zero determinant such that $\sigma(A) = A U_\sigma$ for all $a \in \Gamma$ ; writing down the relation $\sigma(A) = A U_\sigma$ we find

\begin{pmatrix} \sigma(x) & \sigma(y) \\ \sigma(z) & \sigma(t) \end{pmatrix} = \begin{pmatrix} x & x a_\sigma + y \\ z & z a_\sigma + t \end{pmatrix} \quad (\sigma \in \Gamma).

In particular,

x

and

z

belong to

K

and we have

\sigma(y) = x a_\sigma + y, \quad \sigma(t) = z a_\sigma + t \quad (\sigma \in \Gamma).

x \neq 0

we have

a, = \sigma(b) - b

with

b = x^{-1} y

; if

z \neq 0

, we have the same relation with

b = z^{-1} t

. Now

x

and

z

cannot both be zero because

xt - yz = \det A \neq 0.

6. Artin’s theorem

Theorem 2 (Artin). — Let $N$ be a field, $\Gamma$ a group of automorphisms of $N$ and $K$ the field of invariants of $\Gamma$ . Let $V$ be a vector sub-K-space of $N$ of finite dimension over K. Then every K-linear mapping u of V into N is a linear combination with coefficients in N of the restrictions to V of elements of $\Gamma$ .

Let u be a K-linear mapping of V into N and let $V_{(N)} = N \otimes_K V$ be the vector N-space derived from V by extension of scalars; denote by $\tilde{u}$ the N-linear form on $V_{(N)}$ such that $\tilde{u}(x \otimes y) = x \cdot u(y)$ for $x \in N$ and $y \in V$ . For each $a \in \Gamma$ there exists an N-linear form h, on $V_{(N)}$ such that $h_\sigma(x \otimes y) = x \sigma(y)$ for $x \in N$ and $y \in V$ . The canonical mapping of $V_{(N)} = N \otimes_K V$ into $N \otimes_K N$ is injective. Now the Cor. of Prop. 8 (V, p. 64) show that the intersection of the kernels of the linear forms h, on $V_{(N)}$ is reduced to 0. Therefore (II, p. 302, Cor. 1) there exist $a_1, \ldots, \sigma_n$ in $\Gamma$ and $a_1, \ldots, a_n$ in $N$ such that $\tilde{u} = \sum_{i=1}^n a_i h_{\sigma_i}$ whence $u(x) = \sum_{i=1}^n a_i \sigma_i(x)$ for all $x \in V$ .

Let us equip the set $N^N$ of all mappings of N into N with the product topology of the discrete topologies of the factors. Th. 2 means that the set of linear combinations with coefficients in N of elements of $\Gamma$ is dense in the set of K-linear mappings of N into itself.

Theorem 3

— Let N be a field, $\Gamma$ a finite group of automorphisms of N and K the field of invariants of $\Gamma$ . Let n be the cardinal of $\Gamma$ . a) We have $[N : K] = n$ and N is a Galois extension of K with Galois group $\Gamma$ . b) Let $\sigma_1, \ldots, \sigma_n$ be the elements of $\Gamma$ and $(x_1, \ldots, x_n)$ a basis of N over K, then $\det (\sigma_i(x_j)) \neq 0$ . c) Let u be a K-linear mapping of N into N. There exists a unique family $(a_\sigma)_\sigma$ of elements of N such that $u(x) = \sum_{\sigma \in \Gamma} a_\sigma \sigma(x)$ for all $x \in N$ .

We equip the ring $N \otimes_K N$ with the N-algebra structure whose external law is given by $\lambda(x \otimes y) = x \otimes \lambda y$ for $\lambda, x, y$ in N. Then the dimension of the vector N-space $NO, N$ is $[N : K]$ . The dimension of the product vector N-space $N^\Gamma$ is equal to n. The mapping $\psi$ defined in the Cor. of Prop. 8 (V, p. 64) is an N-isomorphism of $NO, N$ onto $N^\Gamma$ , whence $[N : K] = n$ . Let A be the group of K-automorphisms of N. We have $\Gamma \subset A$ , hence K is the field of invariants of A, and N is a Galois extension of K. Further, the order of A is at most equal to $[N : K]$ by Dedekind's theorem (V, p. 27, Cor. 2) and since the order of $\Gamma$ equals $[N : K]$ , we have $\Gamma = A$ . Hence $\Gamma$ is the Galois group of N over K, and this proves a).

With the notation of b) put $f_i = \psi(x_i \otimes 1)$ ; we have $f_i(\sigma) = \sigma(x_i)$ for $1 \leq i \leq n$ and $\sigma \in \Gamma$ . Since $\psi$ is an isomorphism of vector N-spaces, the sequence $(f_1, \ldots, f_n)$ is a basis of N over N, whence $\det (f_j(\sigma_i)) \neq 0$ , that is,

\det (\sigma_i(x_j)) \neq 0

This proves b).

Finally, c) follows from Th. 2 (V, p. 65) which proves the existence of a family $(a_\sigma)_{\sigma \in \Gamma}$ such that $u(x) = \sum_{\sigma \in \Gamma} a_\sigma \sigma(x)$ (for all $x \in N$ ) and from Dedekind's theorem (V, p. 27, Cor. 2) which proves the uniqueness of $(a_\sigma)_{\sigma \in \Gamma}$ .

7. The fundamental theorem of Galois theory

Theorem 4

— Let N be a Galois extension of K and $\Gamma$ its Galois group. Let $\mathscr{K}$ be the set of subextensions of N and $\mathscr{G}$ the set of closed subgroups of $\Gamma$ . For every subgroup $A \in \mathscr{G}$ we denote by $k(\Delta)$ the field of invariants of A and for every subfield $E \in \mathscr{K}$ we denote by $g(E)$ the group of E-*automorphisms* of N. Then $A \mapsto k(\Delta)$ is a bijection of $\mathscr{G}$ onto $\mathscr{K}$ , and $E \mapsto g(E)$ is the inverse bijection.

A) The relation $E = k(g(E))$ (for $E \in \mathscr{K}$ ) is a consequence of the following more precise lemma:

Lemma 1

Let E be a subextension of N. Then N is a Galois extension of E and $\mathrm{Gal}(N/E)$ is a closed subgroup of $\mathrm{Gal}(N/K)$ with the induced topology.

Let $x \in N$ ; the minimal polynomial f of x over E divides in $E[X]$ the minimal polynomial $g$ of x over K (V, p. 17, Cor. 2). Since N is Galois over K, the polynomial g is a product in $N[X]$ of distinct factors of degree 1; hence the same is true of f and so N is Galois over E.

Let $\Gamma$ be the Galois group of N over K and A that of N over E. By definition A is the subgroup of $\Gamma$ consisting of all $\sigma$ such that $\sigma(x) = x$ for all $x \in E$ . Now for each $x \in E$ the mapping $\sigma \mapsto \sigma(x)$ of $\Gamma$ into the discrete space N is continuous, hence D is closed in $\Gamma$ . Let $\sigma \in \Gamma$ ; for $x_1, \ldots, x_n$ in N let $U(x_1, \ldots, x_n)$ be the set of all $\tau \in \Gamma$ such that $\tau(x_i) = \sigma(x_i)$ for $1 \leq i \leq n$ ; put

V(x_1, \ldots, x_n) = U(x_1, \ldots, x_n) \cap \Delta.

Then the family of sets $U(x_1, \ldots, x_n)$ (resp. $V(x_1, \ldots, x_n)$ ) is a base of neighbourhoods of $\sigma$ in $\Gamma$ (resp. A). Hence the topology on A is that induced by $\Gamma$ .

B) The relation $A = g(k(\Delta))$ (for $A \in g$ ) is a consequence of the following more precise lemma:

Lemma 2

Let A be a subgroup of $\Gamma$ . Let E be the field of invariants of A; then the Galois group of N over E is the closure of A in $\Gamma$ .

The Galois group of N over E is closed in $\Gamma$ (Lemma 1) and contains A, hence it contains the closure $\bar{\Delta}$ of A. Let $\sigma$ be an E-automorphism of N and let $x_1, \ldots, x_n$ be in N. Since N is Galois over E (Lemma 1) there exist (V, p. 57, Prop. 2) a subextension $N_0$ of N, Galois of finite degree over E and containing $x_1, \ldots, x_n$ . Let $A_0$ be the image of the subgroup A of $\mathrm{Gal}(N/E)$ under the restriction homomorphism of $\mathrm{Gal}(N/E)$ into $\mathrm{Gal}(N_0/E)$ . Since $[N_0 : E]$ is finite, Dedekind's theorem (V, p. 27, Cor. 2) shows that $\mathrm{Gal}(N_0/E)$ is finite. Hence $A_0$ is finite, and since E is the field of invariants of $A_0$ , we have $\Delta_0 =$

Gal(N_0/E) (V, p. 66, Th. 3). In particular, $\Delta_0$ contains the restriction of $\sigma$ to $N_0$ . Therefore there exists $\tau \in A$ such that $\sigma$ and $\tau$ have the same restriction to $N_0$ , whence $\sigma(x_1) = \tau(x_1), \ldots, \sigma(x_n) = \tau(x_n)$ . It follows that $\sigma$ is a limit point of $A$ in $\Gamma$ , and hence $\mathrm{Gal}(N/E) \subset \overline{\Delta}$ .

Corollary 1

Let E and E' be two subfields of N containing K; then E $\subset$ E' if and only if g(E) $\supset$ g(E'). If A and A' are two closed subgroups of $\Gamma$ , then A $\subset$ A' if and only if k( $\Delta$ ) $\supset$ k( $\Delta'$ ).

For the two inverse bijections $E \rightleftharpoons g(E)$ and $A \rightleftharpoons k(\Delta)$ are inclusion-reversing.

Corollary 2

Let $(E_i)_{i \in I}$ be a family of subfields of N containing K; put $L = \bigcap_{i \in I} E_i$ and $M = K\left( \bigcup_{i \in I} E_i \right)$ . Then g(L) is the smallest closed subgroup of containing $U_{i \in I} g(E_i)$ and we have $g(M) = \bigcap_{i \in I} g(E_i)$ .

The first assertion follows from Cor. 1 and the second is immediate.

Corollary 3

For i = 1, 2 let $E_i$ be a subfield of N containing K and let $\Delta_i = g(E_i)$ . For any $\sigma \in \Gamma$ the relations $\sigma(E_1) = E_2$ and $\sigma \Delta_1 \sigma^{-1} = \Delta_2$ are equivalent.

For we have $\tau \in g(\sigma(E_1))$ if and only if $\tau \sigma(x) = \sigma(x)$ , that is, $\sigma^{-1} \tau \sigma(x) = x$ , for all $x \in E_1$ ; this amounts to saying 'that $\sigma^{-1} \tau \sigma \in \Delta_1$ , whence $g(\sigma(E_1)) = \sigma \Delta_1 \sigma^{-1}$ .

Corollary 4

Let E be a subfield of N containing K and let $A = g(E)$ . For E to be Galois over K it is necessary and sufficient that A should be a normal subgroup of $\Gamma$ . When this is so, the restriction homomorphism of $\Gamma$ into $\mathrm{Gal}(E/K)$ defines by passage to quotients a topological group isomorphism of $\Gamma / \Delta$ onto $\mathrm{Gal}(E/K)$ .

Since N is separable over K, the same is true of E (V, p. 36, Prop. 1). Therefore E is Galois over K if and only if it is quasi-Galois over K; this also means that $\sigma(E) = E$ for every K-automorphism $\sigma$ of N (V, p. 52, Prop. 1 and p. 54, Prop. 3). By Cor. 3 this is equivalent to $\sigma \Delta \sigma^{-1} = A$ for all $\sigma \in \Gamma$ .

The restriction homomorphism $\varphi : \mathrm{Gal}(N/K) \to \mathrm{Gal}(E/K)$ is continuous and surjective (V, p. 60, Prop. 3) and its kernel is clearly equal to $A = \mathrm{Gal}(N/E)$ . Since $\Gamma$ is compact, the homomorphism of $\Gamma / \Delta$ onto $\mathrm{Gal}(E/K)$ derived from $\varphi$ by passage to quotients is an isomorphism of topological groups (Gen. Top., I, p. 87, Cor. 2).

Corollary 5

Let E be a subfield of N containing K. For E to have finite degree over K it is necessary and sufficient that g(E) should be open in $\Gamma$ . When this is so, the index $(\Gamma : g(E))$ is finite and equal to [E : K].

For $g(E)$ to be open it is necessary and sufficient that there should exist a subextension F of N, of finite degree over K, such that in the notation of V, p. 60.

g(E) contains $U_F(\mathrm{Id}_N) = g(F)$ . The relation $g(E) \supset g(F)$ is equivalent to $E \subset F$ by Cor. 1 (V, p. 68), whence the first assertion of Cor. 5.

Suppose that $[E : K]$ is finite. Let $\Omega$ be an algebraic closure of $K$ containing $N$ as subextension (V, p. 23, Th. 2) and let $\mathscr{H}$ be the set of $K$ -homomorphisms of $E$ into $\Omega$ . Every element of $\mathscr{H}$ is induced by a $K$ -automorphism of $\Omega$ (V, p. 52, Prop. 1), and since $N$ is quasi-Galois over $K$ , the mapping $\sigma \mapsto \sigma|_E$ of $\Gamma$ into $\mathscr{H}$ is surjective. For $\sigma$ and $\sigma'$ in $\Gamma$ to have the same restriction to $E$ it is necessary and sufficient that $\sigma^{-1}\sigma' \in g(E)$ , whence Card $A? = (\Gamma : g(E))$ . Finally since $E$ is an etale algebra over $K$ , we have Card $\mathscr{H} = [E : K]$ (V, p. 32, Prop. 4), so in conclusion we have $(\Gamma : g(E)) = [E : K]$ .

Corollary 6

— For $i = 1, 2$ let $E_i$ be a subextension of $N$ and $\Gamma_i$ the Galois group of $N$ over $E_i$ . The following conditions are equivalent: a) The group $\Gamma$ is the direct product of the subgroups $\Gamma_1$ and $\Gamma_2$ . b) The extensions $E_1$ and $E_2$ are Galois over $K$ , we have $E_1 \cap E_2 = K$ and

K(E_1 \cup E_2) = N .

For $\Gamma$ to be the direct product of the subgroups $\Gamma_1$ and $\Gamma_2$ it is necessary and sufficient that the following conditions hold (I, p. 48, Prop. 15): (i) the subgroups $\Gamma_1$ and $\Gamma_2$ are normal in $\Gamma$ ; (ii) $\Gamma_1 \cap \Gamma_2 = \{ \varepsilon \}$ , where $\varepsilon$ is the neutral element of $\Gamma$ ; (iii) $\Gamma = \Gamma_1 . \Gamma_2$ .

Now (i) means that $E_1$ and $E_2$ are Galois over $K$ (Cor. 4). By Cor. 2, condition (ii) is equivalent to $N = K(E_1 \cup E_2)$ , and finally if (i) and (ii) hold, $\Gamma_1 \Gamma_2$ is the least subgroup of $\Gamma$ containing $\Gamma_1 \cup \Gamma_2$ ; it is closed because $\Gamma_1$ and $\Gamma_2$ are compact and the mapping $(a, \tau) \mapsto \sigma \tau$ of $\Gamma_1 \times \Gamma_2$ into $\Gamma$ is continuous (Gen. Top., I, p. 63, Cor. 1). Cor. 2 now shows (iii) to be equivalent to $E_1 \cap E_2 = K$ , and this proves the equivalence of a) and b).

Remark

With the notation of Cor. 6 suppose that conditions a) and b) hold. The restriction homomorphisms $\varphi_i : \Gamma \to \mathrm{Gal}(E_i/K)$ for $i = 1, 2$ induce topological group isomorphisms

\Psi_1 : \Gamma_2 \to \mathrm{Gal}(E_1/K) , \quad \Psi_2 : \Gamma_1 \to \mathrm{Gal}(E_2/K)

By a) we see that the mapping

\sigma \mapsto (\varphi_1(\sigma), \varphi_2(\sigma))

is a topological group isomorphism of

\mathrm{Gal}(N/K)

onto

\mathrm{Gal}(E_1/K) \times \mathrm{Gal}(E_2/K)

8. Change of base field

Let $N$ be a Galois extension of $K$ and $\Gamma$ the Galois group of $N$ over $K$ ; further let $N'$ be a Galois extension of $K'$ with Galois group $\Gamma'$ . We shall identify $K$ (resp. $K'$ ) with its image in $N$ (resp. $N'$ ). Let $u$ be a homomorphism of $K$ into $K'$ and $v$ a homomorphism of $N$ into $N'$ whose restriction to $K$ equals $u$ (cf.

Fig. 1). Let $\sigma \in \Gamma'$ ; since $u(K) \subset K'$ , $a$ is a $u(K)$ -automorphism of $N'$ ; moreover $v(N)$ is a Galois extension of $u(K)$ , hence $a$ induces a $u(K)$ -automorphism of $v(N)$ (V, p. 55, Remark 1). In other words, for every $\sigma \in \Gamma'$ there exists a unique element $v^*(\sigma)$ of $\Gamma$ such that

(8)

v \circ v^*(\sigma) = \sigma \circ v .

![Diagram showing the mappings v and u between N, N', K, K'](fig. 1)

The mapping $v^*$ is a homomorphism of $\mathrm{Gal}(N'/K')$ into $\mathrm{Gal}(N/K)$ . For every $x \in N$ , the mapping $a \mapsto v^*(o)(x) = v^{-1}(\sigma(v(x)))$ of $\Gamma'$ into the discrete space $N$ is continuous, so $v^*$ is continuous.

Three particular cases are of importance:

a) If $F$ is a Galois extension of $K$ and $E$ a subextension of $F$ , we know (V, p. 67, Lemma 1) that $F$ is a Galois extension of $E$ . Let us apply what has been said to the case where $N = F$ , $K' = E$ , $N' = F$ and $v = \mathrm{Id}_F$ . Then $v^*$ is merely the canonical injection

j : \mathrm{Gal}(F/E) \to \mathrm{Gal}(F/K) .

This is sometimes called the inflation homomorphism.

b) Suppose that in addition $E$ is Galois over $K$ . Let us apply what has been said to the case where $N = E$ , $K' = K$ , $N' = F$ and $v$ is the canonical injection of $E$ into $F$ . Then $v^*$ is just the restriction homomorphism

\pi : \mathrm{Gal}(F/K) \to \mathrm{Gal}(E/K)

We know (V, p. 60, Prop. 3) that $\pi$ is surjective, with kernel $\mathrm{Gal}(F/E)$ and that by taking quotients it defines a topological group isomorphism of $\mathrm{Gal}(F/K)/\mathrm{Gal}(F/E)$ onto $\mathrm{Gal}(E/K)$ (V, p. 68, Cor. 4).

c) Suppose that $v^{-1}(K') = K$ and $N' = K'(v(N))$ ; let us show that the homomorphism

v^* : \mathrm{Gal}(N'/K') \to \mathrm{Gal}(N/K) ,

is a topological group isomorphism, sometimes called translation. For the group $\mathrm{Gal}(N'/K')$ is compact, the group $\mathrm{Gal}(N/K)$ is separated and $v^*$ is continuous; so it is enough (Gen. Top., I, p. 87, Cor. 2) to prove that $v^*$ is bijective. Now every element $\sigma$ of the kernel of $v^*$ is an automorphism of $N'$ which induces the identity on $K'$ and on $v(N)$ , hence $a = \epsilon$ because $N' = K'(v(N))$ ; it follows that $v^*$ is injective. Further, the image of $v^*$ is a closed subgroup $A$ of $\mathrm{Gal}(N/K)$

(Gen. Top., I, p. 81, ibid.) and the field of invariants of $A$ is equal to $v^{-1}(K') = K$ ; therefore we have $A = \mathrm{Gal}(N/K)$ (V, p. 67, Th. 4) and so $v^*$ is surjective.

The general case may be reduced to the preceding ones by composition. To begin with we note that $K'(v(N))$ is the field of invariants in $N'$ of the kernel $A$ of $v^*$ ; since $A$ is a normal subgroup of $\mathrm{Gal}(N'/K')$ , the extension $K'(v(N))$ of $K'$ is Galois (V, p. 68, Cor. 4). Thus $v^*$ is composed of the homomorphisms

\mathrm{Gal}(N'/K') \to \mathrm{Gal}(K'(u(N))/K') \xrightarrow{\psi} \mathrm{Gal}(N/v^{-1}(K')) \xrightarrow{j} \mathrm{Gal}(N/K);

in this sequence $\pi$ is the restriction homomorphism associated with the triple $K' \subset K'(u(N)) \subset N'$ , $\psi$ is the translation isomorphism associated with the central square of the diagram (Fig. 2) and $j$ is the inflation homomorphism associated with the triple $K \subset v^{-1}(K') \subset N$ .

The next theorem gives more detailed information about the structure of translation isomorphisms.

!Diagram showing the extensions and translations

Fig. 2

Theorem 5

Let $N'$ be an extension of $K$ generated by two subextensions $K'$ and $N$ . Suppose that $N$ is Galois over $K$ , with Galois group $\Gamma$ and that $K' \cap N = K$ . Then the extension $N'$ of $K'$ is Galois and the canonical homomorphism $\varphi$ of $K' \otimes_K N$ into $N'$ is an isomorphism. Let $\sigma \in \mathrm{Gal}(N/K)$ and let $\sigma'$ be the element of $\mathrm{Gal}(N'/K')$ which corresponds to it under the translation isomorphism; then we have $\sigma' \circ \varphi = \varphi \circ (\mathrm{Id}_{K'} \otimes u)$ .

We have $N' = K'(N)$ and $N$ is algebraic and separable over $K$ ; hence (V, p. 42, Prop. 10), the extension $N'$ of $K'$ is algebraic and separable. By Cor. 4 of V, p. 54 the extension $N'$ of $K'$ is quasi-Galois. Therefore the extension $N'$ of $K'$ is Galois. By c) above the mapping $\sigma \mapsto \sigma|_N$ is an homomorphism $\lambda$ of $\mathrm{Gal}(N'/K')$ onto $\mathrm{Gal}(N/K)$ .

We have $N = K'[N]$ because $N$ is algebraic over $K$ (V, p. 18, Cor. 1), hence $\varphi$ is surjective. If $a$ belongs to $\mathrm{Gal}(N/K)$ , we have

\lambda^{-1}(\sigma) \circ \varphi = \varphi \circ (\mathrm{Id}_{K'} \otimes \sigma).

Therefore the kernel of $\varphi$ is stable under the mappings $\mathrm{Id}_{K'} \otimes \sigma$ , hence of the form $K' \otimes_K N_0$ with $N_0 \subset N$ (V, p. 63, Cor.). For $x$ in $N_0$ we have $x = \varphi(1 \otimes x) = 0$ , hence $N_0 = 0$ and so $\varphi$ is injective.

Corollary 1

Let $E'$ be a subfield of $N'$ containing $K'$ . There exists a unique subfield $E$ of $N$ containing $K$ and such that $E' = K'(E)$ . We have $E = E' \cap N$ .

Put $E = E' \cap N$ , then $E' \supset K'(E)$ . Now put $\Gamma = \mathrm{Gal}(N/K)$ and $A = \mathrm{Gal}(N/E)$ , and define $\Gamma'$ and $A'$ similarly. The mapping $A : a \mapsto a|_N$ is an isomorphism of $\Gamma'$ onto $\Gamma$ and also of $A'$ onto $A$ ; in other words, $A'$ consists of those $\sigma \in \Gamma'$ for which $\lambda(\sigma)$ belongs to $A$ . If $\sigma \in \Gamma'$ leaves the elements of $K'(E)$ fixed, we have $\lambda(\sigma) \in A$ , whence $a \in A'$ and $a$ leaves the elements of $E'$ fixed; by Cor. 1 of V, p. 68 we thus have $K'(E) \supset E'$ .

We have proved the equality $E' = K'(E)$ , whence $\varphi^{-1}(E') = K' \otimes_K E$ . If $F$ is a subfield of $N$ containing $K$ and such that $E' = K'(F)$ , we have likewise $\varphi^{-1}(E') = K' \otimes_K F$ , whence $F = E$ .

Corollary 2

Let $N$ be a Galois extension of $K$ . Suppose that the Galois group $\Gamma$ of $N$ over $K$ is the direct product of two closed subgroups $\Gamma_1$ and $\Gamma_2$ and denote by $E_i$ the field of invariants of $\Gamma_i$ for $i = 1, 2$ . Then the canonical homomorphism of $E_1 \otimes_K E_2$ into $N$ is an isomorphism.

We have $E_1 \cap E_2 = K$ and $N = K(E_1 \cup E_2)$ by Cor. 6 (V, p. 69), and so it is enough to apply Theorem 5.

Remark

Let $K$ and $K'$ be two fields and $u$ a homomorphism of $K$ into $K'$ . Let $K_s$ (resp. $K'_s$ ) be a separable closure (V, p. 45, Prop. 14) of $K$ (resp. $K'$ ) and $\Pi$ (resp. $\Pi'$ ) the Galois group of $K_s$ over $K$ (resp. $K'_s$ over $K'$ ). Since $K_s$ is a separable algebraic extension of $K$ and the extension $(K'_s, u)$ of $K$ is separably closed, there exists (V, p. 45, Cor.) a homomorphism $v$ of $K_s$ into $K'_s$ extending $u$ . From $v$ we obtain a continuous homomorphism $v^*$ of $\Pi'$ into $\Pi$ . Let $v_1$ be another extension of $u$ ; since $K_s$ is a quasi-Galois extension of $K$ , there exists an element $a_s$ of $\Pi$ such that $v_1 = v \circ a_s$ . We conclude that $v_1^*(\tau) = \sigma_0^{-1} v^*(\tau) \sigma_0$ for all $\tau \in \Pi$ .

9. The normal basis theorem

Let $N$ be a Galois extension of $K$ , with Galois group $\Gamma$ . We identify $\Gamma$ with the canonical basis of the group algebra $K^{(\Gamma)}$ (III, p. 446); then $N$ may be considered as a left $K^{(\Gamma)}$ -module (III, p. 447, Example), so that

u \cdot x = \sum_{\sigma \in \Gamma} a_\sigma \sigma(x) \quad \text{for} \quad x \in N \quad \text{and} \quad u = \sum_{\sigma \in \Gamma} a_\sigma \sigma \quad \text{in} \quad K^{(\Gamma)}.

If $N$ is of finite degree over $K$ , the group $\Gamma$ is finite by Dedekind's theorem (V, p. 27, Cor. 2) and we can define the element $t = \sum_{\sigma \in \Gamma} a$ in $K^{(\Gamma)}$ ; then we have

\mathrm{Tr}_{N/K}(x) = \sum_{\sigma \in \Gamma} \sigma(x),

that is, $\mathrm{Tr}_{N/K}(x) = t \cdot x$ for all $x \in N$ .

Let us define an action on the right by $\Gamma$ on $N$ by $x^\sigma = \sigma^{-1}(x)$ . In a similar way we can consider the multiplicative group $N^*$ as a right $Z^{(\Gamma)}$ -module, the external law being written $(x, u) \mapsto x^u$ . For example, the notation $x^{2a + 3r + \pi}$ , where $a, r, \pi$ are elements of $\Gamma$ , indicates the product $(x^a)^2 \cdot (x^r)^3 \cdot x^\pi$ . If $N$ is of finite degree over $K$ and $t = \sum_{\sigma \in \Gamma} a$ as above, then we have $N_{N/K}(x) = \prod_{\sigma \in \Gamma} x^\sigma$ , that is, $N_{N/K}(x) = x'$ for all $x \in N^*$ .

Suppose henceforth that $N$ is of finite degree over $K$ . Given $x \in N$ , for $\{x\}$ to be a basis of the $\mathbf{K}^{(\Gamma)}$ -module $N$ it is necessary and sufficient that the family $(\sigma(x))_{\sigma \in \Gamma}$ should be a basis of $N$ over $K$ . Such a basis is called a normal basis of $N$ over $K$ .

Theorem 6

Let $N$ be a Galois extension of finite degree over $K$ and let $\Gamma$ be the Galois group of $N$ over $K$ . Then there exists a normal basis of $N$ over $K$ ; in other words, the $\mathbf{K}^{(\Gamma)}$ -module $N$ is free of rank 1.

We shall give two proofs of this statement. The first uses the following lemma which will be proved in Chapter VIII (§ 2, No. 5).

\* Lemma 3. — Let $A$ be a $K$ -algebra, $M_1$ and $M_2$ two $A$ -modules of finite rank over $K$ and suppose that there exists an extension $L$ of $K$ such that the modules $L \otimes_K M_1$ and $L \otimes_K M_2$ over the ring $L \otimes_K A$ are isomorphic. Then the $A$ -modules $M_1$ and $M_2$ are isomorphic.

We shall apply Lemma 3 in the case where $A = \mathbf{K}^{(\Gamma)}$ , $M_1 = N$ , $M_2 = A$ , and $L = N$ . By the Cor. of V, p. 64 there exists a $K$ -isomorphism $\varphi$ of $N \otimes_K N$ onto $N \otimes_K \mathbf{K}^{(\Gamma)}$ which maps $x \otimes y$ to $\sum x u^1(y) \otimes \sigma$ . It is clear that $\varphi$ is an isomorphism of $N \otimes_K \mathbf{K}^{(\Gamma)}$ -modules and so the theorem follows from Lemma 3.

For the second proof we shall use the following proposition:

Proposition 10

Let $x \in N$ , then for $\{x\}$ to form a basis of the $\mathbf{K}^{(\Gamma)}$ -module $N$ it is necessary and sufficient that $\det(\sigma \tau(x))_{\sigma, \tau \in \Gamma} \neq 0$ .

Since $\mathbf{K}^{(\Gamma)}$ and $N$ have the same dimension over $K$ , to say that $\{x\}$ is a basis of $N$ over $\mathbf{K}^{(\Gamma)}$ means that the mapping $a \mapsto ax$ of $\mathbf{K}^{(\Gamma)}$ into $N$ is injective. This in turn means that the mapping $b \mapsto b(1 \otimes x)$ of $N \otimes_K \mathbf{K}^{(\Gamma)}$ into $N \otimes_K N$ is injective (II, p. 306, Prop. 14). Now there is an $N \otimes_K \mathbf{K}^{(\Gamma)}$ -module isomorphism of $N \otimes_K N$ onto $N \otimes_K \mathbf{K}^{(\Gamma)}$ which maps $1 \otimes x$ to $\sum \sigma^{-1}(x) \otimes \sigma$ . It follows that $\{x\}$ is a basis of $N$ over $\mathbf{K}^{(\Gamma)}$ if and only if, for every non-zero family of elements $(n_\tau)$ , of $N$ we have $\left( \sum n_\tau \otimes \tau \right) \left( \sum \sigma^{-1}(x) \otimes \sigma \right) \neq 0$ . But this last relation means that there exists $\sigma \in \Gamma$ such that $\sum n_\tau \sigma^{-1}(x) \neq 0$ , whence the proposition.

\* A) Suppose that $K$ is infinite; the mapping $x \mapsto \det(\sigma \tau(x))$ of $N$ into $N$ is a polynomial mapping over $K$ (IV, p. 54). By extension of scalars from $K$ to $N$ we get a corresponding mapping for the vector $N$ -space $N \otimes_K N$ and we have just seen that the latter is not identically zero (because $N \otimes_K N$ is free of rank 1 over $N \otimes_K \mathbf{K}^{(\Gamma)}$ ). So there exists $x \in N$ such that $\det(\sigma \tau(x)) \neq 0$ (IV, p. 18, Th. 2); more generally, from the same reference we have:

Proposition 11

— Suppose that K is infinite and let P : N → K be a polynomial mapping which is non-zero on K. There exists x ∈ N such that P(x) ≠ 0 and {x} is a basis of N over $\mathbf{K}^{(\Gamma)}$ .

B) Suppose that K is finite. By Prop. 4 (V, p. 95) $^1$ every extension of finite degree over K has a cyclic Galois group. We shall therefore more generally consider the case where the group $\Gamma$ is cyclic of order n; we denote by $\gamma$ a generator of $\Gamma$ .

The following lemma is a particular case of more general results proved in Chapter VII. The ring $A$ is either the ring $\mathbf{Z}$ of rational integers or the ring $K[X]$ of polynomials over the field $K$ .

Lemma 4

Let M be a torsion A-module generated by a finite number $x_1, ..., x_h$ of elements; then there exists an element x of M whose annihilator (II, p. 219) is equal to the annihilator of M.

In both cases $A$ is an integral domain and every ideal of $A$ is principal. When $A = \mathbf{Z}$ (resp. $A = K[X]$ ), we denote by $\mathscr{P}$ the set of prime numbers (resp. the set of irreducible monic polynomials in $K[X]$ ). For every element $a \neq 0$ of $A$ there exists then an invertible element $u$ of $A$ and a family $(v_p(a))_{p \in \mathscr{P}}$ , with finite support, of positive integers such that $a = u \prod_{p \in \mathscr{P}} p^{v_p(a)}$ and $u$ and the integers $v_p(a)$ are uniquely determined (I, p. 51 and IV, p. 13, Prop. 13).

Let $a_i$ be the annihilator of $x_i$ , (for $1 \leq i \leq h$ ) and $a$ the annihilator of M; let $a_1, ..., a_h, a$ be non-zero elements of $A$ such that $a_i = Aa_i$ , and $a = Aa$ ; since $a = a_1 \cap ... \cap a_h$ , it follows from what has been said that

v_p(a) = \sup_{1 \leq i \leq h} v_p(a_i) \quad \text{for all } p \in \mathscr{P}.

Let us write $a$ in the form $up_1^{n(1)} ... p_r^{n(r)}$ , with $p_1, ..., p_r$ distinct in $\mathscr{P}$ , $n(1) > 0, ..., n(r) > 0$ and $u$ an invertible element of $A$ . Let $j = 1, ..., r$ ; by (10) there exists an integer $c(j)$ such that $1 \leq c(j) \leq h$ and $v_{p_j}(a_{c(j)}) = n(j)$ ; there exists $b_j$ in $A$ with $a_{c(j)} = p_j^{n(j)} b_j$ and the element $y_j = b_j x_{c(j)}$ has as annihilator the ideal $Ap_j^{n(j)}$ .

Let us show that the annihilator b of $y = y_1 + ... + y_r$ is equal to the annihilator a of M. In any case we have $a \subset b$ , so b is of the form $Ap_1^{m(1)} ... p_r^{m(r)}$ with $0 \leq m(j) \leq n(j)$ for $1 \leq j \leq r$ . If we had $a \neq b$ , there would exist an integer jsuch that $1 \leq j \leq r$ and $m(j) < n(j)$ , and hence $d_j = a/p_j$ would annihilate $y$ . Now we have $d_j y_k = 0$ for $k \neq j$ , whence we obtain $d_j y_j = 0$ ; but the annihilator of $y_j$ is $Ap_j^{m(j)}$ and $d_j$ is not a multiple of $p_j^{m(j)}$ . So the hypothesis $a \neq b$ is absurd.

$^1$ The reader can easily convince himself that the normal basis theorem is not used anywhere before the proof of that proposition.

We shall apply Lemma 4 to the case where $A$ is the polynomial ring $K[X]$ and $M$ the abelian group $N$ with the external law defined by $a \cdot x = \sum_{k=0}^m c_k \gamma^k(x)$ for $a = \sum_{k=0}^\infty c_k X^k$ in $K[X]$ and $x \in N$ . Let $a$ be the annihilator of $M$ , then $\gamma^n = 1$ , hence the polynomial $X^n - 1$ lies in $a$ . Let $F \in a$ ; by IV, p. 11, Cor. there exist elements $c_0, c_1, \ldots, c_{n-1}$ of $K$ and $G \in K[X]$ such that

F(X) = c_0 + c_1 X + \cdots + c_{n-1} X^{n-1} + (X^n - 1) G(X).

Thus we have $c_0 + c_1 \gamma + \ldots + c_{n-1} \gamma^{n-1} = 0$ in $\operatorname{Hom}_K(N, N)$ and since the automorphisms $1, \gamma, \gamma^2, \ldots, \gamma^{n-1}$ of $N$ are distinct, Dedekind's theorem (V, p. 27, Cor. 2) implies that $c_0 = c_1 = \ldots = c_{n-1} = 0$ . Finally, we have

F(X) = (X^n - 1) G(X), \quad \text{that is, } a = (X^n - 1) K[X].

By Lemma 4 there exists an element $x$ of $N$ whose annihilator in $K[X]$ is equal to $(X^n - 1) K[X]$ . Since the monomials $1, X, \ldots, X^{n-1}$ form a basis of a vector subspace of $K[X]$ supplementary to $(X^n - 1) K[X]$ (IV, p. 11, Cor.), the elements $x, \gamma(x), \ldots, \gamma^{n-1}(x)$ of $N$ are linearly independent over $K$ . Since $[N : K] = n$ (V, p. 66, Th. 3), the sequence $(x, \gamma(x), \ldots, \gamma^{n-1}(x))$ is therefore a (normal) basis of $N$ over $K$ .

10. Finite $\Gamma$ -sets and etale algebras

Let $K_s$ be a separable closure of $K$ (V, p. 45, Prop. 14) and $\Gamma$ the Galois group of $K$ , over $K$ . By a $\Gamma$ -*set* we understand a set $X$ with an action $(\sigma, x) \mapsto \sigma x$ of the group $\Gamma$ such that the stabilizer of each point of $X$ is an open subgroup of $\Gamma$ . It comes to the same to say that the mapping $(\sigma, x) \mapsto \sigma x$ of $\Gamma \times X$ into $X$ is continuous when $X$ is equipped with the discrete topology.

Let $X$ be a finite $\Gamma$ -set. We define an action of $\Gamma$ on the $K$ -algebra $K_s^X$ of mappings of $X$ into $K_s$ by the formula

u_\sigma f(x) = \sigma(f(\sigma^{-1}x))

for $\pi \in \Gamma, f \in K_s^X$ and $x \in X$ . Let $\Theta(X)$ be the set of invariants of $\Gamma$ in $K_s^X$ ; this is the sub-K-algebra of $K_s^X$ consisting of mappings $f : X \to K_s$ such that $f(\sigma x) = \sigma(f(x))$ for $\sigma \in \Gamma$ and $x \in X$ .

Lemma 5

Let $X$ be a finite $\Gamma$ -set and let $x_1, \ldots, x_n$ be points of $X$ such that the orbits $\Gamma x_1, \ldots, \Gamma x_n$ form a partition of $X$ . For $1 \leq i \leq n$ let $\Delta_i$ be the stabilizer of $x_i$ in $\Gamma$ and let $L_i$ be the field of invariants of $A_i$ . Then $L_1, \ldots, L_n$ are separable extensions of finite degree of $K$ , and the mapping $f \mapsto (f(x_1), \ldots, f(x_n))$ is a K-algebra isomorphism of $\Theta(X)$ onto $L_1 \times \ldots \times L_n$ .

By hypothesis the subgroups $A_i, \ldots, A_n$ of $\Gamma$ are open and Cor. 5 of V, p. 68, shows that the subextensions $L_1, \ldots, L_n$ of $K$ , are of finite degree over $K$ . Clearly they are separable; now the last assertion of Lemma 5 is immediate.

From Lemma 5 and Th. 4 (V, p. 34f.) we obtain immediately the following result.

Proposition 12

— For every finite $\Gamma$ -set $X$ the algebra $\Theta(X)$ is etale over $K$ , of degree equal to the cardinal of $X$ . Moreover, every etale algebra over $K$ is isomorphic to an algebra of the form $\Theta(X)$ .

Remark

a) It is easy to show that for every $K$ -algebra homomorphism $\varphi$ of $\Theta(X)$ into $K$ , there exists a unique element $x$ of $X$ such that $\varphi(f) = f(x)$ for all $f \in \Theta(X)$ .

2) Let $X$ and $Y$ be two finite $\Gamma$ -sets. Let $\mathfrak{T}_\Gamma(X, Y)$ be the set of mappings $u$ of $X$ into $Y$ such that $u(\sigma x) = \sigma u(x)$ for all $\sigma \in \Gamma$ and all $x \in X$ . For $u \in \mathfrak{T}_\Gamma(X, Y)$ we define a $K$ -algebra homomorphism $u^*: \Theta(Y) \to \Theta(X)$ by $u^*(f) = f \circ u$ . For every homomorphism $\Psi$ of $\Theta(Y)$ into $\Theta(X)$ there exists a unique element $u$ of $\mathfrak{T}_\Gamma(X, Y)$ such that $\Psi = u^*$ .

11. The structure of quasi-Galois extensions

Proposition 13

— Let $N$ be a quasi-Galois extension of $K$ . We denote by $N_r$ , the field of invariants of the group of all $K$ -automorphisms of $N$ and by $N_s$ , the relative separable algebraic closure of $K$ in $N$ ( $V$ , p. 44). Then:

a) $N_r$ is the relative $p$ -radical closure of $K$ in $N$ ( $V$ , p. 25).

b) $N$ , is a Galois extension of $K$ and every $K$ -automorphism of $N_s$ extends in a unique fashion to an $N_r$ -automorphism of $N$ .

c) The fields $N_r$ and $N$ , are linearly disjoint over $K$ and we have $N = K[N, \cup N_s]$ ; in other words, the canonical homomorphism of $N_r \otimes_K N$ , into $N$ is an isomorphism.

Let $\Omega$ be an algebraic closure of $K$ containing $N$ as subextension ( $V$ , p. 23, Th. 2). Every $K$ -automorphism of $\Omega$ induces an automorphism of $N$ because $N$ is quasi-Galois. Therefore every element of $N_r$ is invariant under the group of $K$ -automorphisms of $\Omega$ , hence $p$ -radical over $K$ ( $V$ , p. 53, Cor. 3). Conversely every element of $N$ which is $p$ -radical over $K$ is clearly invariant under every $K$ -automorphism of $N$ and hence belongs to $N_r$ . This proves a).

Every $K$ -automorphism of $\Omega$ maps $N$ into $N$ , hence $N_r$ into $N_r$ , so $N_r$ is a quasi-Galois extension of $K$ ( $V$ , p. 54, Prop. 3). It follows that $N_s$ is a Galois extension of $K$ . Every element of $N_r \cap N$ , is separable algebraic and $p$ -radical over $K$ , hence belongs to $K$ ( $V$ , p. Cor. 3); we thus have $K_r \cap K_s = K$ . Now $N$ is $p$ -radical over $N_r$ ( $V$ , p. 44, Prop. 13) and separable algebraic over $N_r$ ( $V$ , p. 56, Th. 1) hence both $p$ -radical and separable over $K(N_r \cup N_s)$ . Hence we have $N = K(N_r \cup N_s)$ ( $V$ , p. 39, Cor. 3) and the assertions b), c) follow from Th. 5 ( $V$ , p. 71).

Corollary

— Let $p$ the characteristic exponent of $K$ , $\overline{K}$ an algebraic closure of $K$ , $K_s$ , the relative separable closure of $K$ in $\overline{K}$ and $K^{p^{-\infty}}$ the perfect closure of $K$ . Then the canonical homomorphism of $K^{p^\infty} \otimes K_s$ into $\overline{K}$ is an isomorphism.

Remark

Let $R$ (resp. $S$ ) be a $p$ -radical (resp. separable algebraic) extension of $K$ . Then the algebra $R \otimes_K S$ is a field : for $R$ (resp. $S$ ) is isomorphic to a subextension of $K^{p^\infty}$ (resp. $K_s$ ) and it suffices to apply the above Cor. and Prop. 1 of $V$ , p. 17.

Exercises

See the exercises for § 10.