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,

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 $(a_\sigma)_{\sigma \in \Gamma}$.