Lemma 9

The morphism $u$ is surjective with kernel $\mathfrak{a}_{\sigma}$. The mapping deduced from $u$ by passing to the quotient is an isomorphism of $(K,G)$-algebras from $L_{\sigma}$ to $F_{\mathfrak{m}}$.

Since the kernel of $\pi_{\mathfrak{m}}$ is equal to $\mathfrak{m}$, that of $u$ is equal to $\bigcap_{g\in G}g^{-1}.\mathfrak{m}=\mathfrak{a}_{\sigma}$. To prove that $u$ is surjective, it suffices to prove that the vector spaces $L_{\sigma}=$ $L/\mathfrak{a}_{\sigma}$ and $F_{\mathfrak{m}}$ have the same dimension over K. Now, all ideals $g\cdot \mathfrak{m}$ have the same codimension in L, and, by Lemma 8, a),

$[L_{\sigma}: K] =$ Card($\sigma$ )$.[L/\mathfrak{m}: K]$.

Moreover, by Lemma 7, b), we have

Since we have Card($\sigma$ ) $= (G : G_{\mathfrak{m}})$, we have proved the equality $[L_{\sigma}: K] =$ $[F_{\mathfrak{m}}: K]$.

We now suppose, furthermore, that the homomorphism $\lambda$ from G to Aut$_K(L)$ is injective. We identify K with its image in L through the mapping $\xi \mapsto \xi \cdot 1$. Let Ω be an algebraically closed extension of K. The set $\mathscr{F}(G,\Omega )$ of mappings from G to Ω coincides with the coinduced $(\Omega$, G)-algebra Coind$^G_{\{e\}}(\Omega )$. It is a free Ω[G]-module of rank 1. Let $\mathscr{H}$ be the set of K-algebra homomorphisms from L to Ω. We define a right action of G on $\mathscr{H}$ by $(g, \chi )\mapsto \chi \circ \lambda (g)$. The Ω-algebra $\mathscr{F}(\mathscr{H},\Omega )$ is then endowed with the structure of an $(\Omega$, G)-algebra deduced from the right action of G on $\mathscr{H}$.