Lemma 8

a) For every $g\in G,\sigma \in \mathscr{O}$, and $\mathfrak{m}\in \sigma$, we have

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

b) The mapping $\pi :x\mapsto (\pi_{\sigma}(x))_{\sigma\in\mathscr{O}}$ is an isomorphism of $(K,G)$-algebras from L to $\prod_{\sigma\in\mathscr{O}}L_{\sigma}$.

c) Denote by $L^G($resp. $L^G_{\sigma})$ the subalgebra of L (resp. $L_{\sigma}$) of elements invariant under the action of G. Then $\pi$ induces an isomorphism from $L^G$ to $\prod_{\sigma\in\mathscr{O}}L^G_{\sigma}$.

Since the algebra L is reduced and of finite degree, the intersection of the maximal ideals of L is reduced to 0 (VIII, p. 173, Corollary 2). Moreover, if $\mathfrak{m}$ and $\mathfrak{m}'$ are two distinct maximal ideals of L, then we have $\mathfrak{m}+\mathfrak{m}'= L$. By Proposition 10 of I, §8, No. 11, p. 110, the canonical mapping from L to $\prod_{\mathfrak{m}\in\mathscr{S}}L/\mathfrak{m}$ is an isomorphism, as is the canonical mapping from $L/\mathfrak{a}_{\sigma}$ to $\prod_{\mathfrak{m}\in\sigma}L/\mathfrak{m}$ for every $\sigma \in \mathscr{O}$. Assertion a) follows. Since $\mathscr{O}$ is a partition of $\mathscr{S}$, assertion b) follows; assertion c) is an immediate consequence of b).

Let us now fix an orbit $\sigma \in \mathscr{O}$ and an element $\mathfrak{m}$ of $\sigma$. Set $F_{\mathfrak{m}}$ = Coind$^G_{G_{\mathfrak{m}}}(L/\mathfrak{m})$, and denote by $\lambda_{F_{\mathfrak{m}}}$ the action of G on $F_{\mathfrak{m}}$. For any $x\in L$, denote by $\overline{x}$ the mapping from G to $L/\mathfrak{m}$ that sends $g$ to $\pi_{\mathfrak{m}}(gx)$. By formula (15) of VIII, p. 305 and the definition of the action of $G_{\mathfrak{m}}$ on $L/\mathfrak{m}$, it is immediate that $\overline{x}$ belongs to $F_{\mathfrak{m}}$, and the mapping $u:x\mapsto \overline{x}$ from L to $F_{\mathfrak{m}}$ satisfies $\lambda_{F_{\mathfrak{m}}}(g)\circ u=u\circ \lambda (g)$ for $g\in G$. In other words, $u$ is a morphism of $(K$, G)-algebras.