Lemma 12

a) There does not exist any ideal of L, other than $\{0\}$ and L that is invariant under G.

b) Let $g$ be an element of G distinct from the identity element, and let $\mathfrak{a}_g$ be the ideal of L generated by the elements of the form $x-\lambda (g)x$, where $x$ runs through L. We have $\mathfrak{a}_g= L$.

Let $\mathscr{S}$ be the set of maximal ideals of L; for any subset S of $\mathscr{S}$, set $\mathfrak{a}(S) =\bigcap_{\mathfrak{m}\in S}\mathfrak{m}$. Since the ring L is commutative and semisimple (VIII, p. 310, Remark 6), the map $W\mapsto \mathfrak{a}(W)$ is a bijection from the set of subsets of $\mathscr{S}$ to the set of ideals of L (VIII, p. 142, Proposition 9). The ideal $\mathfrak{a}(S)$ is invariant under G if and only if S is. Since G acts transitively on $\mathscr{S}$ (VIII, p. 308, Theorem 2, vi)), the only subsets of $\mathscr{S}$ invariant under G are $\emptyset$ and $\mathscr{S}$. Since $\mathfrak{a}(\emptyset ) = L$ and $\mathfrak{a}(\mathscr{S}) =\{0\}$, we have proved assertion a).

Let us prove b) by contradiction; suppose that we have $\mathfrak{a}_g\not= L$. Let $\mathfrak{m}$ be a maximal ideal of L containing $\mathfrak{a}_g$. We then have $\lambda (g)x\equiv x$ mod $\mathfrak{m}$ for every $x$ in L. In particular, $\mathfrak{m}$ is invariant under $g$, and $\lambda (g)$ is induced by the identity on the field $L/\mathfrak{m}$. By property (vi) in Theorem 2 of VIII, p. 308, the element $g$ of $G_{\mathfrak{m}}$ is trivial, which contradicts the assumption of b).

Let us now prove Theorem 3. The vector space $\mathbf{A}[\mathscr{E}; L]$ has finite dimension Card(G)[L$: K] =n^2$ over K, by VIII, p. 316 and the assumed equality Card(G) $= [L : K] =n$.

Let $\mathfrak{a}$ be a nonzero two-sided ideal of the algebra $\mathbf{A}[\mathscr{E}; L]$. We use the notation of VIII, p. 316. Every element $a$ of $\mathbf{A}[\mathscr{E}; L]$ can be written uniquely as $a=\sum_{g\in G}a_g\varepsilon_g$ with $a_g\in L$ for every $g\in G$; we denote by $\Phi (a)$ the set of elements $g$ of G such that $a_g\not= 0$. By formula (32) of VIII, p. 316, we have the relation

for all $g, g'\in G$ and consequently

for every $g\in G$ and $a\in \mathbf{A}[\mathscr{E}; L]$.

Let $a$ be a nonzero element of $\mathfrak{a}$ for which $\Phi (a)$ is minimal for the inclusion; by formula (39), if necessary replacing $a$ with an element of the form $a\varepsilon_{g^{-1}}$ with $g\in \Phi (a)$, we may assume that we have $e\in \Phi (a)$. Let $s$ be an element of $\Phi (a)$ distinct from $e$. By Lemma 12, b), there exists an element $x$ of L such that $a_s(x-\lambda (s)\cdot x)\not= 0$. But we have the relation

where the sum is taken over the elements $g$ of $\Phi (a)$ distinct from $e$. Because of the minimality of $\Phi (a)$, we then have $xa-ax= 0$, but this contradicts the assumption on $x$. We have, therefore, proved by contradiction that $\Phi (a)$ contains only the identity element $e$ of G, so we have $a\in L$.

Hence $L\cap \mathfrak{a}$ is an ideal of L not reduced to $\{0\}$. Moreover, for every $x$ in L, we have $\varepsilon_gx\varepsilon^{-1}_g=\lambda (g)\cdot x$, so $L\cap \mathfrak{a}$ is invariant under G. By Lemma 12, we therefore have $L\cap \mathfrak{a}= L$, that is, $L\subset \mathfrak{a}$. Since L contains the identity element of $\mathbf{A}[\mathscr{E}; L]$, we have $\mathfrak{a}=\mathbf{A}[\mathscr{E}; L]$.

Since the algebra $\mathbf{A}[\mathscr{E}; L]$ has nonzero finite dimension over K and its only two-sided ideals are $\{0\}$ and $\mathbf{A}[\mathscr{E}; L]$, it is simple.

Let us prove that every element $a$ of $\mathbf{A}[\mathscr{E}; L]$ that commutes with L belongs to L. We write $a$ as $\sum_{g\in G}a_g\varepsilon_g$ with coefficients $a_g$ in L. For every $x$ in L, we have $xa-ax= 0$, and relation (40) shows that we have

for every $g\in G$ and $x\in L$. By Lemma 12, we therefore have $a_g= 0$ for $g\not=e$; consequently, $a\in L$.

Let us, finally, determine the center of $\mathbf{A}[\mathscr{E}; L]$. If $z$ belongs to the center, then it commutes with L, so belongs to L. But we then have

for every $g$ of G. Hence $z$ is invariant under the automorphism group $\lambda (G)$ of L. We therefore have $z\in K$ by assumption.