Galois orbits classify simple modules after splitting base change

If $A=0$, $B=0$ and every unital module is zero, since $w=1w=0w=0$. There are no simple classes, so the empty map is the asserted bijection. Henceforth take $A\mathrm{\ne }0$. By F2–F3, each orbit $O$ gives a block $A{e}_O\cong M_{n_O}(D_O)$ with one simple type $S_O=D_O^{n_O}$, where $n_O\ge 1$. F4–F5 show that these exhaust all simple $A$-modules with no repetition. In particular these simples are finite-dimensional, because each is a column left ideal of the finite-dimensional algebra.

For completeness, any finite-dimensional module $M$ over $M_d(E)$ is a sum of column modules by an explicit map. The orthogonal $E_{jj}$ give $M={⨁}_{j=1}^d{E}_{jj}M$. Choose an $E$-basis $(v_t{)}_{t=1}^h$ of $E_{11}M$ by F9, and send the $j$th standard vector in column copy $t$ to $E_{j1}{v}_t$. Matrix multiplication $E_{ab}{E}_{j1}={\delta }_{bj}{E}_{a1}$ proves module linearity. Every $x\in E_{jj}M$ equals $E_{j1}(E_{1j}x)$, proving spanning. Applying $E_{1j}$ to a relation ${\sum }_{k,t}{\lambda }_{kt}{E}_{k1}{v}_t=0$ gives ${\sum }_t{\lambda }_{jt}{v}_t=0$, proving independence. Thus $M\cong (E^d{)}^{\oplus h}$. For a product of matrix rings, first decompose by its coordinate identities and apply this argument to each coordinate. It includes $h=0$.

For the finite-group specialization, let $G,F,E$ satisfy the finite-group hypotheses, and fix $k=F$ or $k=E$. The field inclusion makes $k$ characteristic zero. Let $V$ be a finite-dimensional $k[G]$-module and $U\subseteq V$ a submodule. F9 supplies a finite basis $u_1,\dots ,u_r$ of $U$ extended to a basis $u_1,\dots ,u_r,v_{r+1},\dots ,v_d$ of $V$. Define $P({\sum }_{j\le r}{a}_j{u}_j+{\sum }_{j>r}{a}_j{v}_j)={\sum }_{j\le r}{a}_j{u}_j$. Coordinate uniqueness makes $P$ linear, with image $U$ and $P{\mathrm{\mid }}_U=\mathrm{id}$. This includes $U=0$ and $U=V$.

Separating a matrix into its $n_O$ columns gives $A{e}_O\cong S_O^{\oplus n_O}$ as left $A$-modules. Scalar extension yields $B{c}_O\cong (E{\otimes }_F{S}_O{)}^{\oplus n_O}$: the first identification sends $e\otimes a{e}_O$ to $(e\otimes a)c_O$; it is onto and injective by regrouping the tensor basis along a basis of $A{e}_O$. The direct-sum isomorphism is F6, and its coordinate formula commutes with the $B$-action. F7 similarly gives ${\dim }_E(E{\otimes }_F{S}_O)={\dim }_F{S}_O$.

Since $G$ contains its identity, $N=\mathrm{\mid }G\mathrm{\mid }\ge 1$. F10 gives $N\cdot 1_k\mathrm{\ne }0$, and F11 gives its inverse $N^{-1}$. In the additive commutative monoid of linear endomorphisms set $Q=N^{-1}{\sum }_{g\in G}gP{g}^{-1}$. This finite sum is defined by F12. Composition and evaluation distribute over it: for an empty list both sides are zero, and adjoining one term follows from binary distributivity; induction on the list length proves the finite identity. Thus $Q$ is a linear map.

Put $M=E{\otimes }_F{S}_O$. By step 1.2 it has a finite simple decomposition. The regular module $B{c}_O$ consists of exactly the split factors indexed by $O$, each with its nonzero column multiplicity. Comparing step 2.1 with these decompositions, F8 gives $n_O$ times the multiplicity of each class in $M$ equal to its regular-block multiplicity. Since $n_O>0$, precisely the classes in $O$ occur in $M$, each positively; no other class occurs. The use of F8 is legitimate because ordering the summands of a finite simple direct sum gives a composition series.

For $h\in G$, distributing composition and reindexing by $g\mapsto hg$ gives $hQ{h}^{-1}=N^{-1}{\sum }_g(hg)P(hg{)}^{-1}=Q$. Hence $Q$ is $G$-equivariant and is a module map by F15. Since $U$ is $G$-stable, each $gP{g}^{-1}$ maps into $U$ and is identity on $U$. Therefore $Q(V)\subseteq U$ and $Q(u)=N^{-1}Nu=u$ for $u\in U$.

The canonical map $T_{\sigma }(e\otimes s)=\sigma (e)\otimes s$ is an isomorphism ${}^{\sigma }M\to M$: for $b\in B$, $T_{\sigma }({\sigma }_B^{-1}(b)m)=b{T}_{\sigma }(m)$, and its inverse is $T_{{\sigma }^{-1}}$. Twisting a direct-sum decomposition therefore permutes the simple types of $M$ without changing their multiplicities, by F8. Any two classes in $O$ are related by a twist, so all the positive multiplicities of step 3.1 equal one integer $m_{S_O}$. F8 also proves its uniqueness.

For $v\in V$, the vector $v-Qv$ lies in $\ker Q$ because $Qv\in U$ and $Q$ fixes $U$. Thus $v=Qv+(v-Qv)$ belongs to $U+\ker Q$. If $u\in U\cap \ker Q$, then $u=Qu=0$, proving $V=U\oplus \ker Q$. The kernel is a submodule since $Q$ is a module map.

The assignment $[S_O]\mapsto O$ is independent of the displayed matrix decomposition: F2 defines the orbit intrinsically via central idempotents, and F3 gives its unique simple class downstairs. Step 3.1 shows different orbits have disjoint supports and that every simple $B$-class is in the support assigned to its orbit. Conversely that support recovers $O$ from $S_O$. This proves bijectivity in both directions. If $E=F$, all orbits are singletons and $E{\otimes }_{F}S\cong S$, so the multiplicity is $1$.

Apply strong induction on $d={\dim }_{k}V$ to the assertion that every such $V$ is a finite direct sum of simple submodules. For $d=0$ use the empty sum. For $d>0$, a simple $V$ is already one summand. Otherwise the definition of simplicity supplies a nonzero proper submodule $U$. Step 4.2 splits it off; both $U$ and $\ker Q$ are proper, the latter because $Q$ is identity on the nonzero $U$. F9 gives dimensions strictly below $d$ for both. The induction hypothesis decomposes each into finitely many simple submodules, whose combined direct sum is $V$. F14 proves the assertion for every $d$, and F17 identifies these modules as semisimple.

The group ring has the finite group basis $[g]$, with $[g][h]=[gh]$; its regular action is one of the actions in F15. Apply step 5.2 to $V=k[G]$, whose dimension is $N$, and then F18 to obtain semisimplicity of $k[G]$ for both $k=F,E$. The basis map $E{\otimes }_{F}F[G]\to E[G]$, $e\otimes [g]\mapsto e[g]$, is bijective by F7 and multiplicative by the group-basis product. For the trivial group this is just $E{\otimes }_{F}F\cong E$, and averaging above is $Q=P$.

Apply F4 to the nonzero semisimple algebra $E[G]$. For a factor $M_n(D)$, its column module $W=D^n$ is simple by F5. To compute its endomorphisms without assuming splitting, let $f$ commute with the factor action. Commuting with $E_{11}$ forces $f(e_1)=e_{1}d$ for some $d\in D$; commuting with $E_{j1}$ gives $f(e_j)=e_{j}d$, and commuting with diagonal left multiplications gives $f(x)=xd$ for every column $x$. Conversely each right multiplication commutes with all matrices. Their compositions reverse multiplication in $D$, so $\mathrm{End}(W)\cong D^{\mathrm{o}\mathrm{p}}$ as an $E$-algebra. F15 identifies this endomorphism ring with ${\mathrm{End}}_G(W)$; by the given splitting-field condition F16 it is exactly the scalar field $E$. Taking the opposite ring leaves $E$ unchanged, so every division factor is $E$ as an $E$-algebra. Hence $E[G]$ is split semisimple. The general classification in step 5.1 now applies to $F[G]$ and the given finite Galois extension and prove the specialization. All selections in this argument concern finite-dimensional spaces or finite decompositions; no additional arbitrary-index choice assumption is used. [F4, F5, F15, F16, step 5.1, step 6.1, algebra] QED