Galois Orbits and Descent of Simple Finite-Group Modules

Put $n=[E:F]=\mathrm{\mid }\mathrm{\Gamma }\mathrm{\mid }\ge 1$, choose an $F$-basis $a_1,\dots ,a_n$ of $E$, and form its trace Gram matrix $H_{ij}=\mathrm{Tr}(a_i{a}_j)$. Separability and nondegeneracy make $H$ invertible. Set $b_j={\sum }_k(H^{-1}{)}_{kj}{a}_k$. Then $\mathrm{Tr}(a_i{b}_j)={\delta }_{ij}$. The invertible coefficient matrix also shows that $(b_j)$ is a basis.

Any finite $F$-independent list $v_1,\dots ,v_r$ in $W^{\mathrm{\Gamma }}$ is $E$-independent. Indeed, if a relation exists, select one with the least positive number of nonzero coefficients and normalize one of those coefficients to $1$. Applying $T_{\sigma }$ and subtracting produces a relation with that coefficient zero; minimality forces every other coefficient to be fixed by every $\sigma$. They all lie in $F$, contradicting $F$-independence. A one-term relation is already impossible since a nonzero scalar cannot annihilate a nonzero vector.

For the canonical action, choose a finite $F$-basis $(v_j)$ of $V$. Each tensor has a unique expression ${\sum }_j{e}_j\otimes v_j$: the product $F$-basis in F6, regrouped by the $v_j$, proves existence and uniqueness of its coefficients in $E$. Such a tensor is fixed exactly when $\sigma (e_j)=e_j$ for every $\sigma ,j$, equivalently each $e_j\in F$. It then equals $1\otimes {\sum }_j{e}_j{v}_j$. Conversely every $1\otimes v$ is fixed. This also proves that $v\mapsto 1\otimes v$ is injective, including the empty-basis case.

Let $X_{\sigma i}=\sigma (a_i)$ and $Y_{\sigma j}=\sigma (b_j)$. A relation among the rows of $X$ vanishes on the $a_i$, hence by $F$-linearity on every $x\in E$; restricting to $E^{\times }$ and using independence of the distinct automorphisms as multiplicative characters makes every coefficient zero. Thus $X$ is invertible. The trace formula gives $X^{\mathsf{T}}Y=I$, so $X{Y}^{\mathsf{T}}=I$. Its row at $1\in \mathrm{\Gamma }$ yields ${\sum }_i{a}_i\sigma (b_i)={\delta }_{1,\sigma }$.

Any tensor in the kernel is a finite sum ${\sum }_j{e}_j\otimes w_j$. By eliminating dependent members of the finite list $(w_j)$, rewrite it as ${\sum }_{k=1}^r{f}_k\otimes v_k$ with the $v_k$ $F$-independent and invariant. Its image is ${\sum }_k{f}_k{v}_k=0$, so step 1.2 forces every $f_k=0$. Hence the tensor is zero and $\mu$ is injective. For $W=0$ the domain and codomain are zero and the same argument uses the empty list.

For $w\in W$ define $P_i(w)={\sum }_{\sigma \in \mathrm{\Gamma }}\sigma (b_i)T_{\sigma }(w)$. For $\tau \in \mathrm{\Gamma }$, semilinearity gives $T_{\tau }(P_i(w))={\sum }_{\sigma }(\tau \sigma )(b_i)T_{\tau \sigma }(w)=P_i(w)$, since $\sigma \mapsto \tau \sigma$ permutes the finite group. Moreover ${\sum }_i{a}_i{P}_i(w)={\sum }_{\sigma }{\delta }_{1,\sigma }{T}_{\sigma }(w)=w$. Thus $P_i(w)\in W^{\mathrm{\Gamma }}$, and $\mu$ is onto. The formula $\mu (e\otimes w)=ew$ is $F$-balanced and $E$-linear.

If $W$ is an algebra, its fixed space is an $F$-subalgebra containing $1$. For invariant $u,v$ and $e,f\in E$, $\mu ((e\otimes u)(f\otimes v))=efuv=(eu)(fv)$ and $\mu (1\otimes 1)=1$. Distributing over finite sums proves multiplicativity on all tensors. Together with bijectivity, this proves both algebra assertions. If $E=F$, $\mathrm{\Gamma }=\{1\}$ and the map is the usual multiplication $F{\otimes }_{F}W\cong W$; the proof divides by no group order in any characteristic. [F1, step 3.1, step 2.2, step 1.3, algebra] QED

For $r>0$, a central element has a scalar matrix in each coordinate, since commuting can be tested one coordinate at a time. The equation ${\lambda }^2=\lambda$ in the field $E$ forces $\lambda =0$ or $1$. Hence all central idempotents are unique subset sums of the coordinate identities $c_1,\dots ,c_r$. Such a nonzero sum is primitive exactly when its subset is a singleton: a larger subset splits into two nonempty subsets, whereas a singleton cannot.

The simple module for factor $j$ is $E^{d_j}$, on which $c_j$ acts as identity and every other $c_k$ as zero. F4 says these give every class exactly once. An automorphism ${\sigma }_B$ preserves centrality, idempotence, nonzeroness and orthogonal splittings in both directions (use its inverse), so it permutes primitive central idempotents. The twist equation gives ${\sigma }_B(c)\star w=cw$. Thus identity action of $c$ on $W$ is equivalent to identity action of ${\sigma }_B(c)$ on ${}^{\sigma }W$, proving both directions of the claimed compatibility.

Each orbit $O$ is a nonempty subset of this finite family. The sum $c_O$ is nonzero, central and idempotent, since its terms are nonzero orthogonal coordinate identities. Every ${\sigma }_B$ permutes its terms, so it is fixed. F2 gives a unique $e_O\in A$ such that $1\otimes e_O=c_O$.

The identities $1\otimes (e_O^2-e_O)=c_O^2-c_O=0$ and $1\otimes (e_{O}a-a{e}_O)=c_O(1\otimes a)-(1\otimes a)c_O=0$ for every $a\in A$ imply idempotence and centrality by injectivity. Nonzeroness follows from $c_O\mathrm{\ne }0$. Disjoint orbits have disjoint coordinate supports, so $c_O{c}_{O^{\mathrm{\prime }}}=0$ for $O\mathrm{\ne }{O}^{\mathrm{\prime }}$, and all orbit sums add to $1_B$. Injectivity reflects these two equalities to $A$.

If $r=0$, $B=0$ and injectivity gives $A=0$. A unital module over the zero ring is zero because $w=1w=0w=0$, so there are no simple modules and no nonzero primitive central idempotents. The families are empty, with sum $0=1_A$. If an orbit has one element, step 3.1 descends that element itself, with the same nonzeroness and uniqueness proof. [F2, step 3.1, step 4.1, algebra] QED

Suppose $e_O=f+g$ with nonzero orthogonal central idempotents $f,g\in A$. Their scalar extensions are nonzero: F6 makes the canonical scalar-extension map injective. Since $f{e}_O=f$ and $g{e}_O=g$, the subset description in F1 writes $1\otimes f$ and $1\otimes g$ as disjoint nonempty subsets partitioning $O$. Each subset is $\mathrm{\Gamma }$-stable because its sum is a tensor $1\otimes a$ and the subset representation is unique. But a nonempty invariant subset of one orbit equals the entire orbit: for any of its points, every point of the orbit is a translate. Two such disjoint subsets cannot exist. Thus $e_O$ is primitive.

For any primitive central idempotent $e$ of $A$, the products $e{e}_O$ are central orthogonal idempotents summing to $e$. At least one is nonzero since $e\mathrm{\ne }0$; exactly one is nonzero by primitivity of $e$. For that orbit, $e=e{e}_O$. Now $e_O=e+(e_O-e)$ is a sum of orthogonal central idempotents: $(e_O-e{)}^2=e_O-e$ and $e(e_O-e)=0$. Primitivity from step 1.1 forces $e_O-e=0$. Thus every primitive central idempotent is exactly one $e_O$.

For $A\mathrm{\ne }0$, choose $A\cong {\prod }_j{M}_{n_j}(D_j)$ by F2. A matrix commuting with all $E_{kk}$ has zero off-diagonal entries, and commuting with all $E_{kl}$ makes its diagonal entries equal. Commuting also with matrices having any $d\in D_j$ in position $(1,1)$ forces that common entry into $Z(D_j)$. Conversely such scalar matrices commute with everything. Since a division ring has no idempotents except $0,1$, the primitive central idempotents in this product are exactly its factor identities. By step 2.1, each $A{e}_O$ is therefore one factor $M_n(D)$.

A nonzero two-sided ideal $J\subset M_n(D)$ contains a matrix $X$ with entry $x_{pq}\mathrm{\ne }0$. For any $a,b$, $E_{ap}X{E}_{qb}=x_{pq}{E}_{ab}\in J$. Left multiplying by the diagonal matrix having $x_{pq}^{-1}$ in position $(a,a)$ gives $E_{ab}\in J$. Hence $I={\sum }_a{E}_{aa}\in J$ and $J=M_n(D)$. Also every left or right ideal is an $F$-subspace of the finite-dimensional algebra $A{e}_O$; a strict descending chain strictly decreases dimension and so terminates. Thus this ring is simple and Artinian, without asserting that its regular left module is simple when $n>1$.

F3 gives its unique simple left-module class $S=D^n$. A commuting endomorphism of this column module preserves the first coordinate line, and its value there determines all other coordinates by the matrix units; commuting with left multiplication by $D$ forces it to be right multiplication by a fixed $d\in D$. Composition reverses the order of these right multiplications, so ${\mathrm{End}}_A(S)\cong D^{\mathrm{o}\mathrm{p}}$, or $D\cong {\mathrm{End}}_A(S{)}^{\mathrm{o}\mathrm{p}}$. F4 ensures that using another product description does not change the factor data up to the stated equivalences. For $A=0$, F1 supplies empty families and there is no primitive idempotent or simple module; for $n=1$ the same calculations give the division ring itself. [F1, F3, F4, step 3.1, step 4.1, algebra] QED

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