Complements on linear representations

a) Assume first that L is an extension of K of finite degree n. Since V \otimes_K L and W \otimes_K L are isomorphic as (A \otimes_K L)-modules, they are isomorphic as A-modules; but, as A-modules, they are isomorphic to V^n and W^n, respectively. Now V and W are A-modules of finite length; thus V (resp. W) is the direct sum of a family (M_i^{r_i})_{1 \leq i \leq p} (resp. (N_j^{s_j})_{1 \leq j \leq l}) of submodules such that the M_i (resp. N_j) are indecomposable, and that two of the M_i (resp. N_j) with distinct indices are not isomorphic (Algebra, Chap. VIII, § 2, no. 2, Th. 1). Then V^n (resp. W^n) is the direct sum of the M_i^{n r_i} (resp. N_j^{n s_j}); it follows (loc. cit.) that p = q and, after a suitable permutation of the N_j, that M_i is isomorphic to N_i and n r_i is equal to n s_i for 1 \leq i \leq p. Thus, V is isomorphic to W.

b) Assume that K is an infinite field. The assumption implies that V and W have the same dimension over K. Let (e_i)_{1 \leq i \leq m} and (e'_i)_{1 \leq i \leq m} be bases of V and W over K, and (a_\lambda) a basis of A over K. An isomorphism u : V \otimes_K L \to W \otimes_K L is a bijective L-linear map and at the same time an (A \otimes_K L)-homomorphism, in other words it satisfies the conditions

Put $a_\lambda e_i = \sum_j \gamma_{\lambda ij} e_j, \ a_\lambda e'_i = \sum_j \gamma'_{\lambda ij} e'_j,$ where the $\gamma_{\lambda ij}$ and $\gamma'_{\lambda ij}$ belong to K, and $u(e_i) = \sum_j \xi_{ij} e'_j,$ where the $\xi_{ij}$ belong to L. The conditions (1) can be written

for all $\lambda, i, k$. By assumption, the homogeneous linear equations (2) have a solution $(\xi_{ij}) \in L^{m^2}$ such that $\det(\xi_{ij}) \neq 0$. Since the coefficients of the system (2) belong to K, we know (Algebra, Chap. II, § 8, no. 5, Prop. 6) that this system also admits non-trivial solutions in $K^{m^2}$; let E be the vector subspace of $M_m(K) = K^{m^2}$, not reduced to zero, formed by these solutions. Let $(c_l)_{1 \leq l \leq p}$ be a basis of $E$, and put $(\xi_{ij}) = \sum_l \eta_l c_l$ for any matrix $(\xi_{ij}) \in E$; then $\det(\xi_{ij})$ is a polynomial $P(\eta_1, \ldots, \eta_p)$ with coefficients in $K$. On the other hand, we know (loc. cit.) that the solutions of (2) in $L^{m^2}$ are of the form $\sum_l \zeta_l c_l$, this time with $\zeta_l \in L$; for such a solution, $\det(\xi_{ij})$ is equal to $P(\zeta_1, \ldots, \zeta_p)$. Granting this, if we had $P(\eta_1, \ldots, \eta_p) = 0$ for all $\eta_1, \ldots, \eta_p \in K$, the coefficients of $P$ would be zero since $K$ is infinite; we would then have $P(\zeta_1, \ldots, \zeta_p) = 0$ for all $\zeta_1, \ldots, \zeta_p \in L$, which is contrary to our assumption. We can therefore find a matrix $(\xi_{ij}) \in E$ such that $\det(\xi_{ij}) \neq 0$, and the corresponding linear map $V \to W$ is an isomorphism.

c) General Case. Let $\Omega$ be an algebraically closed extension of $L$, and $K_0$ the algebraic closure of $K$ in $\Omega$. The assumption implies that $V \otimes_K \Omega$ and $W \otimes_K \Omega$ are isomorphic $(A \otimes_K \Omega)$-modules. Since $K_0$ is infinite, part b) shows that $V \otimes_K K_0$ and $W \otimes_K K_0$ are isomorphic $(A \otimes_K K_0)$-modules. Retaining the notation in b), the system (2) admits a solution $(\xi_{ij}) \in K_0^{m^2}$ such that $\det(\xi_{ij}) \neq 0$. But the $\xi_{ij}$ all belong to some algebraic extension $K_1$ of finite degree over $K$. The $(A \otimes_K K_1)$-modules $V \otimes_K K_1$ and $W \otimes_K K_1$ are isomorphic, and the proof is completed by using a).

Let $p$ be a projection of $E$ onto $F$. For all $x \in E$, put

We have $f(x) \in F$ and $f(y) = y$ for all $y \in F$, so $f$ is a projection of $E$ onto $F$. On the other hand, if $t \in G$, Thus $f$ commutes with $\rho(G)$, so $\ker(f)$ is a complement of $F$ stable under $G$.

Indeed, by Prop. 2, every module over this algebra is semi-simple.

Indeed, let $Q$ be the projection of $M$ onto $M^G$ defined by $Q(x) = q^{-1} \sum_{g \in G} g(x)$ for all $x \in M$. If $i$ denotes the canonical injection of $M^G$ into $M$, $Q \circ i$ is the identity map of $M^G$, so $(Q \otimes 1_{A'}) \circ (i \otimes 1_{A'})$ is the identity map of $M^G \otimes_A A'$. Since $Q \otimes 1_{A'} = q^{-1} \sum_{g \in G} (g \otimes 1_{A'})$, the image of $i \otimes 1_{A'}$ is $(M \otimes A')^G$. On the other hand, $i \otimes 1_{A'}$ is injective by what has been said before.