Proposition 13

Let $K$ be a field, $K_1$ an extension of $K$, $g$ a Lie $K$-algebra and $M, N$ two $g$-modules of finite dimension over $K$. If $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic $g_{(K_1)}$-modules, $M$ and $N$ are isomorphic $g$-modules.

The proof is in two steps.

(1) Suppose first that $K_1$ is an extension of $K$ of finite degree $n$. Let $U$ be the enveloping algebra of $g$, so that the enveloping algebra of $g_{(K_1)}$ is $U_{(K_1)} = U \otimes_K K_1$ (§ 2, no. 9). As $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic as $U_{(K_1)}$-modules they are a fortiori isomorphic as $U$-modules; but as $U$-modules they are respectively isomorphic to $M^n$ and $N^n$. Now $M$ and $N$ are $U$-modules of finite length; $M$ (resp. $N$) is therefore the direct sum of a family $(P_i^{r_i})_{1 \leq i \leq p}$ (resp. $(Q_j^{s_j})_{1 \leq j \leq q}$) of submodules such that the $P_i$ (resp. $Q_j$) are indecomposable and two $P_i$ (resp. $Q_j$) of different indices are not isomorphic (Algebra, Chapter VIII, § 2, no. 2, Theorem 1). Then $M^n$ (resp. $N^n$) is isomorphic to the direct sum of the $P_i^{n r_i}$ (resp. $Q_j^{n s_j}$); it follows (loc. cit.) that $p = q$ and that after permuting the $Q_j$ if necessary $n r_i = n s_i$ and $P_i$ is isomorphic to $Q_i$ for $1 \leq i \leq p$, hence $M$ is isomorphic to $N$.

(2) General case. Let $P$ be the $g$-module $\mathscr{L}_K(M, N)$ and $Q$ the subspace of invariants of $P$, that is the set of homomorphisms of the $g$-module $M$ into the $g$-module $N$. In the $g_{(K_1)}$-module $\mathscr{L}_{K_1}(M_{(K_1)}, N_{(K_1)}) = (\mathscr{L}_K(M, N))_{(K_1)}$, the subspace of invariants is $Q_{(K_1)}$. The hypothesis that $M_{(K_1)}$ and $N_{(K_1)}$ are isomorphic implies that $M$ and $N$ have the same dimension over $K$ and that there exists in $Q_{(K_1)}$ an element $g$ which is an isomorphism of $M_{(K_1)}$ onto $N_{(K_1)}$. Let $(f_1, \ldots, f_d)$ be a basis of $Q$ over $K$ and choose bases of $M$ and $N$ over $K$. If $\lambda_k \in K_1$ for $1 \leq k \leq d$, the matrix of $f = \sum_{k=1}^d \lambda_k f_k$ with respect to these bases has determinant which is a polynomial $D(\lambda_1, \ldots, \lambda_d)$ with coefficients in $K$. When $f = g$, this determinant is non-zero and hence the coefficients of $D$ are not all non-zero. Therefore, if $\Omega$ is the algebraic closure of $K$, there exists (since $\Omega$ is infinite) elements $\mu_k \in \Omega$ $(1 \leq k \leq d)$ such that $D(\mu_1, \ldots, \mu_d) \neq 0$ (Algebra, Chapter IV, § 2, no. 5, Proposition 8). If $K_2$ is the algebraic extension of $K$ generated by the $\mu_k$ $(1 \leq k \leq d)$, it follows that $\sum_{k=1}^d \mu_k f_k$ is an isomorphism of $M_{(K_2)}$ onto $N_{(K_2)}$; but $K_2$ is of finite degree over $K$ (Algebra, Chapter V, § 3, no. 2, Proposition 5) and hence $M$ and $N$ are isomorphic by the first part of the argument.

Again let $K, K_1$ and $\phi$ be as the beginning of this no. Let $\rho$ be a representation of $g$ on a $K$-module $M$ with a finite basis $(x_1, \ldots, x_n)$. Then the bilinear form on $g_{(K_1)}$ associated with $\rho_{(K_1)}$ is derived from the bilinear form associated with $\rho$ by extending the base ring to $K_1$ (for, if $u \in \mathscr{L}_K(M)$, $u$ has the same matrix with respect to $(x_1, \ldots, x_n)$ as $u \otimes 1$ with respect to $(x_1 \otimes 1, \ldots, x_n \otimes 1)$ and hence $u$ and $u \otimes 1$ have the same trace). In particular, if the $K$-module $g$ has a finite basis, the Killing form of $g_{(K_1)}$ is derived from that of $g$ by extending the base ring to $K_1$.