Proposition 1

Let A be an algebra of finite degree n over the field K ; then the following conditions are equivalent : a) The algebra A is diagonalizable. b) There is a basis $(e_1, \ldots, e_n)$ of A such that $e_i^2 = e_i$ and $e_i e_j = 0$ for $i \neq j$. c) The K-algebra homomorphisms of A into $\mathbf{K}$ generate the dual of the vector K-space A. d) Every A-module is a direct sum of submodules which are of dimension 1 over K.

The equivalence of a) and b) follows from Formula (6); on the other hand the n projections $K^n \to K$ are algebra homomorphisms, hence a) implies c). If c) holds, the algebra homomorphisms of A into $\mathbf{K}$ form a basis of the dual of A (V, p. 27, Th. 1), we denote them by $u_1, \ldots, u_n$; then $a \mapsto (u_i(a))$ is an isomorphism of A onto the algebra $K''$, whence a). We have thus established the equivalence of the conditions a), b) and c).

Suppose that b) holds and let M be an A-module; then the homotheties $(e_i)_M$ of ratio $e_i$ are projectors of $M$, and $M$ is a direct sum of the $e_i M$, which are sub-A-modules. We may thus suppose that there exists an index $i$ such that $(e_j)_M = 0$ for $j \neq i$. Hence every vector subspace of $M$ is a sub-A-module, whence d).

Conversely suppose that d) holds and consider the A-module A,. There exists then a basis $(f_i)$ of the vector K-space A such that $Af_i = Kf_i$ for $i = 1, \ldots, n$. After replacing each $f_i$ by a suitable scalar multiple, if necessary, we may suppose that $1 = f_i + \ldots + f_n$. If $i \neq j$, then $f_i f_j$ belongs to $Af_i \cap Af_j = Kf_i \cap Kf_j$ hence it is zero. Now $f_i = f_i f_1 + \ldots + f_i f_n = f_i^2$, whence b).