Remark

— Suppose the conditions of Prop. 16 are satisfied and let $L$ be an extension of $K$. By Prop. 13 the set $\mathfrak{F}_{(L)}$ is diagonalisable if and only if the algebra $L \otimes_K A$ is diagonalisable. It follows from V, p. 30, Prop. 2 that there exists a finite extension $L$ of $K$ such that $\mathfrak{F}_{(L)}$ is diagonalisable. In fact $L$ may be taken to be Galois; indeed, taking a finite subset $\mathfrak{F}'$ of $\mathfrak{F}$ which generates $A$, we may take $L$ to be a splitting field for the minimal polynomials of the elements of $\mathfrak{F}'$ (Prop. 12 and 13).