Proposition 5

*Let A be a K-algebra and m a two-sided ideal of A. Suppose that $A_0 = A/m$ is a free K-module of finite dimension n, that there exists an integer $r > 0$ such that $m^r = \{0\}$ and that $m^{i-1}/m^i$ is a free $A_0$-module of finite dimension $s_i$ for $1 \leq i \leq r$. Let $s = s_1 + \cdots + s_r$ and for all $a \in A$ let $a_0$ denote the class of $a$ mod. m. Then A is a free K-module of dimension $ns$ and, for all $a \in A$,

By virtue of II, § 1, no. 13, Proposition 25, $m^{i-1}/m^i$ is a free K-module of dimension $ns_i$. Hence Proposition 1 of no. 2 can be applied with $P_i = m^{i-1}/m^i$;

this shows in the first place that $A$ is a free $K$-module of dimension $n(s_1 + \cdots + s_r) = ns$. Moreover, the hypothesis implies that the $A$-module $P_i$ is isomorphic to a direct sum of $s_i$ submodules isomorphic to the $A$-module $A_0$; by Proposition 1 of no. 2, therefore $N_{P_i}(a) = N_{A_0}(a)^{s_i}$; finally therefore

In this formula $N_{A_0}(a)$ is defined by considering $A_0$ as a left $A$-module and it is equal to the determinant of the $K$-linear mapping $x \mapsto ax$ of $A_0$ into itself; but, as $ax = a_0x$ for $x \in A_0$, $N_{A_0}(a) = N_{A_0}(a_0)$, which completes the proof of formula (23) for the norm. The two others are shown analogously.