Proposition 1

Tag 065T, permanent. Cite it as [Bourbaki, Tag 065T].

Algebra, chapter III, § 9. Norms and traces

Let $M = M_0 \supset M_1 \supset \ldots \supset M_r = \{0\}$ be a decreasing sequence of submodules of an $A$ -module $M$ such that each of the $K$ -modules $P_i = M_{i-1}/M_i$ ( $1 \leq i \leq r$ ) admits a finite basis. Then the $K$ -module $M$ admits a finite basis and

\operatorname{Tr}_M(a) = \sum_{i=1}^r \operatorname{Tr}_{P_i}(a), \qquad N_M(a) = \prod_{i=1}^r N_{P_i}(a)

(15)

P_{C_M}(a; X) = \prod_{i=1}^r P_{C_{P_i}}(a; X).

Let $B'_i$ be a basis of $P_i$ over $K$ ; then a system of representatives $B_i$ of $B'_i$ (mod. $M_i$ ) is a basis of a supplementary submodule of the $K$ -module $M_i$ in the $K$ -module $M_{i-1}$ (II, § 1, no. 11, Proposition 21). The union $B$ of the $B_i$ ( $1 \leq i \leq r$ ) is a basis of $M$ over $K$ . Let $X_{tt}$ be the matrix of the endomorphism $a_{P_i}$ with respect to the basis $B'_i$ . It is immediate that the matrix of $a_M$ with respect to $B$ is of the form

\begin{pmatrix} X_{rr} & X_{r,r-1} & \cdots & X_{r,1} \\ 0 & X_{r-1,r-1} & \cdots & X_{r-1,1} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & X_{11} \end{pmatrix}

and the proposition follows from formulae (4), (5) and (6) of no. 1 and formula (31) of § 8, no. 6.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Proposition 1

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