Lemma 1

Let p be an integer such that $0 \leqslant p \leqslant m$. For any w in $\wedge^p(M)$, let $h_p(w)$ be the linear mapping $w' \mapsto w \wedge w'$ from $\wedge^{m-p}(M)$ to $\wedge^m(M)$. The linear mapping $h_p : w \mapsto h_p(w)$ from $\wedge^p(M)$ to $\operatorname{Hom}_L(\wedge^{m-p}(M), \wedge^m(M))$ is an isomorphism.

Let $(e_i)_{i \in I}$ be a basis of M; we endow the set I with a total order. For any subset J of I, set $e_J = e_{i_1} \wedge \cdots \wedge e_{i_r}$, where $(i_1, \ldots, i_r)$ is the sequence of elements of J in increasing order. The L-module $\wedge^{m-p}(M)$ admits as a basis the elements $e_S$, where S runs through the set of subsets of I with $m - p$ elements; $\wedge^m(M)$ has $\{e_I\}$ as a basis. Consequently, there exists a basis of $\operatorname{Hom}_L(\wedge^{m-p}(M), \wedge^m(M))$ consisting of linear mappings $e_J^*$ characterized by the formula

where J runs through the set of subsets of I with p elements. It follows from formula (20) of III, §7, No. 8, p. 519 that for every subset J of I with p elements, we have $h_p(e_J) \in \{e_J^*, -e_J^*\}$; since the elements $e_J$ form a basis of $\wedge^p(M)$, the linear mapping $h_p$ is bijective.

Let us now prove Proposition 1. Let u and $\tilde{u}$ be endomorphisms of M. Relation (3) is equivalent to

the mapping $h_1$ is an isomorphism from M to $\operatorname{Hom}_L(\wedge^{m-1}(M), \wedge^m(M))$ by Lemma 1. Consequently, for every endomorphism u of M, there exists a unique endomorphism $\tilde{u}$ of M satisfying relation (3).

Let $x_1, . . . , x_m$ be elements of M. Let us replace $x$ with $u(x_1)$ and $w$ with $x_2\wedge \cdots \wedge x_m$ in (3); we obtain

$\widetilde{u}(u(x_1))\wedge x_2\wedge \cdots \wedge x_m=u(x_1)\wedge \cdots \wedge u(x_m) =$ det($u$)$x_1\wedge \cdots \wedge x_m$.

Consequently, $h_1(\widetilde{u}\circ u(x_1)) =h_1$(det($u$)$x_1$), which gives the relation $\widetilde{u}\circ u=$ det($u$)$_M$ by Lemma 1.

We denote by U the endomorphism $X-\overline{u}$ of the L[X]-module M[X] (VIII, p. 335). By the above applied to U, there exists an endomorphism $\widetilde{U}$ of the L[X]-module M[X] that satisfies the relations

for $x_1, . . . , x_m$ in M and

(10) $\widetilde{U}\circ U =$ det(X $-\overline{u}$)$_{M[X]}$.

Let us view $\widetilde{U}$ as an element of End(M)[X] (VIII, p. 9); by formula (9) and Lemma 1, it has degree $\leqslant m-1$, so we can write it as

where the $u_r$ are endomorphisms of M. By formula (2), relation (10) gives the equality

in the ring End(M)[X]. By identifying the coefficients of the monomials in X on each side, we obtain the relations

(13) $u_r+u_{r-1}\circ u=c_r(u)_M$ for $1\leqslant r\leqslant m-1$

and

From this, we deduce

Now, when we identify the constant terms, equalities (9) and (11) imply $u_{m-1}=\widetilde{u}$; formula (6) follows.

In particular, $\widetilde{u}$ belongs to the subalgebra of End(M) generated by $u$ and therefore commutes with $u$. We have already established the relation $\widetilde{u}\circ u=$ det($u$)$_M$; formula (4) follows.

Finally, let $x_1, . . . , x_m$ be elements of M. We replace $x$ with $x_1$ and $w$ with $\widetilde{u}(x_2)\wedge \cdots \wedge \widetilde{u}(x_m)$ in formula (3). This gives

= det($u$)$^{m-1}x_1\wedge x_2\wedge \cdots \wedge x_m$

and therefore formula (5).