Lemma 1

*Let $X_{ij}$ ($1 \leq i \leq n,\ 1 \leq j \leq n$) be $n^2$ indeterminates, $X$ the square matrix $(X_{ij})$ of order $n$ and $D(X_{11}, \ldots, X_{nn}) \in \mathbf{Z}[X_{11}, \ldots, X_{nn}]$ the determinant $\det(X)$. On the other hand let $A$ be a commutative ring, $M_{ij}$ ($1 \leq i \leq n,\ 1 \leq j \leq n$) $n^2$ matrices of order $m$ over $A$, which are pairwise permutable, and $M$ the square matrix of order $mn$ over $A$ which can be expressed as a square matrix of matrices (II, § 10, no. 7)

Then the determinant of $M$ is equal to the determinant of the square matrix of order $m$.

We proceed by induction on $n$, the cases $n = 0$ and $n = 1$ being trivial.

Let $Z$ be a new indeterminate and $N_{ij}$ the matrix $M_{ij} + \delta_{ij} Z I_m$ ($\delta_{ij}$ the Kronecker index). If $D^{ij}(X_{11}, \ldots, X_{nn})$ is the cofactor of $X_{ij}$ in the matrix $X$ (§ 8, no. 6), then

(§ 8, no. 6, formula (28)). We write $N'_{ij} = D^{ij}(N_{11}, \ldots, N_{nn})$, which is a square matrix of order $m$ over $\mathbf{A}[Z]$ and consider the product $N . U$, where

Performing this product in blocks (II, § 10, no. 5) and using formulae (25), we obtain

where we have written $P = D(N_{11}, \ldots, N_{nn})$. Let

which is a matrix of order $m(n-1)$; then (§ 8, no. 6, formula (31)) (det $N$) (det $U$) = (det $P$)(det $Q$) and det $U = \det N'_{11}$. But by the induction hypothesis, det $Q = \det(D^{11}(N_{11}, \ldots, N_{nn})) = \det N'_{11}$ and by virtue of the definition of the $N_{ij}$, clearly det $Q$ is a polynomial in $\mathbf{A}[Z]$ of degree $m(n-1)$, whose term in $Z^{m(n-1)}$ has coefficient 1 ; it follows immediately that det $Q$ is not a divisor of zero in the graded algebra $\mathbf{A}[Z]$. We therefore conclude that det $N = \det(D(N_{11}, \ldots, N_{nn}))$ in $\mathbf{A}[Z]$; if we substitute 0 for $Z$ in these polynomials, then det $M = \det(D(M_{11}, \ldots, M_{nn}))$.

Having shown this lemma, the K-module V is the direct sum of the K-modules $Ae_j$ ($1 \leq j \leq n$); we write $u(e_j) = \sum_{k=1}^n c_{jk} e_k$. For every element $xe_j \in Ae_j$, with $x \in A$, the component of $u(xe_j)$ in $Ae_k$ is $xc_{jk} e_k$; it follows that the matrix of $u_k$ with respect to the basis $(a_i e_j)$ of the K-module V can be expressed in the form of a square matrix of matrices $(M_{jk})$, where $M_{jk}$ is the matrix of the K-linear mapping $xe_j \mapsto xc_{jk} e_k$ of $Ae_j$ into $Ae_k$ with respect to the bases $(a_i e_j)_{1 \leq i \leq m}$ and $(a_i e_k)_{1 \leq i \leq m}$ of these two K-modules (II, § 10, no. 5). If for all $t \in A$, $M(t)$ denotes the matrix, with respect to the basis $(a_i)_{1 \leq i \leq m}$ of A over K, of the endomorphism $x \mapsto xt$ of A, then $M_{jk} = M(c_{jk})$; as $t \mapsto M(t)$ is a ring homomorphism the matrices $M_{jk}$ are permutable with one another. Then

by Lemma 1. But as $t \mapsto M(t)$ is a ring homomorphism, $D(M_{11}, \ldots, M_{nn})$ is the matrix of the K-endomorphism $x \mapsto x.\det(c_{jk})$ of A with respect to basis $(a_i)$; by definition its determinant is therefore $N_{A/K}(\det(u))$, which proves the second formula of (24). On the other hand, and the proof of Proposition 6 is complete.