A) Uniqueness of $\omega :$ Let $\omega$ be an element of Ω(V). Let $(v_1, . . . , v_n)$ be a basis of V, and let $\mu_1, . . . , \mu_n$ be the coordinates of $e$ in this basis. Denote by I the set of integers from 1 to $n$ (inclusive) such that $\mu_i\not= 0$; it is not empty. Let $i$ be an element of I. The sequence $(v_1, . . . ,\widehat{v}_i, . . . , v_n, e)$ is a basis of V, and since $e-v_i\mu_i$ is a linear combination of the sequence $(v_1, . . . ,\widehat{v}_i, . . . , v_n)$, formula (6) implies

It follows from formulas (1) and (3) that the left-hand side of this equality is equal to $\pi (-1)^{n-i}\pi (\mu_i)\omega (v_1, . . . , v_n)$. Denote by $p$ the projector of V with image W and kernel $eD$. The vectors $v_j-p(v_j)$ are proportional to $e$, and repeatedly applying formula (6) shows that the right-hand side of formula (8) is equal to $\omega (p(v_1), . . . ,\widehat{p}(v_i), . . . , p(v_n), e)$.

It follows from this that if $\omega$ satisfies relation (7), then we have

for every $i$ in I, which proves the uniqueness of $\omega$.

B) Construction of $\omega :$ Let $(v_1, . . . , v_n)$ be a basis of V; we define $\mu_1, . . . , \mu_n$ and I as above. For every $i$ in I, the assumption $\mu_i\not= 0$ implies that the sequence $(v_1, . . . ,\widehat{v}_i, . . . , v_n, e)$ generates the space V, so the sequence $(p(v_1), . . . ,\widehat{p}(v_i), . . . , p(v_n))$ generates the space $W =p(V)$. In other words, this last sequence is a basis of W, and

defines an element of $D^*_{ab}$.

Let $i$ and $j$ be two elements of I such that $i < j$; let us prove the equality $t_i=t_j$. By definition, the vector $v_i\mu_i+v_j\mu_j-e$ is a linear combination of the vectors $p(v_k)$ for $k$ different from $i$ and $j$. Consequently, $p(v_i)\mu_i+p(v_j)\mu_j$ is a linear combination of the vectors $p(v_k)$ for $k$ different from $i$ and $j$. By formula (6), we therefore have

By applying formulas (1) and (3), we deduce

and therefore $t_i=t_j$.

Having proved this, we define $\omega (v_1, . . . , v_n)$ as the common value of the $t_i$ for $i$ in I. We have thus constructed a mapping $\omega$ from B(V) to $D^*_{ab}$ that satisfies relation (9). Let $(w_1, . . . , w_{n-1})$ be a basis of W; if we set $v_i=w_i$ for $1\leqslant i\leqslant n-1$ and $v_n=e$, then we have $I =\{n\}$ and $\mu_n= 1$. We moreover have $p(v_i) =w_i$ for $1\leqslant i\leqslant n-1$, and the formula $\omega (w_1, . . . , w_{n-1}, e) =\varphi (w_1, . . . , w_{n-1})$ is a specific case of (9).

We must now prove that $\omega$ belongs to Ω(V).

C) Proof of formula (1): Let $\lambda_1, . . . , \lambda_n$ be elements of $D^*$ and $(v_1, . . . , v_n)$ be a basis of V. We define $\mu_1, . . . , \mu_n$ and I as above. Choose an $i$ in I. Since we have $e=\sum^n_{j=1}(v_j\lambda_j)(\lambda^{-1}_j\mu_j)$, formula (9) implies

But we have $\pi (\lambda^{-1}_i\mu_i)^{-1}=\pi (\mu_i)^{-1}\pi (\lambda_i)$ and

Comparing formulas (9) and (11) leads to the desired relation

D) Proof of the formula (2): Let $(v_1, . . . , v_n)$ be a basis of V and $i, j$ be two distinct integer in the interval $[1, n]$; we define $\mu_1, . . . , \mu_n$ as before. Consider the basis $(v'_1, . . . , v'_n)$ of V defined by $v'_i=v_i+v_j$ and $v'_k=v_k$ for $k\not=i$, and let us introduce the coordinates $\mu'_1, . . . , \mu'_n$ of $e$ with respect to this basis; they satisfy $\mu'_j=\mu_j-\mu_i$ and $\mu'_k=\mu_k$ for $k\not=j$. We set $t=\omega (v_1, . . . , v_n)$ and $t'=\omega (v'_1, . . . , v'_n)$. We must prove that $t$ and $t'$ are equal.

Let us first note that, by the definition of $\omega$, we have

for every $k$ such that $\mu_k$ and $\mu'_k$ are nonzero.

a) If $\mu_i\not=0$, then the sequences $(p(v_1), . . . ,\widehat{p(v_k)}, . . . , p(v_n))$ and $(p(v'_1), . . . ,\widehat{p(v'_k)}, . . . , p(v'_n))$ are equal, and we have $\mu_i$ = $\mu'_i$, and therefore $t=t'$.

b) If there exists an index $k$ different from $i$ and $j$ such that $\mu_k\not= 0$, then we have $p(v'_l) =p(v_l)$ for $l\not=i$ and $p(v'_i) =p(v_i) +p(v_j)$. The elements $p(v_i)$ and $p(v_j)$ both belong to the sequence $(p(v_1), . . . ,p(\widehat{v}_k), . . . , p(v_n))$. Since $\varphi$ belongs to Ω(W), formula (2) of VIII, p. 447 applies; it gives

But we also have $\mu_k=\mu'_k$, and therefore $t=t'$.

c) It remains to examine the case when the only index $k$ such that $\mu_k\not= 0$ is $j$. We then have $\mu'_j=\mu_j,e=v_j\mu_j$, and $p(v_j) = 0$. Set $k=j$ in formulas (12) and (13). Since the sequences $(p(v_1), . . . ,p(\widehat{v}_j), . . . , p(v_n))$ and $(p(v'_1), . . . ,p(\widehat{v}'_j), . . . , p(v'_n))$ are equal, we have $t=t'$.

Let us prove the corollary by induction on $n$. If $n= 0$, then the only basis of V is the empty one, so the assertion is true. Now suppose $n\geqslant 1$; denote by W the subspace of V generated by $e_1, . . . , e_{n-1}$. By Proposition 1, there exists a bijection Λ from Ω(V) to Ω(W) satisfying

for every basis $(w_1, . . . , w_{n-1})$ of W and every $\omega$ in Ω(V). By the induction hypothesis, there exists a unique element $\varphi$ of Ω(W) such that $\varphi (e_1, . . . , e_{n-1}) =t$. The relation $\omega (e_1, . . . , e_n) =t$, for $\omega$ in Ω(V), is equivalent to $\Lambda (\omega ) =\varphi$. Corollary 1 follows.

(15) $\omega (u(v_1), . . . , u(v_n)) =$ (det $u$)$\omega (v_1, . . . , v_n)$

for every basis $(v_1, . . . , v_n)$ of V and every $\omega$ in Ω(V).

b) The mapping $u\mapsto$ det $u$ from $\mathbf{G}\mathbf{L}(V)$ to $D^*_{ab}$ is a group homomorphism.

Let $\omega_0$ be an element of Ω(V). The mapping $(v_1, . . . , v_n)\mapsto$ $\omega_0(u(v_1), . . . , u(v_n))$ from B(V) to $D^*_{ab}$ belongs to Ω(V). By Corollary 2 of VIII, p. 451, there exists a unique element $t$ of $D^*_{ab}$ such that we have

for $(v_1, . . . , v_n)$ in B(V). If $\omega$ is another element of Ω(V), then there exists an element $s$ of $D^*_{ab}$ such that

for every basis $(v_1, . . . , v_n)$ of V (loc. cit.). We immediately deduce the relation

This proves a); assertion b) is an immediate consequence of a).

(16) det($B_{ij}(1)$) $= 1$

for $i\not=j$ and

(17) det(diag($\lambda_1, . . . , \lambda_n$)) $=\pi (\lambda_1\cdots \lambda_n)$,

for $\lambda_1, . . . , \lambda_n\in D^*$

Let A be an element of $\mathbf{G}\mathbf{L}_n(D)$ and $a_1, . . . , a_n$ be its columns. We have

det(A) $=\omega_0(a_1, . . . , a_n)$.

Now, the columns of the matrix A diag($\lambda_1, . . . , \lambda_n$) are $a_1\lambda_1, . . . , a_n\lambda_n$, and those of the matrix AB$_{ij}(1)$ are therefore $a_1, . . . , a_j+a_i, a_{j+1}, . . . , a_n$. Since $\omega_0$ is the unique element of $\Omega (D^n_d)$ such that $\omega_0(\varepsilon_1, . . . , \varepsilon_n) = 1$, we see that the determinant is the unique mapping $\varphi :\mathbf{G}\mathbf{L}_n(D)\rightarrow D^*_{ab}$ that satisfies the relations

(18) $\varphi ($A diag($\lambda_1, . . . , \lambda_n$)) $=\varphi ($A$)\pi (\lambda_1\cdots \lambda_n)$,

(19) $\varphi ($AB$_{ij}(1)) =\varphi ($A) for $i\not=j$,

This proves, to begin with, that the determinant mapping satisfies relations (16) and (17). We already know that this mapping is a homomorphism from $\mathbf{G}\mathbf{L}_n(D)$ to $D^*_{ab}$. Conversely, if $\varphi$ is a homomorphism from $\mathbf{G}\mathbf{L}_n(D)$ to $D^*_{ab}$ such that $\varphi (B_{ij}(1)) = 1$ for $i\not=j$ and $\varphi$(diag($\lambda_1, . . . , \lambda_n$)) $=\pi (\lambda_1\cdots \lambda_n)$, then $\varphi$ satisfies relations (18) through (20) and is therefore equal to det.