Proposition 3

The mapping det is the unique homomorphism from $\mathbf{G}\mathbf{L}_n(D)$ to $D^*_{ab}$ that satisfies the relations

(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.