Proposition 1

Let $\Phi$ be a sesquilinear form for $J$ on $E$, $(\tilde{x}_1, \ldots, x_n)$ a system of $n$ elements of $E$, and $(a_{ij})_{(i,j = 1, \ldots, n)}$ a family of $n^2$ elements of $A$; put $y_i = \sum_{i=1}^n a_{ji}x_i$. We then have

Indeed, since $\Phi$ is a sesquilinear form, we have $\Phi(y_i, y_j) = \sum_{k,m} a_{ik} \Phi(x_k, x_m) a_{jm}^j$. Hence, if we denote by $A$ the matrix $(a_{ij})$, the matrix $(\Phi(y_i, y_j))$ is equal to $A . (\Phi(x_i, x_j)). {}^t A^j$. Since $\det({}^t A) = \det(A)$, and $\det(A^j) = \det(A)^j$, our assertion is proved.

In particular, if $(e_i)$ and $(e'_i)$ are two bases of $E, D$ and $D'$ the discriminants of $\Phi$ with respect to these bases, and $a$ the determinant of the passing matrix from the basis $(e_i)$ to the basis $(e'_i)$, we have

It follows from prop. 1 that, if $(e_i)$ is a basis of $E$ and $(x_i)$ an arbitrary system of $n$ elements of $E$, $D_\Phi(e_1, \ldots, e_n)$ divides $D_\Phi(x_1, \ldots, x_n)$. In particular the discriminants of $\Phi$ with respect to any two bases of $E$ generate the same principal ideal of $A$.

Let $(E_i)_{i \in I}$ be a finite family of finite-dimensional free $A$-modules, $\Phi_i$ a sesquilinear form for $J$ on $E_i$, and $B_i$ a basis of $E_i$. If $\Phi$ denotes the direct sum of the $\Phi_i$ (§ 1, no 3) and $B$ the basis of $\prod_{i \in I} E_i$ obtained by joining the $B_i$, then obviously

Let $\Phi$ be a sesquilinear form for $J$ on $E$, $h$ a homomorphism of $A$ into a commutative ring $A'$, $\Phi'$ the sesquilinear form on $A' \otimes_A E$ obtained by extension of $\Phi$ (§ 1, no 4) and $(x_1, \ldots, x_n)$ an arbitrary system of elements of $E$. Since $A' \otimes_A E$ is a free $A'$-module, $D_{\Phi'}(1 \otimes x_1, \ldots, 1 \otimes x_n)$ is defined, and we have obviously