Proposition 2

Suppose that $A$ is an integral domain. Let $\Phi$ be a sesquilinear form for $J$ on $E$ and $(e_1, \ldots, e_n)$ a basis of $E$, such that $D_\Phi(e_1, e_2, \ldots, e_n) \neq 0$.

a) For a system $(x_1, \ldots, x_n)$ of $n$ elements of $E$ to be free, it is necessary and sufficient that $D_\Phi(x_1, \ldots, x_n)$ be $\neq 0$.

b) For a system $(x_1, \ldots, x_n)$ of $n$ elements of $E$ to be a basis of $E$, it is necessary and sufficient that $D_\Phi(x_1, \ldots, x_n)$ and $D_\Phi(e_1, \ldots, e_n)$ be associated elements in $A$ (cf. Chapter VI, § 1, No. 5).

Let $x_j = \sum_{i=1}^n a_{ji} e_i$ ($a_{ji} \in A$). Let us first prove a). If $D_\Phi(x_1, \ldots, x_n) = 0$, we have $\det(a_{ji}) \cdot \det(a_{ji})' = 0$ (Prop. 1) since $D_\Phi(e_1, \ldots, e_n) \neq 0$ and $A$ is integral; we therefore have $\det(a_{ji}) = 0$, and the vectors $x_j$ are linearly dependent (Chapter III, § 7, No. 1, Th. 1, applied to the vector space $K \otimes_A E$, where $K$ denotes the field of fractions of $A$). Conversely, if these vectors are linearly dependent we have $\det(a_{ji}) = 0$ (ibid.), whence $D_\Phi(x_1, \ldots, x_n) = 0$ (Prop. 1).

Let us now prove b). If $D_\Phi(x_1, \ldots, x_n)$ and $D_\Phi(e_1, \ldots, e_n)$ are associated in $A$, Prop. 1 shows that $\det(a_{ij}) \cdot \det(a_{ij})'$ is invertible in $A$. Thus $\det(a_{ij})$ is itself invertible in $A$; hence the matrix $(a_{ij})$ over $A$ is invertible (Chapter III, § 6, No. 5, Th. 2), and the endomorphism $g$ of $E$ defined by $g(e_i) = x_i$ ($i = 1, \ldots, n$) is an automorphism; consequently $(x_1, \ldots, x_n)$ is a basis of $E$. The converse follows immediately from Prop. 1.