Proposition 4

Suppose that $A$ is integral. Let $S$ be a basis of $E$. A necessary and sufficient condition for a sesquilinear form $\Phi$ on $E$ to be nondegenerate is that one has $D_\Phi(S) \neq 0$.

Indeed, let K be the field of fractions of A, and let $\Phi'$ be the extension of $\Phi$ to the K-vector space $K \otimes_A E$; identify E with a subset of this vector space. The relation $D_{\Phi}(S) \neq 0$ is then equivalent to $D_{\Phi'}(S) \neq 0$ (formula (6)), which itself expresses that $s_{\Phi'}$ is bijective (prop. 3), that is to say that $\Phi'$ is nondegenerate (§ 1, No. 6, prop. 6). Now, for every $x \in K \otimes_A E$, there exists $a \in A$ such that $ax \in E$; consequently, for $\Phi$ to be degenerate, it is necessary and sufficient that $\Phi'$ be so. This proves our assertion.