Proposition 6

Let $\Phi$ be an A-bilinear form on E, and K a subring of A such that A is a K-module free of finite dimension q. If $(e_i)_{i=1,\ldots,n}$ is a basis of E over A and $(a_j)_{j=1,\ldots,q}$ a basis of A over K, then $(a_j e_i)$ is a basis of E over K. The mapping $\Phi'$ of $E \times E$ into K defined by $\Phi'(x, y) = \mathrm{Tr}_{A/K}(\Phi(x, y))$ is a K-bilinear form on E, and one has

The first two assertions being evident, it is enough to prove (7). By definition the first member is the determinant of the K-linear endomorphism $u$ of E defined by

Consider the A-linear endomorphism $v$ of E defined by $v(e_i) = \sum_s \Phi(e_i, e_s) e_s$, and the K-linear endomorphism $w$ of E defined by $w(a_j e_i) = (\sum_r \mathrm{Tr}_{A/K}(a_j a_r) a_r) e_i$. We have

since $w(a e_s) = \sum_r \mathrm{Tr}_{A/K}(a a_r) a_r e_s$ for all $a \in A$; thus $u$ is the composite mapping $w \circ v$. Therefore, denoting by $v_K$ the mapping $v$ considered as a K-linear mapping, we have $\det(u) = \det(v_K) \det(w)$. Now $\det(v_K) = N_{A/K}(\det(v))$ (chap. VIII, § 12, no 2, prop. 7), and it is clear that $\det(v) = D_\Phi(e_1, \ldots, e_n)$. On the other hand, since each of the A-modules $Ae_i (i = 1, \ldots, n)$ is stable under $w$, and the determinant of the restriction of $w$ to $Ae_i$ is $\det(\mathrm{Tr}_{A/K}(a_j a_r)) = D_{A/K}(a_1, \ldots, a_q)$, we have $\det(w) = (D_{A/K}(a_1, \ldots, a_q))^n$. Formula (8) therefore reduces to $\det(u) = \det(v_K) \det(w)$, a formula proved above.