Lemma 3

Under the hypotheses of Proposition 18, let (w_1, ..., w_n) be a basis of K' over K contained in A' (no. 5, Corollary 2 to Proposition 16); then there is a unique basis (w_1^*, ..., w_n^*) of K' over K for which Tr_{K'/K}(w_i w_j^*) = \delta_{ij} (Kronecker index); if d = D_{K'/K}(w_1, ..., w_n) is the discriminant of the basis (w_1, ..., w_n) (Algebra, Chapter IX, § 2), then d \neq 0 and

In particular, if d is an invertible element of A, A' is a free A-module with basis (w_1, ..., w_n).

As K' is a separable K-algebra, d \neq 0 (Algebra, Chapter IX, § 2, Proposition 5) and the K-bilinear form

over K' is therefore non-degenerate (loc. cit., Proposition 4); this shows the existence and uniqueness of the basis $(w_i^*)_{1 \leq i \leq n}$ (Algebra, Chapter IX, § 1, no. 6, Corollary to Proposition 6). This being so, the first inclusion of (2) is obvious. Let $x$ be an element of $A'$; let us write $x = \sum_{i=1}^n \xi_i w_i^*$ where $\xi_i \in K$; for all $i$, $\xi_i = \mathrm{Tr}_{K'/K}(xw_i)$, hence $\xi_i$ is integral over $A$ (Corollary 2 to Proposition 17) and, as $A$ is integrally closed, $\xi_i \in A$ for $1 \leq i \leq n$; this shows the second inclusion (2). Finally, let us write $w_j^* = \sum_{i=1}^n \alpha_{ji} w_i$ where $\alpha_{ji} \in K$; then

for all $j$ and $k$; Cramer's formulae show that the $\alpha_{ji}$ belong to $d^{-1}A$, whence the third inclusion (2). The last assertion follows immediately from (2), which in this case gives $A' = \sum_{i=1}^n Aw_i$.

In the two corollaries which follow the hypotheses and notations are those of Proposition 18.