Remark

Tag 0DK2, permanent. Cite it as [Bourbaki, Tag 0DK2].

Algebra, chapter IX, § 2. Discriminant d'une forme sesquilinéaire

Soient $(e_1, \ldots, e_n)$ a basis of $B$ over $A$ , and

e_ie_j = \sum_{k=1}^n c_{ijk}e_k \quad (c_{ijk} \in A)

(chap. II, § 7, no 2). Since the matrix of the $A$ -linear endomorphism $x \to e_kx$ of $B$ with respect to $(e_r)$ is $(c_{ksr})$ , we have $\mathrm{Tr}_{B/A}(e_k) = \sum_{r=1}^n c_{krr}$ , whence $\mathrm{Tr}_{B/A}(e_ie_j) = \sum_{k,r} c_{ijk}c_{krr}$ . It follows that

D_{B/A}(e_1, \ldots, e_n) = \det_{i,j} (\sum_{k,r} c_{ijk}c_{krr}).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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