Remark

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

Algebra, chapter IX, § 1. Formes sesquilinéaires

Let E be a free A-module, and let θ be the canonical isomorphism of $\bigwedge^m E$ onto the submodule of antisymmetrized tensors of order m (Chap. III, § 5, No. 6, Prop. 6). Let $\Phi$ be a sesquilinear form on E, $\Phi_{(m)}$ the extension of $\Phi$ to $\bigwedge^m E$ , and $\Theta$ the sesquilinear form on $\bigwedge^m E$ which is the inverse image by $\theta$ of the extension of $\Phi$ to $\bigotimes^m E$ . From the definition of $\theta$ and of the antisymmetrized tensor, and from (35), we have

\Theta(x_1 \wedge \ldots \wedge x_m, y_1 \wedge \ldots \wedge y_m) = \sum_{\sigma, \tau} \varepsilon_\sigma \varepsilon_\tau \Phi(x_{\sigma(1)}, y_{\tau(1)}) \ldots \Phi(x_{\sigma(m)}, y_{\tau(m)})

where $\sigma$ and $\tau$ run through the symmetric group $S_m$ . From the formula for calculating determinants and formula (37), this expression can be written

\sum_{\tau \in S_m} \varepsilon_\tau \det(\Phi(x_i, y_{\tau(k)})) = m! \det(\Phi(x_i, y_k));

in other words, we have $\Theta = m! \Phi_{(m)}$ .

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

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