Remark 2

Tag 076D, permanent. Cite it as [Bourbaki, Tag 076D].

— By IV, p. 80, Cor. 1 and formula (8), we have, for each polynomial $q \in K[X]$ , that

\det q(u) = \operatorname{res}(\chi_u, q) ,

where $\operatorname{res}(\chi_u, q)$ is the resultant of the polynomials $\chi_u$ and $q$ . In particular if we take $q = \chi_u'$ we obtain

\det \chi_u'(u) = (-1)^{n(n-1)/2} \operatorname{dis}(\chi_u),

where $\operatorname{dis}(\chi_u)$ is the discriminant of the polynomial $\chi_u$ (IV, p. 82, formula (47)). Furthermore:

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

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