Remark 2

Tag 06CZ, permanent. Cite it as [Bourbaki, Tag 06CZ].

Algebra, IV, p. 46, § 5. Symmetric tensors and polynomial mappings

Let $\mathscr{M}$ be the set of mappings of $(1, p)$ into I. We define a mapping $\rho \mapsto \rho^*$ of $\mathscr{M}$ into $\mathbf{N}^{(I)}$ by putting

\rho^*(i) = \operatorname{Card} \rho^{-1}(i)

For two elements $\rho_1, \rho_2$ of $\mathscr{M}$ to satisfy $\rho_1^* = \rho_2^*$ it is necessary and sufficient that there should exist $\sigma \in \mathfrak{S}_p$ such that $\rho_2 = \rho_1 \circ \sigma$ (I, p. 95). By Prop. 3 (iii) we have, for $|\nu| = p$ ,

x_\nu = \sum_{\rho \in \mathscr{M}, \rho^* = \nu} x_{\rho(1)} \otimes x_{\rho(2)} \otimes \ldots \otimes x_{\rho(p)}

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

Cited by

Proposition 4, § 5

Cites

I, p. 95
Prop. 3