Proposition 10

Tag 06D7, permanent. Cite it as [Bourbaki, Tag 06D7].

Algebra, chapter IV, § 5. Symmetric tensors and polynomial mappings

— Let M be a free A-module, then the primitive elements (III, p. 602) of the bigebra $\mathbf{TS}(M)$ are the elements of M.

Let $(e_i)_{i \in I}$ be a basis of M, and for $\nu \in \mathbf{N}^{(I)}$ put $e_\nu = \prod_{i \in I} \gamma_{\nu_i}(e_i)$ . Let $z = \sum_{\nu \in \mathbf{N}^{(I)}} \lambda_\nu e_\nu$ be an element of $\mathbf{TS}(M)$ , then by (12) we have

c(z) = \sum_\nu \lambda_\nu \sum_{\rho, \sigma \in \mathbf{N}^{(I)}, \rho + \sigma = \nu} e_\rho \otimes e_\sigma = \sum_{\rho, \sigma} \lambda_{\rho + \sigma} e_\rho \otimes e_\sigma

hence

c(z) - 1 \otimes z - z \otimes 1 = \sum_{\rho \neq 0, \sigma \neq 0} \lambda_{\rho + \sigma} e_\rho \otimes e_\sigma - \lambda_0 e_0 \otimes e_0

and so

zprimitive $\Leftrightarrow \lambda_{\rho + \sigma} = 0$ when $\rho \neq 0$ and $\sigma \neq 0$ and $\lambda_0 = 0$ $\Leftrightarrow \lambda_\nu = 0$ when $|\nu| \neq 1$ $\Leftrightarrow z \in M$ .

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

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Cites

III, p. 602