Proposition 15

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

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

— Let $M$ be a free $A$ -module, $N$ an $A$ -module and $q$ an integer $\geqslant 0$ . We suppose that the mapping $y \mapsto q!y$ is an automorphism of $N$ . Let $f \in \mathrm{Pol}^q(M, N)$ , then there exists one and only one symmetric $q$ -linear mapping $h$ of $M^q$ into $M$ such that $f(x) = h(x, x, \ldots, x)$ for all $x \in M$ . For any $x_1, \ldots, x_q \in M$ we have

h(x_1, x_2, \ldots, x_q) = \frac{(-1)^q}{q!} \sum_{H \subset \{1, 2, \ldots, q\}} (-1)^{\mathrm{Card}\ H} f\left( \sum_{i \in H} x_i \right).

a) There exists a $q$ -linear mapping $g$ of $M^q$ into $N$ such that $f(x) = g(x, x, \ldots, x)$ for all $x \in M$ . Let us define a $q$ -linear mapping $h$ of $M$ into $N$ by

h(x_1, x_2, \ldots, x_q) = \frac{1}{q!} \sum_{\sigma \in S_q} g(x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(q)}).

Then $h$ is symmetric and $f(x) = h(x, x, \ldots, x)$ for all $x \in M$ .

b) Let $h$ be a symmetric $q$ -linear mapping of $M^q$ into $N$ such that f(x) = g(x, x, ..., x). Let l be the linear mapping of T^q(M) into N such that h(x_1, ..., x_q) = l(x_1 \otimes ... \otimes x_q) for any x_1, ..., x_q \in M. We have

(-1)^q q! h(x_1, ..., x_q) = (-1)^q \sum_{\sigma \in S_q} h(x_{\sigma(1)}, ..., x_{\sigma(q)}) = = (-1)^q l(s(x_1 \otimes ... \otimes x_q)) = \sum_{H \subset \{1, ..., q\}} (-1)^{\text{Card } H} l \left( \gamma_q \left( \sum_{i \in H} x_i \right) \right)

by Prop. 3, (v) (IV, p. 45). Now

l \left( \gamma_q \left( \sum_{i \in H} x_i \right) \right) = h \left( \sum_{i \in H} x_i, ..., \sum_{i \in H} x_i \right) = f \left( \sum_{i \in H} x_i \right)

and this proves formula (16) and the uniqueness of h.

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

Cited by

Proposition 2, § 12

Cites

Prop. 3, (v) (IV, p. 45