Proposition 14

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

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

— Let $M, N, P$ be $A$ -modules, $q$ and $r$ integers $\geqslant 0$ , and assume that $M$ and $N$ are free. If $f \in \mathrm{Pol}^q(M, N)$ , $f' \in \mathrm{Pol}^r(N, P)$ , then $f' \circ f \in \mathrm{Pol}^{qr}(M, P)$ .

There exists a $q$ -linear mapping $g$ of $M^q$ into $N$ and an $r$ -linear mapping $g'$ of $N^r$ into $P$ such that

\begin{aligned} f(x) &= g(x, x, \ldots, x) & \text{for all } x \in M, \\ f'(y) &= g'(y, y, \ldots, y) & \text{for all } y \in N. \end{aligned}

Hence for every $x \in M$ we have

f'(f(x)) = g'(f(x), f(x), \ldots, f(x)) = g'(g(x, x, \ldots, x), \ldots, g(x, x, \ldots, x))

and the mapping $(x_1, \ldots, x_{qr}) \mapsto g'(g(x_1, \ldots, x_q), \ldots, g(x_{q(r-1)+1}, \ldots, x_{qr}))$ of $M^{qr}$ into $P$ is $qr$ -linear.

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

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Cites

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