Proposition 7

Tag 06D4, permanent. Cite it as [Bourbaki, Tag 06D4].

Algebra, IV, p. 50, § 5. Symmetric tensors and polynomial mappings

Let $M$ be an $A$ -module, $u$ the mapping $(x, y) \mapsto x + y$ of $M \oplus M$ into $M$ and $f$ the composite mapping

\mathbf{TS}(M) \otimes \mathbf{TS}(M) \xrightarrow{h} \mathbf{TS}(M \oplus M) \xrightarrow{\mathbf{TS}(u)} \mathbf{TS}(M)

where $h$ is the canonical homomorphism. If $z, z' \in \mathbf{TS}(M)$ , then $f(z \otimes z') = zz'$ .

For let $i$ be the mapping $x \mapsto (x, 0)$ of $M$ into $M \oplus M$ . We have $u \circ i = \mathrm{Id}_M$ , hence $\mathbf{TS}(u) \circ \mathbf{TS}(i) = \mathrm{Id}_{\mathbf{TS}(M)}$ ; therefore

f(z \otimes 1) = \mathbf{TS}(u)(h(z \otimes 1)) = \mathbf{TS}(u)(\mathbf{TS}(i)(z)) = z .

Likewise $f(1 \otimes z') = z'$ , whence $f(z \otimes z') = f(z \otimes 1) f(1 \otimes z') = zz'$ .

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

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