Lemma 2

Tag 060W, permanent. Cite it as [Bourbaki, Tag 060W].

Algebra, chapter III, § 4. Tensor products of algebras

Let $A$ be a commutative ring, $E$ an $A$ -algebra and $(x_i)_{1 \leq i \leq n}$ and $(y_i)_{1 \leq i \leq n}$ two finite sequences of $n$ elements of $E$ ; suppose that for $1 \leq j \leq i \leq n$ ,

x_i y_j = \lambda_{ij} y_j x_i \quad \text{with } \lambda_{ij} \in A.

Then

(x_1 x_2 \ldots x_n)(y_1 y_2 \ldots y_n) = \left( \prod_{i > j} \lambda_{ij} \right) (x_1 y_1)(x_2 y_2) \ldots (x_n y_n).

The lemma being trivial for $n = 1$ , we again argue by induction on $n$ for $n \geq 2$ . By virtue of Lemma 1,

(x_1 \ldots x_n)(y_1 \ldots y_n) = x_1 (x_2 \ldots x_n) y_1 (y_2 \ldots y_n) = \left( \prod_{i > 1} \lambda_{i1} \right) (x_1 y_1)(x_2 \ldots x_n)(y_2 \ldots y_n)

and it then suffices to apply the induction hypothesis to obtain (16).

For every family $\lambda = (\lambda_{ij})$ of elements of $\mathbf{A}$ , with $1 \leq j < i \leq n$ , and for every permutation $\sigma \in \mathfrak{S}_n$ , we write

\varepsilon_\sigma(\lambda) = \prod_{i > j,\ \sigma^{-1}(i) < \sigma^{-1}(j)} \lambda_{ij} = \prod_{i < j,\ \sigma(i) > \sigma(j)} \lambda_{\sigma(i),\ \sigma(j)}.

Observe that, when $\mathbf{A} = \mathbf{Z}$ and $\lambda_{ij} = -1$ for every ordered pair $(i, j)$ such that $1 \leq j < i \leq n$ , $\varepsilon_\sigma(\lambda)$ is just the signature $\varepsilon_\sigma$ of the permutation $\sigma$ (I, § 5, no. 7).

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

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Proposition 10, § 4

Lemma 2

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