Theorem 3

Tag 055Z, permanent. Cite it as [Bourbaki, Tag 055Z].

Algebra, chapter I, § 1. Laws of composition; associativity; commutativity

Let T be an associative law on E and $(x_\alpha)_{\alpha \in A}$ a non-empty finite family of elements of E which are pairwise permutable. If A is a union of non-empty subsets $(B_i)_{i \in I}$ which are pairwise disjoint, then

\prod_{\alpha \in A} x_\alpha = \prod_{i \in I} \left( \prod_{\alpha \in B_i} x_\alpha \right).

This follows from Theorem 2 if $A$ and $I$ are totally ordered so that the $B_i$ satisfy the conditions of Theorem 1.

We single out two important special cases of this theorem:

1. If $(x_{\alpha \beta})_{(\alpha, \beta) \in A \times B}$ is a finite family of permutable elements of an associative magma whose indexing set is the product of two non-empty finite sets $A, B$ (a "double family"), then

\prod_{(\alpha, \beta) \in A \times B} x_{\alpha \beta} = \prod_{\alpha \in A} \left( \prod_{\beta \in B} x_{\alpha \beta} \right) = \prod_{\beta \in B} \left( \prod_{\alpha \in A} x_{\alpha \beta} \right)

as follows from Theorem 3 by considering $A \times B$ as the union of the sets $\{\alpha\} \times B$ on the one hand and of the sets $A \times \{\beta\}$ on the other.

For example, if $B$ has $n$ elements and for each $\alpha \in A$ all the $x_{\alpha \beta}$ have the same value $x_\alpha$ , then

\prod_{\alpha \in A} \left( \prod^n_{\beta=1} x_\alpha \right) = \prod^n_{\alpha=1} \left( \prod_{\alpha \in A} x_\alpha \right).

If $B$ has two elements, we obtain the following results: let $(x_\alpha)_{\alpha \in A}, (y_\alpha)_{\alpha \in A}$ be two non-empty families of elements of $E$ . If the $x_\alpha$ and the $y_\beta$ are pairwise permutable, then

\prod_{\alpha \in A} x_\alpha \prod_{\alpha \in A} y_\alpha = \left( \prod_{\alpha \in A} x_\alpha \right) \prod_{\alpha \in A} y_\alpha.

Because of formula (7) the composition of a double sequence $(x_{ij})$ whose indexing set is the product of two intervals $\{p, q\}$ and $\{r, s\}$ in $\mathbf{N}$ is often denoted for a commutative associative law written additively by

\sum_{i=p}^q \sum_{j=r}^s x_{ij} \quad \text{or} \quad \sum_{j=r}^s \sum_{i=p}^q x_{ij}

and similarly for laws denoted by other symbols.

2. Let $n$ be an integer $> 0$ and let $A$ be the set of ordered pairs of integers $(i, j)$ such that $0 \leq i \leq n, 0 \leq j \leq n$ and $i < j$ ; the composition of a family $(x_{ij})_{(i, j) \in A}$ (under a commutative associative law) is also denoted by $\prod_{0 \leq i < j \leq n} x_{ij}$ (or simply $\prod_{i < j} x_{ij}$ , if no confusion arises); Theorem 3 here gives the formulae

\prod_{0 \leq i < j \leq n} x_{ij} = \prod_{i=0}^{n-1} \left( \prod_{j=i+1}^n x_{ij} \right) = \prod_{j=1}^n \left( \prod_{i=0}^{j-1} x_{ij} \right).

There are analogous formulae to (7) for a family whose indexing set is the product of more than two sets and analogous formulae to (10) for a family whose indexing set is the set $S_p$ of strictly increasing sequences $(i_k)_{1 \leq k \leq p}$ of $p$ integers such that $0 \leq i_k \leq n$ ( $p \leq n + 1$ ): in the latter case the composition of the family $(x_{i_1 i_2 \ldots i_p})_{(i_1, \ldots, i_p) \in S_p}$ is denoted by

\prod_{0 \leq i_1 < i_2 < \cdots < i_p < n} x_{i_1 i_2 \ldots i_p} \quad \text{or simply} \quad \prod_{i_1 < i_2 < \cdots < i_p} x_{i_1 i_2 \ldots i_p}.

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

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Definition 3, § 2

Theorem 3

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