Proposition 1

Tag 05H3, permanent. Cite it as [Bourbaki, Tag 05H3].

Algebra, chapter I, § 8. Rings

Let A be a commutative ring and $(x_\lambda)_{\lambda \in L}$ a finite family of elements of A. For every family of positive integers $\beta = (\beta_\lambda)_{\lambda \in L}$ , let $|\beta| = \sum_{\lambda \in L} \beta_\lambda$ . Then

\left( \sum_{\lambda \in L} x_\lambda \right)^n = \sum_{|\beta|=n} \frac{n!}{\prod_{\lambda \in L} \beta_\lambda!} \prod_{\lambda \in L} x_\lambda^{\beta_\lambda}.

We apply formula (13) with

L_i = L

and

x_{i,\lambda} = x_\lambda

for

1 \leq i \leq n

. Then

\left( \sum_{\lambda \in L} x_\lambda \right)^n = \sum_{\alpha_1, \ldots, \alpha_n} x_{\alpha_1} \cdots x_{\alpha_n},

the sum extending over all sequences

\alpha = (\alpha_1, \ldots, \alpha_n) \in L^n

Let $\alpha$ be in $L^n$ ; for all $\lambda \in L$ let $U_\lambda^\alpha$ denote the set of integers $i$ such that $1 \leq i \leq n$ and $\alpha_i = \lambda$ and let $\Phi(\alpha) = (U_\lambda^\alpha)_{\lambda \in L}$ . It is immediate that $\Phi$ is a bijection of $L^n$ onto the set of partitions of $\{1, 2, \ldots, n\}$ indexed by $L$ . For all $\beta \in \mathbf{N}^L$ such that $|\beta| = n$ , let $L_\beta^n$ denote the set of $\alpha \in L^n$ such that $\mathrm{Card}\ U_\beta^\alpha = \beta_\lambda$ for all $\lambda \in L$ . It follows that the family $(L_\beta^n)_{|\beta|=n}$ is a partition of $L^n$ and that

\mathrm{Card}\ L_\beta^n = \frac{n!}{\prod_{\lambda \in L} \beta_\lambda!}

(§ 5, no. 5).

Finally, for $\alpha \in L_\beta^n$ ,

x_{\alpha_1} \cdots x_{\alpha_n} = \prod_{\lambda \in L} \prod_{i \in U_\lambda^\alpha} x_{\alpha_i} = \prod_{\lambda \in L} \prod_{i \in U_\lambda^\alpha} x_\lambda = \prod_{\lambda \in L} x_\lambda,

whence

\sum_{(\alpha_1, \ldots, \alpha_n)} x_{\alpha_1} \cdots x_{\alpha_n} = \sum_{|\beta|=n} \sum_{\alpha \in L_\beta^n} x_{\alpha_1} \cdots x_{\alpha_n} = \sum_{|\beta|=n} \sum_{\alpha \in L_\beta^n} \prod_{\lambda \in L} x_\lambda^{\beta_\lambda} = \sum_{|\beta|=n} \frac{n!}{\prod_{\lambda \in L} \beta_\lambda!} x_\lambda^{\beta_\lambda}

and formula (14) thus follows from (15).

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

Cited by

Nothing.

Cites

§ 5, no. 5