Bourbaki

Proposition 1

Tag 0582, permanent. Cite it as [Bourbaki, Tag 0582].

Algebra, chapter I, § 3. Actions

Let E1,E2,…,EnE_1, E_2, \ldots, E_n and F be commutative monoids written additively and let u be a mapping of E1×⋯×EnE_1 \times \cdots \times E_n into F, which is distributive with respect to all the variables. For i=1,2,…,ni = 1, 2, \ldots, n, let LiL_i be a non-empty finite set and (xi,λ)λ∈Li(x_{i,\lambda})_{\lambda \in L_i} a family of elements of EiE_i. Let yi=∑λ∈Lixi,λy_i = \sum_{\lambda \in L_i} x_{i,\lambda} for i=1,2,…,ni = 1, 2, \ldots, n. Then

(2)

u(y1,…,yn)=∑αu(x1,α1,…,xn,αn) u(y_1, \ldots, y_n) = \sum_\alpha u(x_1, \alpha_1, \ldots, x_n, \alpha_n)

the sum being taken over all sequences α=(α1,…,αn)\alpha = (\alpha_1, \ldots, \alpha_n) belonging to L1×⋯×LnL_1 \times \cdots \times L_n.

We argue by induction on n, the case n=1n = 1 following from formula (2) of § 1, no. 2. From the same reference

(3)

u(y1,…,yn−1,yn)=∑αn∈Lnu(y1,…,yn−1,xn,αn) u(y_1, \ldots, y_{n-1}, y_n) = \sum_{\alpha_n \in L_n} u(y_1, \ldots, y_{n-1}, x_n, \alpha_n)

for yn=∑αn∈Lnxn,αny_n = \sum_{\alpha_n \in L_n} x_{n,\alpha_n} and the mapping z↦u(y1,…,yn−1,z)z \mapsto u(y_1, \ldots, y_{n-1}, z) of EnE_n into FF is a magma homomorphism. By the induction hypothesis applied to the distributive mappings (z1,…,zn−1)↦u(z1,…,zn−1,xn,αn)(z_1, \ldots, z_{n-1}) \mapsto u(z_1, \ldots, z_{n-1}, x_n, \alpha_n) from E1×⋯×En−1E_1 \times \cdots \times E_{n-1} to FF,

(4)

u(y1,…,yn−1,xn,αn)=∑α1,…,αn−1u(x1,α1,…,xn−1,αn−1,xn,αn), u(y_1, \ldots, y_{n-1}, x_n, \alpha_n) = \sum_{\alpha_1, \ldots, \alpha_{n-1}} u(x_1, \alpha_1, \ldots, x_{n-1}, \alpha_{n-1}, x_n, \alpha_n),
the sum being taken over the sequences (α1,…,αn−1)(\alpha_1, \ldots, \alpha_{n-1}) belonging to M=L1×⋯×Ln−1M = L_1 \times \cdots \times L_{n-1}. Now L1×⋯×Ln=M×LnL_1 \times \cdots \times L_n = M \times L_n; writing
tα1,…,αn=u(x1,α1,…,xn,αn), t_{\alpha_1, \ldots, \alpha_n} = u(x_1, \alpha_1, \ldots, x_n, \alpha_n),
we have (5)
∑α1,…,αntα1,…,αn=∑αn(∑α1,…,αn−1tα1,…,αn−1,αn) \sum_{\alpha_1, \ldots, \alpha_n} t_{\alpha_1, \ldots, \alpha_n} = \sum_{\alpha_n} \left( \sum_{\alpha_1, \ldots, \alpha_{n-1}} t_{\alpha_1, \ldots, \alpha_{n-1}, \alpha_n} \right)
by formula (7) of § 1, no. 5. (2) follows immediately from (3), (4) and (5).

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

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Cites

Transcribed from the printed edition, 1998, Springer. Not translated.