Bourbaki

Proposition 3

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

Algebra, IV, p. 45, § 5. Symmetric tensors and polynomial mappings

(i) If x∈Mx \in M, the pth power of xx, calculated in TS(M)\mathbf{TS}(M), is equal to p!  γp(x)p! \; \gamma_p(x). (ii) Let x1,…,xn∈Mx_1, \ldots, x_n \in M; then

γp(x1+x2+⋯+xn)=∑p1+p2+⋯+pn=pγp1(x1)  γp2(x2)…γpn(xn). \gamma_p(x_1 + x_2 + \cdots + x_n) = \sum_{p_1 + p_2 + \cdots + p_n = p} \gamma_{p_1}(x_1) \; \gamma_{p_2}(x_2) \ldots \gamma_{p_n}(x_n) .
(iii) Let x1,…,xn∈Mx_1, \ldots, x_n \in M, let p1,…,pnp_1, \ldots, p_n be integers ⩾0\geqslant 0 and p=p1+…+pnp = p_1 + \ldots + p_n. Let EE be the set of mappings φ\varphi of {1,…,p}\{1, \ldots, p\} into (1,…,n)(1, \ldots, n) such that
Card⁡φ−1(1)=p1,…,Card⁡φ−1(n)=pn. \operatorname{Card} \varphi^{-1}(1) = p_1, \ldots, \operatorname{Card} \varphi^{-1}(n) = p_n .
Then
γp1(x1)  γp2(x2)…γpn(xn)=∑φ∈Exφ(1)⊗xφ(2)⊗…⊗xφ(p). \gamma_{p_1}(x_1) \; \gamma_{p_2}(x_2) \ldots \gamma_{p_n}(x_n) = \sum_{\varphi \in E} x_{\varphi(1)} \otimes x_{\varphi(2)} \otimes \ldots \otimes x_{\varphi(p)} .
(iv) Let x∈Mx \in M and let q,rq, r be integers ⩾0\geqslant 0. Then
γq(x)  γr(x)=(q+r)!q!  r!  γq+r(x). \gamma_q(x) \; \gamma_r(x) = \frac{(q + r)!}{q! \; r!} \; \gamma_{q + r}(x) .
(v) Let x1,…,x,∈Mx_1, \ldots, x, \in M, and for H⊂(1,…,n)H \subset (1, \ldots, n) put xH=∑i∈Hxix_H = \sum_{i \in H} x_i, then
(−1)nx1x2…xn=∑H⊂{1,…,n}(−1)Card⁡Hγn(xH). (-1)^n x_1 x_2 \ldots x_n = \sum_{H \subset \{1, \ldots, n\}} (-1)^{\operatorname{Card} H} \gamma_n(x_H) .

The assertion (i) follows at once from Prop. 2 (ii).

Let us prove (ii); by an induction on n we see that it is enough to consider the case n=2n = 2. Then we have

γp(x1+x2)=(x1+x2)⊗(x1+x2)⊗...⊗(x1+x2)@factors) \gamma_p(x_1 + x_2) = (x_1 + x_2) \otimes (x_1 + x_2) \otimes ... \otimes (x_1 + x_2) \quad @\text{factors})
=∑p1+p2=p∑σ∈Sp1,p2σ(x1⊗x1⊗...⊗x1⊗x2⊗x2⊗...⊗x2) = \sum_{p_1 + p_2 = p} \sum_{\sigma \in S_{p_1, p_2}} \sigma(x_1 \otimes x_1 \otimes ... \otimes x_1 \otimes x_2 \otimes x_2 \otimes ... \otimes x_2)
=∑p1+p2=p∑σ∈Sp1,p2σ(γp1(x1)⊗γp2(x2)) = \sum_{p_1 + p_2 = p} \sum_{\sigma \in S_{p_1, p_2}} \sigma(\gamma_{p_1}(x_1) \otimes \gamma_{p_2}(x_2))
=∑p1+p2=pγp1(x1)γp2(x2). = \sum_{p_1 + p_2 = p} \gamma_{p_1}(x_1) \gamma_{p_2}(x_2).

To prove (iii), let Sp1,...,pnS_{p_1, ..., p_n} be the set of permutations of (1,p1+...+pn)(1, p_1 + ... + p_n) whose restrictions to the intervals

(1,p1),(p1+1,p1+p2),...,(p1+...+pn−1+1,p1+...+pn)(1, p_1), (p_1 + 1, p_1 + p_2), ..., (p_1 + ... + p_{n-1} + 1, p_1 + ... + p_n)
are increasing. By I, p. 60, example 2 and Prop. 2, (ii) we have
γp1(x1)γp2(x2)...γpn(xn)=∑ρ∈Sp1,p2,...,pnρ(x1⊗x1⊗...⊗x1⊗x2⊗x2⊗...⊗x2⊗...⊗xn⊗xn⊗...⊗xn) \gamma_{p_1}(x_1) \gamma_{p_2}(x_2) ... \gamma_{p_n}(x_n) = \sum_{\rho \in S_{p_1, p_2, ..., p_n}} \rho(x_1 \otimes x_1 \otimes ... \otimes x_1 \otimes x_2 \otimes x_2 \otimes ... \otimes x_2 \otimes ... \otimes x_n \otimes x_n \otimes ... \otimes x_n)
(with pip_i factors xix_i) and this sum is equal to
∑φ∈Exφ(1)⊗xφ(2)⊗...⊗xφ(p). \sum_{\varphi \in E} x_{\varphi(1)} \otimes x_{\varphi(2)} \otimes ... \otimes x_{\varphi(p)}.

In (iii) let us put n=2,x1=x2=x,p1=qn = 2, x_1 = x_2 = x, p_1 = q and p2=rp_2 = r, then we obtain (iv) (I, loc. cit.).

Finally (v) follows from Prop. 2, (ii) and Prop. 2 of I, p. 100, applied to the elements xix_i of the ring T(M)\mathbf{T}(M).

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

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Transcribed from the printed edition, 2003, Springer. Not translated.