Bourbaki

Remark

Tag 0629, permanent. Cite it as [Bourbaki, Tag 0629].

Algebra, chapter III, § 6. Symmetric algebras

(1) Suppose that EE is a graded AA-algebra of type Z\mathbf{Z}, with graduation (En)(E_n), and suppose also that the linear mapping ff (assumed to satisfy (1)) is such that (2)

f(M)⊂E1. f(M) \subset E_1.
Then the relation g(x1x2…xp)=f(x1)f(x2)…f(xp)g(x_1 x_2 \ldots x_p) = f(x_1) f(x_2) \ldots f(x_p) with the xi∈Mx_i \in M shows that g(Sp(M))⊂Epg(S^p(M)) \subset E_p for all p≥0p \geq 0 and hence gg is a graded algebra homomorphism.

(2) Every element of S(M)S(M) is a sum of products of the form x1x2…xnx_1 x_2 \ldots x_n, where the xix_i belong to MM; care should be taken not to confuse such products taken in S(M)S(M) with the analogous products taken in T(M)T(M).

(3) If n!.1n!.1 is invertible in AA, the AA-module Sn(M)S^n(M) is generated by the elements of the form xnx^n, where x∈Mx \in M; this follows from the above remark and I, § 8, no. 2, Proposition 2.

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

Cited by

Nothing.

Cites

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