Bourbaki

Proposition 4

Tag 062B, permanent. Cite it as [Bourbaki, Tag 062B].

Algebra, chapter III, § 6. Symmetric algebras

If u:M→Nu: M \to N is a surjective AA-linear mapping, the homomorphism S(u):S(M)→S(N)S(u): S(M) \to S(N) is surjective and its kernel is the ideal of S(M)S(M) generated by the kernel P⊂M⊂S(M)P \subset M \subset S(M) of uu.

We write v=T(u):T(M)→T(N)v = T(u): T(M) \to T(N); we know (§ 5, no. 2, Proposition 3) that vv is surjective and hence it follows from the definitions that v(JM′)=JN′v(\mathfrak{J}'_M) = \mathfrak{J}'_N; if R\mathfrak{R} is the kernel of vv, then v−1(JN′)=R+JM′v^{-1}(\mathfrak{J}'_N) = \mathfrak{R} + \mathfrak{J}'_M. As S(u):T(M)/JM′→T(N)/JN′S(u): T(M)/\mathfrak{J}'_M \to T(N)/\mathfrak{J}'_N is derived from vv by passing to the quotients, it is a surjective homomorphism whose kernel is R′=(R+JM′)/JM′\mathfrak{R}' = (\mathfrak{R} + \mathfrak{J}'_M)/\mathfrak{J}'_M. As R\mathfrak{R} is generated by the kernel PP of uu (§ 5, no. 2), so is R′\mathfrak{R}'.

If u:M→Nu: M \to N is an injective linear mapping, it is not always true that S(u)S(u) is an injective mapping (Exercise 1). However it is so when uu is an injection such that u(M)u(M) is a direct factor in NN and then the image of S(u)S(u) (isomorphic to S(M)S(M)) is a direct factor of S(N)S(N); the proof is the same as that for the analogous assertions for T(u)T(u) (§ 5, no. 2) replacing TT by SS.

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

Cited by

Cites

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