Bourbaki

Proposition 16

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

Algebra, chapter IV, § 5. Symmetric tensors and polynomial mappings

— *Let M be a free A-module, N an A-module, q a positive integer and u the canonical homomorphism of Hom (TSq(M)\mathbf{TS}^q(M), N) into PolY(M,N)\mathbf{Pol}^Y(M, N).

(i) If A is an infinite integral domain and N is torsion-free, then u is an isomorphism.

(ii) If the mapping y↦q!yy \mapsto q!y in N is injective, u is an isomorphism.

In the two cases of the proposition we have to prove that u is injective, that is, every linear mapping f of TSq(M)\mathbf{TS}^q(M) into N which is zero on γq(M)\gamma_q(M), vanishes.

Assume A to be an infinite integral domain and N torsion-free. For every z∈TSq(M)z \in \mathbf{TS}^q(M) there exists a∈A−{0}a \in A - \{0\} such that azaz is an A-linear combination of elements of y, (M) (IV, p. 47, Prop. 5). Hence af(z)=f(az)=0af(z) = f(az) = 0, and so f(z)=0f(z) = 0.

Suppose next that the mapping y↦q!yy \mapsto q!y in N is injective, then by IV, p. 45, Prop. 3, (v), f vanishes on s . T^q(M). Hence if z∈TSq(M)z \in \mathbf{TS}^q(M), we have q!f(z)=f(sz)=0q!f(z) = f(sz) = 0, and so f(z)=0f(z) = 0.

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.