Bourbaki

Remark

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

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

— 2) The composite homomorphism A,=u∘0:S(M∗)→PolA(M,A)A, = u \circ 0 : S(M^*) \to \mathrm{Pol}_A(M, A) is the unique unital homomorphism of algebras inducing the inclusion of

M∗=Pol1(M,A) M^* = \mathrm{Pol}^1(M, A)

in Pol(M,A)\mathrm{Pol}(M, A). If MM is finitely generated free and AA is an infinite integral domain, then λM\lambda_M is bijective (Prop. 20 and Cor. of Prop. 19). In particular if AA is an infinite integral domain, then the canonical homomorphism f↦f~f \mapsto \tilde{f} of A[X1,…,Xn]A[X_1, \ldots, X_n] into Pol(An,A)\mathrm{Pol}(A^n, A) (IV, p. 4) is an isomorphism.

3) Consider the coproduct cS:S(M∗)→S(M∗×M∗)c_S : S(M^*) \to S(M^* \times M^*) (III, p. 575, Example 6). For any v∈S(M∗)v \in S(M^*), x,y∈Mx, y \in M, the polynomial mapping λM×M(cS(v)):M×M→A\lambda_{M \times M}(c_S(v)) : M \times M \to A maps (x,y)(x, y) to A,(v)(x+y)A, (v)(x + y). For the two algebra homomorphisms of S(M∗)S(M^*) into Map(M×M,A)\mathrm{Map}(M \times M, A) defined in this way agree on M∗M^*, in virtue of the relation

(v⊗1+1⊗v)(x,y)=v(x+y)(v∈M∗) (v \otimes 1 + 1 \otimes v)(x, y) = v(x + y) \quad (v \in M^*)

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, 2003, Springer. Not translated.