Bourbaki

Proposition 7

Tag 04DC, permanent. Cite it as [Bourbaki, Tag 04DC].

Commutative Algebra, chapter III, § 1. Finitely generated graded algebras

Let pp be a graded prime ideal of AA, dd an integer ≥1\geq 1 and p′p' the graded prime ideal p∩A(d)p \cap A^{(d)} of A(d)A^{(d)}; for every graded AA-module MM, the homomorphism (M(d))(p′)→M(p)(M^{(d)})_{(p')} \to M_{(p)} derived from the canonical injection M(d)→MM^{(d)} \to M is bijective.

If SS is the set of homogeneous elements of AA not belonging to pp and S′=S∩A(d)S' = S \cap A^{(d)}, the canonical homomorphism ϕ:S′−1M(d)→S−1M\phi : {S'}^{-1}M^{(d)} \to S^{-1}M is a homogeneous homomorphism of degree 0 and it is injective, for, if x∈Mndx \in M_{nd} satisfies sx=0sx = 0 for s∈An,s∉ps \in A_n, s \notin p, then also sdx=0s^dx = 0 and sd∈Amd,sd∉p′s^d \in A_{md}, s^d \notin p'. It remains to show that the image under ϕ\phi of (M(d))(p′)(M^{(d)})_{(p')} is the whole of M(p)M_{(p)}; but if x∈Mn,s∈Anx \in M_n, s \in A_n and s∉ps \notin p, then also x/s=(xsd−1)/sdx/s = (xs^{d-1})/s^d where xsd−1∈And,sd∈Anxs^{d-1} \in A_{nd}, s^d \in A_n and sd∉p′s^d \notin p', whence our assertion.

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 volume. Not translated.