Bourbaki

Proposition 21

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

Commutative Algebra, chapter V, § 1. Notion of an integral element

Let A=⨁i∈ZAi\mathbf{A} = \bigoplus_{i \in \mathbf{Z}} \mathbf{A}_i be a graded integral domain and SS the set of homogeneous elements ≠0\neq 0 of A\mathbf{A}. The integral closure A′\mathbf{A}' of A\mathbf{A} is then a graded subring of S−1AS^{-1}\mathbf{A}. If further Ai=0\mathbf{A}_i = 0 for i<0i < 0, then Ai′=0\mathbf{A}'_i = 0 for i<0i < 0.

If A=A0\mathbf{A} = \mathbf{A}_0, the proposition is trivial. Otherwise we may apply Lemma 4; the ring S−1AS^{-1}\mathbf{A} is an integrally closed domain and therefore A⊂S−1A\mathbf{A} \subset S^{-1}\mathbf{A}; as S−1AS^{-1}\mathbf{A} is graded, so is A′\mathbf{A}' by Proposition 20; the latter assertion also follows from Proposition 20.

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

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