Bourbaki

Proposition 3

Tag 04N4, permanent. Cite it as [Bourbaki, Tag 04N4].

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

Let (Ri)1≤i≤n(R_i)_{1 \leq i \leq n} be a finite family of (not necessarily commutative) AA-algebras and let R=∏i=1nRiR = \prod_{i=1}^n R_i be their product. For an element x=(xi)1≤i≤nx = (x_i)_{1 \leq i \leq n} of RR to be integral over AA, it is necessary and sufficient that each of the xix_i be integral over AA. For RR to be integral over AA, it is necessary and sufficient that each of the RiR_i be integral over AA.

It is obviously sufficient to prove the first assertion. The condition is necessary by Proposition 2. Conversely, if each of the xix_i is integral over AA, the subalgebra A[xi]A[x_i] of RiR_i is a finitely generated AA-module and hence so is the subalgebra ∏i=1nA[xi]\prod_{i=1}^n A[x_i] of RR; as A[x]A[x] is contained in this subalgebra, xx is integral over AA by Proposition 1.

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.