Bourbaki

Proposition 16

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

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

Let AA be a ring, RR an AA-algebra, A′A' the integral closure of AA in RR and SS a multiplicative subset of AA. Then the integral closure of S−1AS^{-1}A in S−1RS^{-1}R is S−1A′S^{-1}A'.

Let b/sb/s be an element of S−1A′S^{-1}A' (s∈S, b∈A′s \in S,\ b \in A'). Since the diagram

A→iASS−1A↓h↓S−1hR→iRSS−1R \begin{array}{ccc} A & \xrightarrow{i_A^S} & S^{-1}A \\ \downarrow h & & \downarrow S^{-1}h \\ R & \xrightarrow{i_R^S} & S^{-1}R \end{array}

is commutative, b/1b/1 is integral over S−1AS^{-1}A (no. 1, Proposition 2). As 1/s∈S−1A1/s \in S^{-1}A, b/s=(b/1)(1/s)b/s = (b/1)(1/s) is integral over S−1AS^{-1}A.

Conversely, let r/t (r∈R, t∈S)r/t \ (r \in R,\ t \in S) be an element of S−1RS^{-1}R which is integral over S−1AS^{-1}A; then r/1=(t/1)(r/t)r/1 = (t/1)(r/t) is integral over S−1AS^{-1}A. Then there is a relation of the form

(r/1)n+(a1/s)(r/1)n−1+⋯+(an/s)=0, (r/1)^n + (a_1/s)(r/1)^{n-1} + \cdots + (a_n/s) = 0,

where ai∈A (1≤i≤n)a_i \in A \ (1 \leq i \leq n) and s∈Ss \in S. This relation may also be written

(srn+a1rn−1+…+an)/s=0 (sr^n + a_1 r^{n-1} + \ldots + a_n)/s = 0

and therefore there exists s′∈Ss' \in S such that s′n(srn+a1rn−1+…+an)=0{s'}^n(sr^n + a_1 r^{n-1} + \ldots + a_n) = 0; we deduce that (s′sr)n+s′a1(s′sr)n−1+…+s′nsn−1an=0(s'sr)^n + s'a_1(s'sr)^{n-1} + \ldots + {s'}^ns^{n-1}a_n = 0. By definition therefore s′sr∈A′s'sr \in A', whence r/1∈S−1A′r/1 \in S^{-1}A' and r/t∈S−1A′r/t \in S^{-1}A'.

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.