Bourbaki

Exercise 9, § 7

Commutative Algebra, VIII

(a) Let A and a be as above and let b be an ideal of A containing a. Now prove that the following conditions are equivalent: (i) Every element of b is integral over a. (ii) The A(a)\mathscr{A}(a)-graded algebra A(b)\mathscr{A}(b) is integral. (iii) The P(a)-graded algebra P(b) is integral. (iv) There exists an integer k⩾0k \geqslant 0 such that a.bk=bk+1a.b^k = b^{k+1} and hence an.bk=bk+na^n.b^k = b^{k+n} for all positive integers n. We say that b is integral over a. (b) Now prove that there exists a greatest ideal among those which are integral over a. It is denoted a‾\overline{a} and is called the integral closure of a in A. Verify that if a⊂ba \subset b then a‾⊂b‾\overline{a} \subset \overline{b} and that a‾=a‾\overline{a} = \overline{a}. Now prove that if the integral closure P(a)‾\overline{P(a)} is a P(a)-finitely generated module, then a.ak=ak+1a.a^k = a^{k+1} for k sufficiently large. (c) Now prove that if a∈Aa \in A satisfies the integral dependence relation

ak+b1ak−1+⋯+bk=0withk⩾1andbi∈ai, a^k + b_1 a^{k-1} + \cdots + b_k = 0 \quad \text{with} \quad k \geqslant 1 \quad \text{and} \quad b_i \in a^i,
then a.(a+Aa)k−1=(a+Aa)ka.(a + Aa)^{k-1} = (a + Aa)^k. Deduce that the integral closure of a in A coincides with the set of elements of A which are integral over a.

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.