Bourbaki

Exercise 8, § 7

Commutative Algebra, VIII, p. 104

Let AA be a ring and aa an ideal of AA. Put ai=Aa^i = A for i≤0i \leq 0 and consider the AA-algebra A(a)=∑i∈Za−ivi⊂A[v,v−1]\mathscr{A}(a) = \sum_{i \in \mathbf{Z}} a^{-i} v^i \subset A[v, v^{-1}] and the AA-algebra P(a)=∑i≥0aivi⊂A[v]P(a) = \sum_{i \geq 0} a^i v^i \subset A[v]. These are two graded AA-algebras having the same total ring of fractions. Now prove that for an element a∈Aa \in A the following conditions are equivalent:

(i) The element avav of the A(a)\mathscr{A}(a)-algebra A[v,v−1]A[v, v^{-1}] is integer over A(a)\mathscr{A}(a).

(ii) The element avav of the P(a)P(a)-algebra A[v]A[v] is integer over P(a)P(a).

(iii) There exists an integral dependence relation:

ak+b1ak−1+⋯+bk=0withbj∈aj. a^k + b_1 a^{k-1} + \cdots + b_k = 0 \quad \text{with} \quad b_j \in a^j.

One will say that aa is integer over the ideal aa.

Solution

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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.