Bourbaki

Exercise 13, § 4

Commutative Algebra, IX, p. 80

Let AA be a noetherian integral ring, KK its field of fractions, A‾\overline{A} its integral closure, ff a nonzero element of A, P\mathfrak{P} a prime ideal of height 1 of A‾\overline{A} containing ff, pp the prime ideal

1^1 For every ring B, we denote by BredB_{\text{red}} the quotient of B by the ideal of nilpotent elements of B.

A∩PA \cap \mathfrak{P} of A. Prove that one has p∈Ass⁡A(A/fA)p \in \operatorname{Ass}_A(A/fA) by reducing to the case where A is local, with maximal ideal pp, and by establishing successively under this assumption the following assertions: a) P\mathfrak{P} is divisoriel (use Exer. 10, c)). b) pp is divisoriel (use Exer. 12, f)). c) If a1,...,ara_1, ..., a_r are nonzero elements of K such that

A:p=A+∑i=1rAai,one has p=⋂i=1r(A∩ai−1A) A : p = A + \sum_{i=1}^r Aa_i, \quad \text{one has } p = \bigcap_{i=1}^r (A \cap a_i^{-1}A)
and there exists an index ii such that p=A∩ai−1Ap = A \cap a_i^{-1}A. d) There exists a nonzero element gg of A such that p∈Ass⁡A(A/gA)p \in \operatorname{Ass}_A(A/gA). e) p∈Ass⁡A(A/fA)p \in \operatorname{Ass}_A(A/fA).

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.