Bourbaki

Exercise 12, § 7

Commutative Algebra, X, p. 172

Let AA be a noetherian local ring. In order that AA be a Grothendieck ring, it is necessary and sufficient that, for every finite integral AA-algebra BB, every maximal ideal mm of BB and every prime ideal qq of the completed ring Bm^\widehat{B_m} satisfying Bm∩q=(0)B_m \cap q = (0), the local ring (Bm^)q(\widehat{B_m})_q be regular (to show that this condition is necessary, it suffices, thanks to the preceding exercise, to show that, for every prime ideal p\mathfrak{p} of AA, and every finite extension KK of κ(p)\kappa(\mathfrak{p}), the ring K⊗AA^K \otimes_A \widehat{A} is regular; one may introduce a finite integral AA-algebra BB with field of fractions K, denote by m_1, ..., m_r its maximal ideals, and interpret the local rings of K ⊗_A Ā as local rings of one of the Bmi^\widehat{B_{m_i}}.

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