Bourbaki

Exercise 5, § 8

Algebra, X, p. 202

Let (Mi)i∈E(M_i)_{i \in E} be a family of A-modules. (a) Now prove that diA(∏i∈EMi)=sup⁡diA(Mi)\mathrm{di}_A(\prod_{i \in E} M_i) = \sup \mathrm{di}_A(M_i). (b) If A is noetherian, now prove that diA(⨁i∈EMi)=sup⁡diA(Mi)\mathrm{di}_A(\bigoplus_{i \in E} M_i) = \sup \mathrm{di}_A(M_i). (c) If diA(⨁i∈EMi)=sup⁡diA(Mi)\mathrm{di}_A(\bigoplus_{i \in E} M_i) = \sup \mathrm{di}_A(M_i) for every family (Mi)i∈E(M_i)_{i \in E} of A-modules, A is noetherian (use Exercise 21, p. 170).

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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.