On the Categorical Interpretation of the Ring Cohomology

This leads to two ways to construct the ring cohomology: cohomology for rings which are regarded as algebras due to Shukla [Sh], and ring cohomology due to Mac Lane [Mac]. The ring structure has been categorized by the definition of an Ann-category [4]. Each congruence class of Ann-categories is characterized by a cohomology class of structures (Theorem 4.3[4]). For regular Ann-categories, this class is bijective to the group H Sh(R,M) (Theorem 4.4 [5]). In the general case, this group is replaced with the group H Mac(R,M) (Theorem 7.6 [7]). In [1], authors have modified the definition of an Ann-category to be the one of a categorical ring, where the condition (Ann-1) is omitted, and the compatibility of the operation ⊗ with the associativity and commutativity constraints respect to + is replaced with the compatibility of the operation ⊗ with the “associativity commutativity” constraint. Categorical rings have been classified by the Mac Lane cohomology group H Mac(R,M). There are 2 problems emerged: Firstly, do the sets of categorical rings and Ann-categories coincide? Secondly, in the case that these sets can not be proved to coincide, is it possible to prove the existence of the bijection between the set of congruence classes of pre-sticked Ann-categories of the type (R,M) and the group H Mac(R,M)? We have completely solved both of these problems. In [6], authors have showed that the set of Ann-categories is a subset of the set of categorical rings, and these two sets coincide if and only if the operation ⊗ in each categorical ring is compatible with the unitivity constraints of the operation ⊕. (This condition is also called the condition (U)). A question is raised: May the condition (U) be deduced from the axiomatics of a categorical ring? The second problem has been solved in [7]. After that, in [2], authors have proved that the condition (U) is always satisfied in a categorical ring, and therefore the two sets of the categories coincide. However, authors have made a mistake in the proof of the Lemma.