Let be a ring with 1 and let be any set of (left) -modules. Prove that the relation “is isomorphic to” is an equivalence relation on .
Recall that we say if there exists an -module isomorphism . We need to show that this relation is reflexive, symmetric, and transitive.
- (Reflexive) Let . Note that the identity mapping is an -module isomorphism since for all and . Thus , and so is reflexive.
- (Symmetric) Suppose such that . That is, there exists an -module isomorphism . Because is a bijection, the converse relation is well-defined, and thus a bijective function . We claim that is in fact an -module homomorphism. To see this, let and . Note that and for some . Now . So is an -module isomorphism, and we have .
- (Transitive) Suppose such that and . Then there exist -module isomorphisms and . Recall that is a bijection; we claim that it is also an -module homomorphism. To see this, let and . Then . So is an -module isomorphism, and we have .
Thus is an equivalence relation on .
Note also that is a unital -module homomorphism, that if is a unital -module homomorphism then so is , and if and are unital -module homomorphisms (and the composition makes sense) then so is . Thus, if we replace “-modules” by “unital -modules”, the result still holds.