“Is isomorphic to” is an equivalence relation on any set of R-modules

Let R be a ring with 1 and let \mathcal{M} be any set of (left) R-modules. Prove that the relation “is isomorphic to” is an equivalence relation on \mathcal{M}.


Recall that we say M \cong N if there exists an R-module isomorphism \theta : M \rightarrow N. We need to show that this relation is reflexive, symmetric, and transitive.

  1. (Reflexive) Let M \in \mathcal{M}. Note that the identity mapping \mathsf{id} : M \rightarrow M is an R-module isomorphism since \mathsf{id}(x + r \cdot y) = x + r \cdot y = \mathsf{id}(x) + r \cdot \mathsf{id}(y) for all x, y \in M and r \in R. Thus M \cong M, and so \cong is reflexive.
  2. (Symmetric) Suppose M,N \in \mathcal{M} such that M \cong N. That is, there exists an R-module isomorphism \theta : M \rightarrow N. Because \theta is a bijection, the converse relation \theta^{-1} is well-defined, and thus a bijective function N \rightarrow M. We claim that \theta^{-1} is in fact an R-module homomorphism. To see this, let r \in R and x,y \in N. Note that x = \theta(a) and y = \theta(b) for some a,b \in M. Now \theta^{-1}(x + r \cdot y) = \theta^{-1}(\theta(a) + r \cdot \theta(b)) = \theta^{-1}(\theta(a + r \cdot b)) = a + r \cdot b = \theta^{-1}(x) + r \cdot \theta^{-1}(y). So \theta^{-1} : N \rightarrow M is an R-module isomorphism, and we have N \cong M.
  3. (Transitive) Suppose M,N,T \in \mathcal{M} such that M \cong N and N \cong T. Then there exist R-module isomorphisms \theta : M \rightarrow N and \zeta : N \rightarrow T. Recall that \zeta \circ \theta : M \rightarrow T is a bijection; we claim that it is also an R-module homomorphism. To see this, let x,y \in M and r \in R. Then (\zeta \circ \theta)(x + r \cdot y) = \zeta(\theta(x + r \cdot y)) = \zeta(\theta(x) + r \cdot \theta(y)) = \zeta(\theta(x)) + r \cdot \zeta(\theta(y)) = (\zeta \circ \theta)(x) + r \cdot (\zeta \circ \theta)(y). So \zeta \circ \theta : M \rightarrow T is an R-module isomorphism, and we have M \cong T.

Thus \cong is an equivalence relation on \mathcal{M}.

Note also that \mathsf{id} is a unital R-module homomorphism, that if \theta is a unital R-module homomorphism then so is \theta^{-1}, and if \theta and \zeta are unital R-module homomorphisms (and the composition makes sense) then so is \zeta \circ \theta. Thus, if we replace “R-modules” by “unital R-modules”, the result still holds.

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