## “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.