Isomorphisms define an equivalence relation between modules

Question

I have recently started theory of modules and I have the following doubt:

Is "being isomorphic" an equivalence relation among $R$-modules?

Concretely, I am not sure that if $f$ is an $R$-module isomorphism, then $f^{-1}$ exists and it is also a $R$-module isomorphism.

Thanks.

Answer 1

I am very late but I would like to answer this question.

Definition: Let $R$ be a ring and let $M$ and $N$ be $R-$modules. A map $\psi:M\to N$ is an $R$-module isomorphism if $\psi$ is bijective and respects the $R$-module structures of $M$ and $N$, i.e.,

• $\psi(x+y)=\psi(x)+\psi(y), \forall x,y\in M$
• $\psi(rx)=r\psi(x),\forall r\in R,\forall m\in M.$ I'll denote traditionally 'isomorphism' by $'\cong'$. Here we will show that $R$-module isomorphism is an equivalence relation. To prove this, we will show that$R$-module isomorphism is reflexive, symmetric and transitive.

Reflexivity: Let $M$ be an $R$-module. Let $i:M\to M$ be an identity map. $i$ is bijective and $i(x+y)=x+y=i(x)+i(y), \forall x,y\in M$ and $i(rx)=rx=r(x),\forall r\in R, \forall x\in M$. Thus $M\cong M.$

Symmetry: Let $M$ and $N$ be $R-$modules and $M$ is $R$-module isomorphic to $N$. The map $\psi:M\to N $ is bijective and $R$-module homomorphism. Now we claim that $N$ is $R$-module isomorphic to $M$. Since $\psi$ is bijective therefore $\psi$ is invertible. That is, $\psi^{-1}:N\to M$. It remains to show that $\psi^{-1}$ is $R$-module homomorphism. \begin{equation*}\psi^{-1}(n_1+n_2)=\psi^{-1}(\psi(m_1)+\psi(m_2))=\psi^{-1}\psi(m_1+m_2)=m_1+m_2=\psi^{-1}(n_1)+\psi^{-1}(n_2). \end{equation*} Transitivity: Let $M,N$ and $O$ be $R$-modules. We have $M$ is $R$-module isomorphic to $N$ and $N$ is $R$-module isomorphic to $O$. We'll claim that $M$ is $R$-module isomorphic to $O$. $\phi:M\to N$, $\psi:N\to O$ and $\psi o\phi:M\to O$. Let $h=\psi o\phi$. We know that composition of bijective maps is bijective. Therefore $h$ is bijective. Now we'll show that $h$ is $R$-module homomorphism, \begin{equation} h(m_1+m_2)=\psi o\phi(m_1+m_2)=\psi(\phi((m_1+m_2)))=\psi(\phi(m_1))+\psi(\psi(m_2))=h(m_1)+h(m_2) \end{equation}. I left condition second to prove in both symmetry and transitivity. That is pretty easy to do.

Answer 2

The answer is morally yes, but technically no, because equivalence relations technically have to be defined on sets and the collection of modules isn't a set. But nevertheless you can verify that all of the equivalence relation axioms hold, as in Ryszard's answer, if you ignore the distinction between sets and proper classes.