Prove that exits two $R-$modules $M,N$ different $0$ such that $Hom(M,N)=0$

Question

Suppose that $R$ is a ring and $1_R$ is its multiplicative identity. Prove that exits two $R-$modules $M,N$ different $0$ such that $Hom(M,N)=0$??

Answer 1

That not true if $R$ is a field. Did you have anything else on your mind?

I think it is also not true for local artinian rings (say $\mathbb{Z}/p^n$ ).

It will be true for semisimple rings with two different classes of irreducible modules.

Answer 2

Let $R=\mathbb{Z}$, $M=\mathbb{Q}$, $N=R=\mathbb{Z}$. Let $\phi\in \mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$. We first show that $\phi(1)=0$.

Suppose that $\phi(1)\neq 0$. Note that $\phi(1)\in\mathbb{Z}$, and so by the Fundamental Theorem of Arithmetic, we can write \begin{equation*} \phi(1)=\prod_{i=1}^{m}p_{i}^{\alpha_{i}} \end{equation*} Let $q$ be any prime such that $q\neq p_{i}$ for all $i$. Then since $\phi$ is a $\mathbb{Z}$-module homomorphism, \begin{equation*} \phi(1)=\phi(q/q)=q\phi(1/q) \end{equation*} so that $q|\phi(1)$, a contradiction. This forces $\phi(1)=0$.

Next we show that $\phi(a)=0$ for all $a\in\mathbb{Q}$. Let $a\in\mathbb{Q}$. Then writing $a=r/s$ for some $r,s\in\mathbb{Z}^{+}$ with $\gcd(r,s)=1$, \begin{equation*} \phi(a)=\phi(r/s)=r\phi(1/s). \end{equation*} It suffices to show that $\phi(1/s)=0$. We have \begin{equation*} \phi(1)=s\phi(1/s) \end{equation*} and since $s\neq 0$, $\phi(1/s)=0$. Thus, $\phi(a)=0$ for all $a\in\mathbb{Q}$, which implies that $\phi=0$. Thus, $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$.