Prove that exits two $R-$modules $M,N$ different $0$ such that $Hom(M,N)=0$ 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 $Hom(M,N)=0$??
 A: 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.
A: 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, $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$.
