Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 1E (Groups, Rings and Modules)

The zero element is torsion. If $m,n\in T$, choose nonzero $r,s\in R$ with $rm=0$ and $sn=0$. Since $R$ is an integral domain, $rs\ne 0$, andThus $m+n\in T$. If $a\in R$, then $r(am)=a(rm)=0$, so $am\in T$. Hence $T$ is closed under addition and scalar multiplication and is an submodule of $M$.

If $m\in T_1$, then $rm=0$ for some nonzero $r$, andThus $\varphi (T_1)\subseteq T_2$. The formulais consequently independent of the representative and defines an $R$-module homomorphism on the quotient module.

If $\varphi$ is injective and $\Phi (m+T_1)=T_2$, then $\varphi (m)$ is torsion, so $r\varphi (m)=0$ for some nonzero $r$. Injectivity gives $rm=0$, hence $m\in T_1$. Therefore

The converse fails. For $R=\mathbb{Z}$, takeThen $\varphi$ is not injective, but quotienting $M_1$ by its torsion submodule leaves $\mathbb{Z}$, and $\Phi :\mathbb{Z}\to \mathbb{Z}$ is the identity.