Let $T$ be the set of torsion elements in R. Prove that the only torsion element of $R/T$ is $0$.

Question

An element $a$ of a ring $R$ is said to be a torsion element if there exists a positive integer $n$ such that $na = 0$; in other words, $a$ has finite order if we view $R$ as a group under addition. For example, if $R = \mathbb{Z}/5$ then every element is torsion; if $R = \mathbb{Z}$, then $0$ is the only torsion element.

Let $T$ be the set of torsion elements in $R$. Prove that the only torsion element of $R/T$ is $0$.

I am trying to suppose that there is a non-zero torsion element in $R/T$, and by showing that the preimage of element is in $T$ hence it is zero. Thus we may get a contradiction.

But how to prove that? May I ask for steps? Am I on the right track?

Thanks in advance!

Answer 1

Let $a+T(\neq T)\in R/T$ be a torsion element .So $a\notin T$

Then for all $n\in \Bbb N$ we have $na\neq 0$.

Hence $\exists m\in \Bbb N$ such that $m(a+T)=T\implies ma+T=T\implies ma\in T\implies k(ma)=0\implies (km)a=0$

which is false as for all $n\in \Bbb N$ we have $na\neq 0$.

Answer 2

Suppose that $\overline{r}\in R/T$ is a torsion element, then there is some positive integer $n$ such that $n\overline{r}=\overline{0}$. So $\overline{nr}=\overline{0}$ and hence $nr\in T$. This means that there is some positive integer $m$ such that $m(nr)=0$, thus $(mn)r=0$. Therefore $r\in T$ and we get $\overline{r}=\overline{0}$.