Giving an example of a ring R such that the torsion satisfies this

Question

Recall: Let $R$ be a (unital, commutative) ring and let $M$ be a $R$-module. An element $m \in M$ is called a torsion element if $rm = 0$ for some non-zero $r \in R$.

I now need to come up with an example of a ring $R$ such that:

(i) Its set of torsion elements $T \subset R$ is a submodule of $R$ (seen as an $R$-module) and

(ii) The quotient $R/T$ is not torsion free (meaning it has non-zero torsion elements)

I know that if $R$ is an integral domain, then this will not work, since then $R/T$ is torsion free (this can be proven).

So I don't want $R$ to be an integral domain. I have no inspiration though. Anyone has some ideas?

Answer 1

We have that $\mathbb{Z}_4$ is a $\mathbb{Z}_4$-module and its torsion set is $T(\mathbb{Z}_4)=\{\overline{0},\overline{2}\}=\overline{2}\mathbb{Z}_4$.

The quotient $^{\mathbb{Z}_4}/_{\overline{2}\mathbb{Z}_4}$ is isomorphic to $\mathbb{Z}_2$, that is not torsion free seen as a $\mathbb{Z}_4$-module.

Answer 2

Let $K$ be a field, and set $R=£[X]/(X^n)$ for $n>1$. $R$ is a local ring, with the single prime ideal $\mathfrak m=(X)/(X^n)$. Its torsion submodule is the ideal $\mathfrak m/(X^n)$, and the quotiont $R/T$ is just the residue field of $R$, which is $K$. Obviously, $R/\mathfrak m$ is not a torsion-free $R$-module.