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I'm looking for examples of a commutative ring $R$ such that its set of torsion elements $T$ is a submodule of $R$ (seen as an $R$-module), but such that $R/T$ is not torsion free. Anyone?

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Take any ring $R$ with zero divisors, but where not every element is a zero divisor, and where the ideal generated by the zero divisors is not everything; for example, the ring of dual numbers $\{a+b\epsilon\mid a,b\in\mathbb{Z}, \epsilon^2=0\}$.

Then $a+b\epsilon$ is a torsion element if and only if there exists a nonzero $x+y\epsilon$ such that $(x+y\epsilon)(a+b\epsilon) = 0$; since $(x+y\epsilon)(a+b\epsilon) = xa + (xb+ya)\epsilon$, this happens if and only if $a=0$: if $a=0$, then take $x+y\epsilon=\epsilon$; if $a\neq 0$, we would need $x=0$, in which case we would require $ya\epsilon=0$, hence $y=0$.

The collection of all zero divisors is then and ideal, so $T=\mathbb{Z}\epsilon$, and $R/T \cong\mathbb{Z}$. But now let $1+T\in R/T$. Then $\epsilon\neq 0$, but $\epsilon(1+T) = \epsilon+T = T$, so $1+T$ is a torsion element. So $R/T$ is not torsion free (in fact, it is a torsion module, since $\epsilon(R/T) = T/T$).

(Basically: if such a ring exists, then there is an $s\neq 0$ for which there exists $r\neq 0$ with $rs$ a nonzero zero divisor. Then there exists $t$ nonzero such that $trs=0$; we need $tr=0$ so that $s$ not lie in the torsion submodule, so $r$ itself must be a nonzero zero divisor. So all we need is to make sure that there are nonzero zero divisors, but that not everything is a zero divisor. Then multiplying any nontrivial element of $R/T$ by a zero divisor will map it to $T$.)

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    In short: Any commutative ring $R$ with zero divisors for which the ideal generated by the zero divisors is not all of $R$ will provide you with an example.2011-09-17