If $R$ is an integral domain one defines an $R$-module $M$ to be torsion-free if the torsion submodule $T(M)=\{x\in M\mid l_R(x)≠ 0\} $ is zero, where $ l_R(x)$ is the left annihilator of $x$ in $R$. In general, one defines the torsion submodule as $\{x\in M \mid sx=0$ for some non-zero divisor $s\in R\}$ and $M$ is named torsion-free if $T(M)=0$.
What is the problem if we remove the condition " for some non-zero divisor $s\in R$ " ? In fact, what happens if we only define a "subset" of $M$ as $S=\{x\in M\mid l_R(x)≠ 0\} $?
Thanks!
You can certainly define this set, and it could be considered a set of "torsion elements" inside $M$, but the set does not have as many nice properties in general. As in the example of $\mathbb Z/6\mathbb Z$ given above, $2$ and $3$ are torsion but $3-2=1$ is not, so it is not even a subgroup.
There are, however, other useful ways of defining torsion in rings with zero divisors.
If $R$ is commutative, you can define $m\in M$ to be a torsion element if it is annihilated by a non-zerodivisor of $R$, and the resulting set is a submodule of $M$.
If $M$ is a right $R$ module, and $R$ is any ring, you can also define $m$ to be torsion if its annihilator is an essential right ideal of of $R$. ($m$ is actually said to be "singular," but singularity is just a version of "torsion.") The result is again a submodule of $M$.