## Definition and basic properties of the set of torsion elements in a module

Let $R$ be a ring with $1 \neq 0$ and let $M$ be a left $R$-module. An element $m \in M$ is called torsion if there exists $r \in R$ such that $r \cdot m = 0$. We define $\mathsf{Tor}_R(M) = \{ m \in M \ |\ r \cdot m = 0\ \mathrm{for\ some\ nonzero}\ r \in R \}$.

1. Prove that if $R$ is an integral domain, then $\mathsf{Tor}(M)$ is a submodule of $M$.
2. Show by an example that if $R$ is not an integral domain, then $\mathsf{Tor}(M)$ need not be a submodule.
3. Prove that if $R$ has zero divisors, then every nontrivial $R$ module has nonzero torsion elements.

1. We use the submodule criterion. Note that $1 \cdot 0 = 0$ in $M$, so that $0 \in \mathsf{Tor}(M)$. Now suppose $a,b \in \mathsf{Tor}(M)$ and $r \in R$. There exist nonzero $u,v \in R$ such that $u \cdot a = v \cdot b = 0$. Since $R$ is an integral domain, $uv \neq 0$. Now $uv \cdot (a+r \cdot b) = v \cdot (u \cdot a) + r \cdot (v \cdot b) = 0$ (since $R$ is commutative), so that $a+r \cdot b \in \mathsf{Tor}(M)$. Thus $\mathsf{Tor}(M)$ is a submodule of $M$.
2. Consider the ring $R = \mathbb{Z}/(6)$, and consider $M = R$ as a left module over itself. Note that neither $\overline{2}$ nor $\overline{3}$ are zero in $R$, but that $\overline{3} \cdot \overline{2} = \overline{2} \cdot \overline{3} = 0$. Thus $\overline{2}, \overline{3} \in \mathsf{Tor}(M)$. However, $\overline{2} + \overline{3} = \overline{5}$, and $r \cdot \overline{5} = 0$ only if $r = 0$. Thus $\mathsf{Tor}(M)$ is not a submodule of $M$, as it is not closed under addition.
3. Suppose $R$ has zero divisors and let $M$ be a left $R$-module. Suppose by way of contradiction that $\mathsf{Tor}(M) = 0$. Now let $r,s \in R$ be nonzero such that $rs = 0$, and let $m \in M$ be nonzero. Now $r \cdot (s \cdot m) = rs \cdot m$ $= 0 \cdot m = 0$, so that $s \cdot m$ is torsion. Thus $s \cdot m = 0$, and so $m = 0$, a contradiction. Thus every nontrivial $R$-module contains nonzero torsion elements.