In my advanced Algebra course, we were told to prove when $R$ is an integral domain and $M$ an $R$-module, $\textrm{Tor}(M)$ is a submodule of $M$ (here $\textrm{Tor}(M)$ denotes the torsion of $M$, which is the elements of $M$ that vanish somewhere on some nonzero element of $R$). As a supplementary exercise, we were told to find a simple example of a ring $R$ and module $M$ where $\textrm{Tor}(M)$ was not an $R$-submodule of $M$, and clearly $R$ couldn't be an integral domain by the previous result.
Now finding an example is simple enough. Personally I used the example $R=M=\mathbb{Z}/6$ and I got an A on the test overall. But I'm wondering, is there an example where $R$ is not an integral domain, but instead of removing the no-zero-divisors property, we instead remove commutativity? I'm thinking it would have to be either a subring of -or constructed from- quaternions, but I'm not very familiar with this.
Note: While $R$ an integral domain is a sufficient condition for $\textrm{Tor}(M)$ to be a submodule, it is not necessary, as $R=M=\mathbb{Z}/4$ has torsion $\{0,2\}$, which is a submodule.
Best Answer
Let $R$ be the free ring on two elements $x$ and $y$. Explicitly, $R=\mathbb{Z}\langle x,y\rangle$ is the ring of noncommutative polynomials in $x$ and $y$ with coefficients in $\mathbb{Z}$. You can easily check that this is a (noncommutative) domain by looking only at top-degree terms.
Now let $I\subset R$ be the left ideal generated by $x$ and let $M=R/I$. Explicitly, $I$ consists of polynomials in which all the monomials end in $x$, so $M$ can by identified with the set of polynomials which contain no monomials that end in $x$, with the action of $R$ on $M$ defined as usual except that $x\cdot 1=0$. So, $1\in\operatorname{Tor}(M)$, but $y\not\in\operatorname{Tor}(M)$. Since $y=y\cdot 1$, this means $\operatorname{Tor}(M)$ is not a submodule of $M$.