Let $R$ be a principal ideal domain, $M$ be a $R$ module and $T(M) = \{x \in M \mid \text{ann}(x) \neq \{0\}\}$ where $\text{ann}(x) = \{r \in R \mid rx = 0\}$. Show that $T(M)$ is an $R$-submodule of $M$.
In order to prove that is asked we need to show that:
1) $T(M)$ is an additive abelian subgroup of $M$,
2) $RT(M) \subset T(M)$.
Proof of 1):
Closure:
If $a,b \in T(M)$ then $\text{ann}(a),\text{ann}(b) \neq \{0\}$. Pick any two elements different than $0$ of those annihilators, for example $r \in \text{ann}(a)$ and $s \in \text{ann}(b)$. Therefore: $$ (rs)(a+b) = rsa + rsb = s(ra) + r(sb) = 0 + 0 = 0 $$
Associativity:
If $a,b,c \in T(M)$ then $a+(b+c) = (a+b)+c$ because $a,b,c \in M$ and $M$ is a group.
Identity:
$0 \in T(M)$ since $rx = 0$ for any $r \in R$, which implies that $\text{ann}(0) = R$.
Inverse:
If $x \in T(M)$ then $-x \in T(M)$ because $\text{ann}(x) = \text{ann}(-x)$.
Proof of 2):
Let $x \in T(M)$ and $r \in R$. It follows that $\text{ann}(x) \neq \{0\}$. Let $d \in \text{ann}(x) \setminus{\{0\}}$. Now see that: $$ drx = rdx = r(dx) = r(0) = 0 $$ Which implies that $d \in \text{ann}(rx)$, and since $d \neq 0$ it follows that $rx \in T(M) \implies RT(M) \subset T(M)$.
I used the fact that a PID is an integral domain and therefore is commutative. Is there a trivial example that shows that the theorem does not hold if the ring is non-commutative?
Also, can someone please check my work?
Thanks!
You have the very strong assumption of $R$ being a PID and haven't really used enough of it. Commutativity is good, but you need the fact that it's a domain as well. For instance, in your proof of closure under addition, you said that $(rs)(a + b) = 0$ so $a + b \in T(M)$. But how do you know that $rs \neq 0$? You should state this.
As for counterexamples, you do need the ring to be a domain. Indeed, take $R = M = \mathbb Z / 6$. Check that $T(M) = \{0, 2, 3, 4\}$, but $2 + 3 = 5 \notin T(M)$.
You also need commutativity. Take $R = k\langle x, y \rangle $ with $k$ a field. Here I mean the polynomial ring in noncommuting variables. Now take $M = R/xR \oplus R / yR$. Take $e_1 = (1, 0), e_2 = (0, 1)$ in $M$. Then $x e_1 = 0$ and $y e_2 = 0$ while $x, y \neq 0$. However, $e_1 + e_2 = (1, 1)$ is not torsion. Indeed, if $(f, f) = 0$ in $M$ then $f \in xR \cap yR$. You can check that $xR \cap yR = 0$.