Definitions:
- $M_i$ are $R$ modules;
- $\text{ann}(x) = \{r \in R \mid rx = 0\}$;
- $T(M)=\{m \in M \mid \text{ann}(m) \neq \{0\}\}$;
Problem
Show that if $M = M_1 \oplus \cdots \oplus M_n$ then $T(M) = T(M_1) \oplus \cdots \oplus T(M_n)$.
My Solution:
Let $x \in T(M) \subseteq M$. It follows that $x = m_1 + \cdots + m_n$ where $m_i \in M_i$. Since $\text{ann}(x) \neq \{0\}$, there is a $r \in R\setminus\{0\}$ such that $rx = 0$, hence: $$ 0 = rx = r(m_1 + \cdots + m_n) = rm_1 + \cdots + rm_n \implies rm_i=0 $$ Therefore if $x \in T(M)$ then $r \in \text{ann}(m_i)$ for $i \in \{1,\cdots,n\}$ which follows that each $m_i$ is a torsion element of $M_i$. Therefore: $$ T(M) = T(M_1) + \cdots + T(M_n) $$ Now to show that this sum is direct, consider two different decompositions of an element $y \in T(M)$: $$ t_1' + \cdots + t_n' = y = t_1 + \cdots + t_n \implies t_1'-t_1 + \cdots + t_n' - t_n = 0 $$ Since $0 = t_1'-t_1 + \cdots + t_n' - t_n \in M_1 \oplus \cdots \oplus M_n = M$ we have that: $$ (t_1'-t_1) + \cdots + (t_n' - t_n) = 0 + \cdots + 0 \implies t_i' - t_i = 0 \implies t_i' = t_i $$ Which shows that the sum is direct.
That is a problem from the Modules chapter of the book Introduction to Abstract Algebra - W.Keith Nicholson.
Can someone please check my solution? Thanks!
This is not true if $R$ is not a domain. For example, take
$$ \mathbb{Z}_6 = \mathbb{Z}_2 \oplus \mathbb{Z}_3 $$
as a $\mathbb{Z}_6$-module. Then $t(\mathbb{Z}_6) = 0$ but $t(\mathbb{Z}_2),t(\mathbb{Z}_3) \neq 0$.
When $R$ is a domain, the other inclusion you have ommitted is proven as follows: if $m_i \in M_i$ are torsion elements, so that $r_i m_i = 0$ for $r_i \neq 0$, then defining $r := r_1 \cdots r_n$ (which is nonzero precisely because $R$ is a domain) we have
$$ r\sum_im_i = \sum_irm_i = \sum_i (r_1 \cdots \widehat{m_i} \cdots r_n)r_im_i = 0, $$
and thus $m_1 + \cdots +m_n \in t(M)$.
Some other remarks: