Let $A$ be an integral domain and $M$ be a finitely generated $A$-module. Denote by $M_{\mathrm{tors}}$ the torsion sub-module of $M$ and $I$ be the annihilator of $M_{\mathrm{tors}}$. Is it true that $$IM \cap M_{\mathrm{tors}}=0\ ?$$
I was trying to prove this by contradiction. $A \to I.D. \& M \to A$-module.
$$M_{tors}=\{m \in M | a.m = 0 \mbox{ for some nonzero } a \in A\}$$
$$I = \{a \in A | a.m=0 \forall m \in M_{tors}\}$$
If $IM \cap M_{tors} \not= 0$, then let $m′ \in IM \cap M_{tors}$. So, if we take $S={m_1,m_2,...,m_n}$ be the minimal generating set of $M$ then $m′=a_1.m_1+a_2.m_2+...+a_n.m_n$ and $a.m′=0 \forall a \in I$. Hence we have, $$a.m′=aa_1.m_1+aa_2.m_2+...+aa_n.m_n=0 \forall a \in I$$ Since $A$ is an I.D. if $a \not= 0$ then at least one of the coefficients in the representation of $a.m′$ must be nonzero.
But this cannot be a contradiction. I try to find example with nontrivial intersection of the annihilator and torsion, but I cannot find any. Can anyone help me?