I was studying Introduction to Module Theory from the book Dummit D., Foote R. Abstract algebra (3ed., Wiley, 2004). A question related to this topic I got from a friend of mine:
Let $A$ be an integral domain and $M$ be a finitely generated $A$-module. Denote by $M_{tors}$ the torsion sub-module of $M$ and $I$ be the annihilator of $M_{tors}$. Is it true that $$IM \cap M_{tors}=0?$$
I tried to prove this by contradiction:
$A\rightarrow I.D.$ & $M\rightarrow A$-module.
$M_{tors}=\{m\in M\colon a.m=0 \text{ for some nonzero } a\in A\}$
$I=\{a\in A\colon a.m=0 \forall m\in M_{tors}\}$
If $IM\cap M_{tors} \ne 0$ 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 $m'=a_1.m_1+a_2.m_2+...+a_n.m_n$ & $a.m'=0 \forall a\in I$.
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 $a \ne 0$ then at least one of the coefficients in the representation of $a.m'$ must be nonzero.
But this cannot be a contradiction.
Then I tried to find example that denies this, but I also failed at that. Can anyone help me out?