Let R be a PID, and let B be a torsion R-module, and let p be a prime in R.
Prove that if pb=0 for b in B, then Ann(B) is in (p), the ideal generated by p.
The Ann(B) is defined as
$\displaystyle \text{Ann(B)} = \{ r \in R | rb = 0, \forall b \in B\}$,
(T(B)=B since B is a torsion R-module, where T(B) is a torsion submodule of B).
We see that Ann(B) is an ideal of R.
Since R is a PID, every non-zero prime ideal is a maximal ideal. Every ideal in R (except R itself) is contained in a maximal ideal. Thus, Ann(B) is contained in a maximal ideal.
Since pb=0 for b in B, we have $\displaystyle p \in Ann(B)$ by the definition of an annhilator. We also have $\displaystyle p \in (p)$, where p is a prime. Since (p) is a maximal ideal, any ideal I(except R itself) containing an element of (p) should be contained in (p). Thus, Ann(B) is in (p).