# Any cyclic submodule $Dz$ such that $\operatorname{ann} z\subset \operatorname{ann} x$ for every $x \in M$, is a pure submodule.

Let $$M$$ be a finitely generated torsion module over a p.i.d. $$D$$. Show that any cyclic submodule $$Dz$$ such that $$\operatorname{ann} z\subset \operatorname{ann} x$$ for every $$x \in M$$, is a pure submodule. [Jacobson, Basic Algebra, P194.9]

My attempt:

Since $$M$$ is a finitely generated torsion module over a p.i.d., then $$M$$ can be decomposed into finitely many $$p$$-components $$M_p$$: $$M=\bigoplus_{i=1}^n M_{p_{i}},\quad p_i \,\text{are distinct primes in}\ D$$ Furthermore, we can decompose $$M_p$$ and write $$M$$ as a direct sum of primary cyclic modules, i.e. $$M\cong\bigoplus_{i=1}^n\bigoplus_{j_i=1}^{m_i} (D/(p_{i}^{j_i}))^{t_{ij_i}} .$$ Since $$\operatorname{ann} z\subset \operatorname{ann} x$$ for every $$x \in M$$, then $$\operatorname{ann} z\subset\bigcap_{i,j_i}(p_{i}^{j_i}) .$$ Note that $$(p_{i_1}^{j_{i_1}})\cap (p_{i_2}^{j_{i_2}})=\emptyset$$ if $$i_1\neq i_2$$. If $$\operatorname{ann}z=\emptyset$$ then we are done; otherwise we must have $$n=1$$ and $$\operatorname{ann}z\subset \bigcap_j (p^{j})=(p^{j_{max}}), \quad j_{max}\, \text{is the greatest j 's} .$$ However, $$z\in M$$. So $$\operatorname{ann}z$$ must appear as one of $$(p^k)$$ which has to be $$(p^{j_{max}})$$.

Hence we proved $$Dz$$ is a direct summand $$D/(p^{j_{max}})$$ of $$M$$ and it is a pure submodule.

Is that correct?