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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?

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