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

Question

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.

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Is that correct?

Answer 1

Recently this exercise occured to me too. The approach from the structure theorem may be misleading. The proof in your attempt is wrong in $(p) \cap (q) = (pq)$ but not $\varnothing$. But even after fixing this with $M$ isomorphic to $D/(r_1) \oplus \dots D/(r_n)$ with $r_1 |r_2 | \dots | r_n$ I can still only prove that $Dz$ is isomorphic to $D/(r_n)$ in this approach and it's hard for me to reach the result $M = D/(r_n) \oplus M/Dz$ from this approach.

Maybe the correct approach is to prove it without the structure theorem (and prove the structure theorem with this result). The following proof is modified from lemma 8 from this note, which I found from this thread.

Statement Let $M$ be a finitely generated torsion module over a PID $D$, and $z \in M$ is an element such that $ann(z) \subset ann(x)$ for all $x \in M$. Show that $Dz$ is a pure module of $M$, i.e., for any $gm = hz$ with $g, h\in D$ and $m \in M$, there is some $n \in Dz$ such that $gn = hz$.

Proof: Since $D$ is a PID, the annihilator $ann(z)$ must be a principal ideal, say $ann(z) = (d)$. Consider the annihilator of $[m]$ in $M/Dz$, it’s also a principal ideal, say $ann_{M/Dz}([m]) = (e)$. Then $e|d$ for $dm = 0$ and $d[m] = 0$. Denote $d = ce$. From $e[m] = [0]$ we suppose $em = fz$. Then

$$ 0= dm=cem = cfz $$

Hence $d|cf$, which means $ce|cf$ and hence $e|f$.

From $gm = hz$ we have $e|g$. Hence

$$ hz = \frac{g}{e}em = \frac{g}{e} f z $$

We can take $n = \frac{f}{e}z$ for $e|f$.

Q.E.D.