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Let $R$ be a PID and $M$ be a torsion R-module with $\mathrm {Ann}_{\ R} M = (c)$. Assume that $c = ab$ in $R$ with $(a,b) = 1$. Show that $M = M_a \oplus M_b$ where $M_r : = \{x \in M : rx = 0 \}$ for $r \in R$.

It is easy to see that $M_a$ and $M_b$ are submodules of $M$ and $M_a \cap M_b = \{0 \}$. Since $(a,b)=1$ there exists $x,y \in R$ such that $ax+by=1$.Now let $m \in M_a \cap M_b$. Then $(ax+by)m=m$. But that will imply $m=0$ since $m \in M_a \cap M_b$. Hence it is clear that $M_a \oplus M_b \subset M$. How do I prove the reverse part which is the key part of this question i.e. how can I express $M$ as the sum of the elements of $M_a$ and $M_b$? Please help me in this regard.

Thank you in advance.

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Let $m\in M$.

Then $cm=abm=0$, hence $bm\in M_a$, and $am\in M_b$.

From $bm\in M_a$, we get $bym\in M_a$.

From $am\in M_b$, we get $axm\in M_b$.

Now simply note that $bym+axm=(ax+by)m=1m =m$.

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  • $\begingroup$ But is the condition $M=\mathrm {Tor}\ M$ of no importance in solving this question? $\endgroup$ – Arnab Chatterjee. Mar 28 '18 at 14:15
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    $\begingroup$ If $R$ is an integral domain, and $c\in R$ is nonzero, the condition $\mathrm {Ann}_{\ R} M = (c)$ implies that $M$ is a torsion-module over $R$ (and in fact, it's a stronger condition). $\endgroup$ – quasi Mar 28 '18 at 14:27
  • $\begingroup$ oh!I see.Thanks @ quasi for your help. $\endgroup$ – Arnab Chatterjee. Mar 28 '18 at 17:38

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