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.


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 $bym+axm=(ax+by)m=1m =m$.

  • $\begingroup$ But is the condition $M=\mathrm {Tor}\ M$ of no importance in solving this question? $\endgroup$ – Arnab Chatterjee. Mar 28 '18 at 14:15
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.