I am interested in the following result:
Let $D$ be a $PID$. Let $d\in D$ such that $d=p^{r}q^{s}$ with $s,r\geq 1$ where $p,q$ are non-associated irreducible elements. If $M=D/(d)$ then $M=t_{p}(M)\oplus t_{q}(M)$.
Note that $t_{p}(M)$ represents the $p$-torsion submodule of $M$.
I have seen some proofs about this result but I have read that it is possible to prove it using the fact that in a DIP the Bezout's identity is fulfilled for coprime elements.
Could someone tell me how to prove it that way?
Thank you so much.
If $p$ and $q$ are coprime then the GCD of $p^r$ and $q^s$ is $1$. By Bezout you can then write $1 = ap^r + bq^s$ for some $a, b \in D$. Now for any $m \in M$ we have
$$m = 1m = ap^rm + bq^sm$$
Note that $q^s(ap^rm) = adm = 0$ so $ap^rm$ is $q$-torsion. Similarly, $bq^sm$ is $p$-torsion.
This gives $M = t_p(M) + t_q(M)$, to get $M = t_p(M) \oplus t_q(M)$ we need $t_p(M) \cap t_q(M) = 0$. So assume $m \in t_p(M) \cap t_q(M)$ and let $\mathrm{ann}_D(m) = \{x \in D \ | \ xm = 0\}$. As $m \in t_p(M)$ there is some power of $p$ in $\mathrm{ann}_D(m)$, say $p^e$. Similarly, $m \in t_q(M)$ so $q^f \in \mathrm{ann}_D(m)$ for some $f$. By Bezout, $1 = a'p^e + b'q^f$ for some $a', b'$, so $1 \in \mathrm{ann}_D(m)$, hence $m = 1m = 0$.