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Let $x$ and $y$ be torsion elements in an R-module $M$ having orders $a$ and $b$. With $a,b \in R$, $a$ and $b$ are relatively prime, and $R$ is a PID. I want to show that $x + y$ has order $ab$.

So if a and b are relatively prime then there exists some $m,n\in R$ s.t.

$am+bn=1$

But i'm not sure where to go from here. Any help would be great.

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    $\begingroup$ My advice: try to first find a proof for the special case $R=\Bbb Z$ and then see if you can adapt that proof to the general case. $\endgroup$ – Marc Paul Feb 15 '16 at 20:38
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Let $I=\{r\in R:r(x+y)=0\}$. $I$ is an ideal, and $abR\subset I$ since $ax=0$ and $by=0$.

On the other hand, suppose that $r(x+y)=0$. Multiplying $1=am+bn$ by $r$ yields $r=ram+rbn$, hence $rx=rbnx=bn(-ry)=0$. Similarly $ry=0$. Therefore $I\subset aR\cap bR=abR$, with the last equality due to the fact that $a$ and $b$ are coprime. So we have shown both inclusions.

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