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So I basically have to prove the following. If $a,b\in R$ non-zero, with $R$ a PID. Then we want to show $R/aR\oplus R/bR\cong R/cR\oplus R/dR$Where $c$ is the least commond multiple of $a$ and $b$, and $d$ is the greatest common divisor of $a$ and $b$.

Here is my proof: Since $R$ is a PID it is also a UFD, so let $a=u_1p_1^{\alpha_1}...p_n^{\alpha_n}$$b=u_2p_1^{\beta_1}...p_n^{\beta_n}$where $u_i$ are units, and some of the $\alpha$'s or $\beta$'s might be zero, but it is written this way for a later convinience.

By the chinese remainder theorem we have that $R/aR\cong R/p_1^{\alpha_1}R\oplus...\oplus R/p_n^{\alpha_n}R$. I do the same with $R/bR$, and then I have $R/aR\oplus R/bR\cong R/p_1^{\alpha_1}R\oplus...\oplus R/p_n^{\alpha_n}R\oplus R/p_1^{\beta_1}R\oplus...\oplus R/p_n^{\beta_n}R$Then for a given if $\alpha_i<\beta_i$, then I interchange the summands of $R/p_i^{\alpha_i}R$ and $R/p_i^{\beta_i}R$. I basically write the high exponents in the front and the low exponents in the back. The combination of the low exponents furnishes the gcd, and the combination of the high exponents yields the lcm.

I was wondering if there is an easier way (which probably there is, I am somewhat tired) that gives an explicit map.

Thanks.

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I'm quite confident that your proof is optimal. In any case I think it's a pretty proof - you should be satisfied!

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    See my answer for one way to make it "explicit".2012-03-22
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Hint $\ $ The Bezout identity $\rm\: ad+bc\:\! =\:\! g\:$ yields the following Smith normal form reduction $\left[\begin{array}{cc}\rm a & \!\!\!0\\0 &\rm \!\!\!b\end{array}\right] \sim \left[\begin{array}{cc} \rm a &\rm \!\!\!b\\0 &\rm\!\!\! b\end{array}\right]\sim \left[\begin{array}{cc} \rm a &\rm\!\!\! b\\0 &\rm\!\!\! b\end{array}\right] \left[\begin{array}{cc} \rm d &\rm \!\!\!\smash{-\!b/g} \\ \rm c &\rm a/g\end{array}\right] = \left[\begin{array}{cc} \rm g &\rm 0\\ \rm bc &\rm \!\!\!ab/g\end{array}\right] \sim \left[\begin{array}{cc} \rm g &\rm 0\\ 0 &\rm \!\!ab/g\end{array}\right] = \left[\begin{array}{cc} \rm \!gcd(a,b) &\rm 0\\0 &\rm \!\!\!\!\!\!\!\smash{lcm(a,b)}\!\end{array}\right] $

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    See also [this later question.](http://math.stackexchange.com/q/612449/242)2016-11-19