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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.

2 Answers 2

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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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    Replacing `-b/g` with `\smash{-b/g}` and `lcm(a,b)` with `\smash{lcm(a,b)}` seems to work in the preview; smash fools LaTeX into thinking that the box in question has standard height and depth. The slash and the l seem to be the ones giving the trouble.2012-03-22
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    @BillDubuque I have no idea how to use your hint (let alone what it means). Could you elaborate on that?2012-03-22
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    @BillDubuque You didnt want to help me to start with. If you elaborate I will turn it into an upvote.2012-03-22
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    @BillDubuque It's alright. Im sure I'll survive.2012-03-22
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    @Daniel For the record, I planned to elaborate when I had enough spare time. Try to be more patient before downvoting. I give hints because I believe they usually provide better learning experiences. I almost always elaborate when asked. If you google "Smith Normal form" you will find many expositions on the web. Of course you can always ask further questions here if such expositions are not clear. Best of luck.2012-03-22
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    Hi Bill! I upvoted your observation but in my opinion this is not more explicit than using the CRT, as the construction depends on the same choice which must be made with the CRT (namely picking a solution to Bezout's identity). Anyways, just nitpicking. Regards,2012-03-24
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    @Bruno Thanks. I interpreted the OP's request to "give an explicit map" to mean to give an explicit (vs. his implicit) specification of some isomorphism, which the above does (algorithmically). It seems your notion of "explicit" is a bit different, perhaps having to do with "canonical" (which was not mentioned by the OP). Why does you notion of explicit imply canonical?2012-03-24
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    See also [this later question.](http://math.stackexchange.com/q/612449/242)2016-11-19