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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 common 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.

user26857
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Daniel Montealegre
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2 Answers2

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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]^{\phantom{|^|}} \!\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] $

See this answer for further detail.

Bill Dubuque
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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. – Arturo Magidin Mar 22 '12 at 03:41
  • @BillDubuque I have no idea how to use your hint (let alone what it means). Could you elaborate on that? – Daniel Montealegre Mar 22 '12 at 06:36
  • @BillDubuque You didnt want to help me to start with. If you elaborate I will turn it into an upvote. – Daniel Montealegre Mar 22 '12 at 19:45
  • @BillDubuque It's alright. Im sure I'll survive. – Daniel Montealegre Mar 22 '12 at 19:46
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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. – Bill Dubuque Mar 22 '12 at 19:58
  • 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, – Bruno Joyal Mar 24 '12 at 22:08
  • @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? – Bill Dubuque Mar 24 '12 at 22:20
  • See also this later question. – Bill Dubuque Nov 19 '16 at 20:49
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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!

Bruno Joyal
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