# Isomorphism of two direct sums

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". –  Bill Dubuque Mar 22 '12 at 2:59

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. –  Arturo Magidin Mar 22 '12 at 3:41
@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 6:36
@Daniel Given the ridiculous downvote, I'm not much motivated to elaborate. –  Bill Dubuque Mar 22 '12 at 19:43
@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
@Daniel Sorry, you've lost that chance, which is too bad, since it explains precisely what you wanted. –  Bill Dubuque Mar 22 '12 at 19:46