4

I've recently noticed a few different posts asking about this problem Isomorphism of two direct sums, all using the CRT. I think I understand it this way. What caught my eye is in the above link a user gives this hint:

The Bezout identity $\rm\: ad+bc\:\! =\:\! g\:$ yields the following Smith normal form reduction $\left[\begin{array}{cc}\rm 1 & \!\!\!0 |& \!\!\!a & \!\!\!0\\ 0 &\rm \!\!\!1| & \!\!\!0 & \!\!\!b\end{array}\right]$ $\sim$$\left[\begin{array}{cc}\rm 1 & \!\!\!1 |& \!\!\!a & \!\!\!b\\ 0 &\rm \!\!\!1| & \!\!\!0 & \!\!\!b\end{array}\right]$ $\sim$$\left[\begin{array}{cc}\rm 1 & \!\!\!1 |& \!\!\!a & \!\!\!b\\ 0 &\rm \!\!\!1| & \!\!\!0 & \!\!\!b\end{array}\right]$ $\left[\begin{array}{cc}\rm 1 & \!\!\!0 |& \rm d &\rm \!\!\!\smash{-\!b/g}\\ 0 &\rm \!\!\!1| & \!\!\!\rm c &\rm a/g\end{array}\right] $= $\left[\begin{array}{cc}\rm 1 & \!\!\!1 |& \!\!\!g & \!\!\!0\\ 0 &\rm \!\!\!1| & \!\!\!bc & \!\!\!ab/g\end{array}\right]$ = $\left[\begin{array}{cc}\rm 1 + bc/g & \!\!\!1 |& \!\!\!gcd(a,b) & \!\!\!0\\ bc/g &\rm \!\!\!1| & \!\!\!0 & \!\!\!lcm(a,b)\end{array}\right]$

The OP became impatient which led the hint giver to not elaborate. To me this seems like a solution. I'm having trouble seeing why it is just a hint. Can anyone let me know what more I would need to correctly justify the isomorphism?

EDIT: Left Identity matrix not in orginal comment. Is adjoining the other matrix more useful?

Community
  • 1
user75667
  • 41
  • To find an explicit isomorphism I think you still need to calculate the row/column transformation matrices. Note also that when $x \neq 0$ divides $y$ in a domain, the notation $y/x$ makes sense, even though $x$ might be a nonunit. If $y = qx = q'x$, then $q = q'$ by cancellation. – spin May 03 '13 at 18:03
  • @spin When you say transformation matrix do you mean like what I have done in the edit above? – user75667 May 03 '13 at 19:51
  • @spin Maybe you want to expand your comment to an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 14 '13 at 18:43

0 Answers0