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Why does one need an extra operation for performing smith normal form over a PID? One might suspect and say that it is because of the lack of Euclidean algorithm or just say that we need the additional operation to get what we need.

But I cannot exactly see what property of Euclidean domain makes the fourth operation unnecessary. I would also like to know how this is justified theoretically? Thanks

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  • $\begingroup$ You might get a better answer sooner if you say what your operations are. $\endgroup$ – Chris Godsil Apr 13 '15 at 12:45
  • $\begingroup$ I believe the operations used for Smith normal form are well known. $\endgroup$ – user114539 Apr 13 '15 at 13:18
  • $\begingroup$ It can be found here for example: math.stackexchange.com/questions/133076/… $\endgroup$ – user114539 Apr 13 '15 at 13:35

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