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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

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

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