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Let $R=Z[x_{1},x_{2},\dots,x_{n}]$ be a multivariate polynomial ring. Is it possible to define a normal form for a general $m \times m$ matrix $M$ with entries from $R$?

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    $\begingroup$ A normal form up to what equivalence relation? $\endgroup$ – Qiaochu Yuan Mar 8 '12 at 15:48
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See this question -- the answer contain all the references you might ever want.

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  • $\begingroup$ @IgorRivin: Thankyou. There are many terminologies I do not understand. Which of those references should I focus on for the case in my question? $\endgroup$ – T.... Mar 9 '12 at 18:00
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    $\begingroup$ You should look at the early papers of Helmer and Kaplansky (cited in the last two answers), and go on from there. $\endgroup$ – Igor Rivin Mar 9 '12 at 20:50
  • $\begingroup$ Theorem $5.2$ in here seems to answer what I am asking is impossible! ams.org/journals/tran/1949-066-02/S0002-9947-1949-0031470-3/… Is my interpretation correct? $\endgroup$ – T.... Mar 11 '12 at 7:25
  • $\begingroup$ I am guessing that's right, but of course I don't know exactly what you are looking for :) $\endgroup$ – Igor Rivin Mar 11 '12 at 19:17
  • $\begingroup$ Hi Igor. Thankyou. The paper says to prove diagonal form for general m x m matrices, it suffices to show for 2 x 2. Say we have [x1 x2; x3 x4] (first row x1 x2 and second row x3 x4), it looks like it is not possible to find 'one' formula for smith form. However, if we extend Z[x1,x2,x3,x4] to Z[x1,x2,x3,x4,x2^-1,x3^-1]. We can premultiply by [1 (x3^z(X3))^-1; 0 1] where z(X3) = 0 if X3=0 else it is 1. That way we can change the given matrix to [x1 x2; 1 x4x3^-1] or [x1 x2; 0 x4] if we define (0^0)^-1 = 1^-1. This should be able to take care of the issue to get 'one' formula. Isn't it? $\endgroup$ – T.... Mar 12 '12 at 17:29

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