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In $Z[x]$, which pairs of $a,b$ commute as modulo operators, such that ($c$ mod $a$) mod $b =$ ($c$ mod $b$) mod $a$ for every $c$?

What I got so far is:

Clearly the equasion holds for every pair where one polynomial divides the other.

Another group of pairs are these with $deg(a)=0$ and lead_coeff$(b)=1$.

However, especially the second group seems pretty arbitrary, and it feels like I am missing the bigger picture. Is there a neccessary and sufficient condition for $a,b$?

Max
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  • What do you mean by "c mod a" preecisely? What is x^2 mod 3x for example? – quid Aug 05 '14 at 12:09
  • @quid c mod a is defined like in Z: The smallest positive r such that c=ba+r. In other words, you do polynomial division and output the rest. 3x never fits into x^2 in Z[X], so x^2 mod 3x=x^2. A better example would be `(7x^2-3) mod (2x+1): (7x^2-3)=(2x+1)(3x-2)+(x^2+x-1) -> result:(x^2+x-1)`. – Max Aug 05 '14 at 12:45
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    $\Bbb Z[x],$ does not enjoy (Euclidean) division-with-remainder. What do you mean by "smallest"? When you divide $x^2$ by $3x$ why is your "remainder" $x^2$ vs. $x^2!-3x?$ Why is the former "smaller"? Perhaps you meant $,\Bbb Q[x],$ vs. $,\Bbb Z[x],?$ – Bill Dubuque Aug 05 '14 at 12:53
  • @BillDubuque By smallest I meant the smallest positive. But I am a little confused right now, because I am unsure if that is the correct definition. I'll try if I can find it, but unfortunately most definitions seem to be given in Q[x], while I infact meant to ask the question for Z[x]. – Max Aug 05 '14 at 13:10
  • @BillDubuque (a mod b) in Z[x] should be the representative of a in Z[x]/b. Can we derive the definition from that? – Max Aug 05 '14 at 13:19
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    What is your procedure for choosing "the representative" element of the coset $, a + b, \Bbb Z[x],?,$ You cannot say choose the "smallest positive" element without first defining what that means. – Bill Dubuque Aug 05 '14 at 13:22
  • I don't know, that's why I'm asking. I figured there is a definition for representatives that can be applied here. – Max Aug 05 '14 at 13:24
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    There are effective ways of computing in quotients of $,\Bbb Z[x],,$ but they are much more complex than the Euclidean algorithm. Look up Hermite(-Smith) normal forms. See e.g. this answer. – Bill Dubuque Aug 05 '14 at 13:27

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