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I believe that I should show the forward direction by first showing the factorization of two polynomials, f and g, such that f=p1 . . . ps and g=q1 . . . qs, where each pi and qj are irreducible polynomials. And I also know that the coefficients can only be 0 or 1 in Z/2Z. I am not sure where I should go from here to show that each of the pi=qj.

mmm
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    Does associate just mean differ by multiplication by a unit? If so, just think about what the units are in Z/2Z[x] – Seth Apr 16 '14 at 23:09
  • Okay, then the only unit in Z/2Z is 1, therefore, for the polynomials to be associates that means that one is just a multiple of the other by 1 and therefore, is the same polynomial. – mmm Apr 16 '14 at 23:11
  • Yes, thats what I was thinking. – Seth Apr 16 '14 at 23:11

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Hint $\ $ If $\,1\,$ is the only unit in the domain $\,D\,$ then it is the only unit in $\,D[x]\,$ since both domains have the same units. More generally see here on the units of $\,R[x].$

Bill Dubuque
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