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Primes and Associates in Rings

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Let a, b be members of a commutative ring with identity R. If a is a a prime and a, b are associates then b is also prime. True/False


    2. Relevant equations
    Definitions: a is prime if a|xy implies a|x or a|y
    a and b are associates if there exists a unit u s.t a=bu


    3. The attempt at a solution
    First off I'm not sure whether its even true or not; so I've tried to both prove it and find a counter example:
    Attempt at a proof:
    let b|xy.
    Then if b = au for some unit u, au|xy, that is for some k in R auk = xy.
    so ak divides xy(u^(-1))
    but k does not necessarily have an inverse so this doesn't really get me anywhere.

    I've shown in an earlier question that if a is irreducible and a, b are associates then b is irreducible. Prime implies irreducible but not vice versa so if there is a counter example it will depend on an element of R that is irreducible but not prime. I know examples of these can be found in rings such as Z(5i) (the set of numbers a+5bi for a, b in Z). But I don't know how to find units of this ring.
    Consider (a+5bi)(c+5di) = 1.
    This gives the simultaneous equations 1+25bd-ac = 0 = ad+bc
    But so far I have not managed to find any integer solutions (and to be honest I don't know any way of doing this other than trial and error).

    Thanks for any help!
     
  2. jcsd
  3. Sep 20, 2011 #2

    micromass

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    OK, so stop here. We know that auk=xy. So this implies that [itex]ak=u^{-1}xy[/itex]. Thus a divides [itex]u^{-1}xy[/itex]. Now use that a is prime.
     
  4. Sep 21, 2011 #3
    Thank you! Done it now :)
     
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