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Is Z[1/5] a PID?

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine if Z[1/5] is a PID (Z is the ring of integers).

    2. Relevant equations

    3. The attempt at a solution
    I'm not sure...I tried showing that it isn't by showing that it is not a UFD but that didn't work out well. Maybe there is some way I can show it is a PID by showing it's a Euclidean domain.
  2. jcsd
  3. Aug 19, 2009 #2
    Is it or isn't it? Usually you would try to show it isn't, which you did, without success, so maybe it is. Another clue: what are the easiest examples (in your book or notes) of integral domains that are not PID's? Are they more complicated than Z[1/5]? I bet they are. So that's a clue that this one is.

    Try to show it is. I would just try to show it directly. Two hints. First, what do elements of Z[1/5] look like? Second, how do you even show (directly) that Z is a PID?
  4. Aug 19, 2009 #3
    Thanks for the reply and hints! I was able to prove it directly without too much trouble: take an ideal I in Z[1/5]. Intersecting it with Z gives an ideal in Z, which is generated by some p in Z (since Z is a PID). Its easy to show that p generates I by taking an element r in I, multiplying by 5^N for some N to clear denominators so 5^N*r is in I intersect Z (so 5^N*r=kp). Then r = 5^(-N)kp is in the ideal generated by p.
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