1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Showing a localization is a principal ideal domain (non-trivial problem)

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]R[/itex] be an integral domain. Say a prime [itex]p \in R[/itex] is small if [itex]a\in\bigcap\limits _{n=1}^{\infty}\left\langle p^{n}\right\rangle = \left\langle 0\right\rangle[/itex]

    Show that if [itex]p[/itex] is a small prime and [itex]D = R \setminus \left\langle p\right\rangle[/itex] then [itex]R_D[/itex] is a principal ideal domain.

    2. Relevant equations

    Some basic facts... ED implies PID, field implies PID, PID implies UFD. Localization is a PID if R is a PID, etc.

    3. The attempt at a solution

    I tried using First Isomorphism Theorem to show [itex]R_D[/itex] is a field; I quickly shot this down with a counterexample by taking R to be the integers, and since every prime is small in the integers, taking p = 2 shows that 2/3 in the localization has no inverse.

    The only other approach I can think of that can be used when I know squat about the ring is to show it is an ED. But I can't find a valid Euclidean function.

    Any ideas? I don't want solutions, obviously... just a tiny hint to push me in a promising direction.
    Last edited: Jun 6, 2012
  2. jcsd
  3. Jun 6, 2012 #2
    Ok, can you show that an element x that is not invertible in [itex]R_D[/itex] can be written as py.

    In general, can you show that if x is not invertible, then it can be written as [itex]p^ny[/itex] with y invertible.
  4. Jun 6, 2012 #3
    Thank you for the hint! Will see where it takes me.
  5. Jun 7, 2012 #4
    I think I have a proof involving finite descent; I eventually show that any ideal I is either equal to the ideal generated by some power of p (with 1 in the denominator), or we reach the point where the ideal generated by p is contained in I. Since I was able to show any proper ideal must be contained in (p), I am done... That is, any ideal is of the form (p^n) or the ring itself.

    Sry for the lack of LaTex, I'm typing on a tablet.

    Was there a more elegant approach?
  6. Jun 7, 2012 #5
    That's indeed what I had in mind.
  7. Jun 7, 2012 #6
    Thanks very much, your hint was very concise and well chosen. All the best.
  8. Jun 7, 2012 #7
    I am still confused. Could you please explain it in steps?
  9. Jun 7, 2012 #8
    msg me if you want some hints
    Last edited by a moderator: Jun 14, 2012
  10. Jun 7, 2012 #9
    or micromass for that matter, he was the architect
  11. Jun 7, 2012 #10
    You will have to show us what you tried. Did you prove my post #2?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook