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Example of PID and its maximal ideal

  1. Mar 21, 2012 #1
    I got one example on my notes about PID and maximal ideal. I feel it is a strange example as it doesn't make sense to me, and there are no explanations. It says:

    For a prime p[itex]\in[/itex][itex]\mathbb{N}[/itex], denote by [itex]\mathbb{Z}_{(p)}[/itex] the subring of [itex]\mathbb{Q}[/itex] given by
    [itex]\mathbb{Z}_{(p)}[/itex]={[itex]\frac{m}{n} \in[/itex][itex]\mathbb{Q}[/itex]|p does not divide n}.
    Then [itex]\mathbb{Z}_{(p)}[/itex] is a PID, and it has exactly one maximal ideal.

    I can't see the reason of this example at all, and I'm not able to imagine what are the ideals like in Z_(p), can anyone please explain to me why it is a PID and only has one maximal ideal? Thanks a lot.
  2. jcsd
  3. Mar 21, 2012 #2
    Show that all ideals are of the form

    [tex]\{\frac{m}{n}\in \mathbb{Z}_{(p)}~\vert~m\in I\}[/tex]

    for I an ideal in [itex]\mathbb{Z}[/itex].
  4. Mar 21, 2012 #3


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    intuitively, the more units there are, i.e. the more denominators are allowed, the fewer ideals there are, since any ideal containing a unit is the whole ring.

    so you started from a pid, namely Z, and added in a lot of denominators, i.e. anything not divisible by p, so you lost a lot of ideals, and the remaining ones are probably still generated by the same principal generators as before.

    i.e. try intersecting an ideal of your ring with Z, and see if the generator of that ideal also generates your original ideal.

    anytime the set of non units itself forms an ideal, that ideal contains all others. what are the non units here?
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