§ In a PID, all prime ideals are maximal, geometrically

Assume RR is Noetherian.
  • By Krull's principal ideal theorem,we have that given a principal ideal I=(α)I = (\alpha), all minimal primeideals p\mathfrak p above II has height at most 1.
  • Recall that a minimal prime ideal p\mathfrak p lying over an ideal IIis the minimal among all prime ideals containing II. That is, ifIqpI \subseteq \mathfrak q \subseteq \mathfrak p, then q=I\mathfrak q = Ior q=p\mathfrak q= \mathfrak p.
  • In our case, we have that RR is a PID. We are trying to show that all primeideals are maximal. Consider a prime ideal pR\mathfrak p \subseteq R.It is a principal ideal since RR is a PID. It is alsoa minimal prime ideal since it contains itself. Thus by Krull's PID theorem, has height at most one.
  • If the prime ideal is the zero ideal (p=0\mathfrak p = 0),then it has height zero.
  • If it is any other prime ideal (p(0))(\mathfrak p \neq (0)), then it has heightat least 1, since there is the chain (0)p(0) \subsetneq \mathfrak p. Thusby Krull's PID theorem, it has height exactly one.
  • So all the prime ideals other than the zero ideal, that is, all the pointsof Spec(R)Spec(R) have height 1.
  • Thus, every point of Spec(R)Spec(R) is maximal, as there are no "higher points"that cover them.
  • Hence, every principal ideal is maximal.
In a drawing, it would look like this:
NO IDEALS ABOVE  : height 2
(p0)  (p1) (p2)  : height 1
      (0)        : height 0
So each pi is maximal. This is a geometric way of noting that in a principal ideal domain, prime ideals are maximal.

§ References