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

Assume $R$ is Noetherian.
• By Krull's principal ideal theorem,we have that given a principal ideal $I = (\alpha)$, all minimal primeideals $\mathfrak p$ above $I$ has height at most 1.
• Recall that a minimal prime ideal $\mathfrak p$ lying over an ideal $I$is the minimal among all prime ideals containing $I$. That is, if$I \subseteq \mathfrak q \subseteq \mathfrak p$, then $\mathfrak q = I$or $\mathfrak q= \mathfrak p$.
• In our case, we have that $R$ is a PID. We are trying to show that all primeideals are maximal. Consider a prime ideal $\mathfrak p \subseteq R$.It is a principal ideal since $R$ 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 ($\mathfrak p = 0$),then it has height zero.
• If it is any other prime ideal $(\mathfrak p \neq (0))$, then it has heightat least 1, since there is the chain $(0) \subsetneq \mathfrak p$. 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 $Spec(R)$ have height 1.
• Thus, every point of $Spec(R)$ is maximal, as there are no "higher points"that cover them.
• Hence, in a PID, every prime 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.