Assume $R$ is Noetherian.

• By Krull's principal ideal theorem , we have that given a principal ideal $I = (\alpha)$, all minimal prime ideals $\mathfrak p$ 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 prime ideals are maximal. Consider a prime ideal $\mathfrak p \subseteq R$. It is a principal ideal since $R$ is a PID. It is also a 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 height at least 1, since there is the chain $(0) \subsetneq \mathfrak p$. Thus by Krull's PID theorem, it has height exactly one.
• So all the prime ideals other than the zero ideal, that is, all the points of $Spec(R)$ 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.

§ References

• Andreas Gathmann's notes