- 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.

```
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.