Prime ideals are maximal among principal ideals: geometry? The claim is for a domain $R$, among principal ideals of the form $(r)$ for $r \in R$, the principal ideals which are prime are maximal among principal ideals.
That is, we have $(p)$ a principal ideal which is also prime, $p \neq $0. If $(p) \subseteq (a)$ then either $(a) = (p)$ or $(a) = R$.
The proof is quite short:

*

*Since $(p) \subseteq a$ we have $p = ar$.

*Since $ar = p \in (p)$ and $(p)$ is prime, either $a \in (p)$ or $r \in (p)$.

*Case 1: $a \in (p)$. we get $(a) \subseteq (p)$. Combined with the assumption that $(p) \subseteq (a)$ we get $(a) = (p) ~ \square$

*Case 2: $r \in (p)$. This means that $r = ps$. Hence $p = ar = a(ps) = (as)p$. Thus $p - (as)p = 0$, or $p(1 - as) = 0$. Since $p \neq 0$, $R$ is a domain, we have $as = 1$: $a$ is a unit in $R$. So $(a) = R ~ \square$
I wish to understand the above proof in terms of $\operatorname{Spec}(R)$.

*

*We have that $(p)$ is a generic point of $\operatorname{Spec}(R)$. it also corresponds to the equation $p = 0$

*We next take the ideal $(a)$, which corresponds to the equation $(a) = 0$. But this ideal need not be prime, and is thus not part of the prime spectrum $\operatorname{Spec}(R)$. How to we proceed from here?

In general, I want to re-learn all basic ideal theory in terms of algebraic geometry and spectrum. Is this always possible?
 A: 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, 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.
