# 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?

• Please use \operatorname{Spec} to format $\operatorname{Spec}$. I've made the upgrade in this post for you. Commented Jul 27, 2020 at 9:09
• Is the statement about maximality true? Consider in $\mathbb{Z}[x]$ the ideal generated by an irreducible polynomial. But probably the question is not clear to me... Commented Jul 27, 2020 at 9:11
• Your statement is true in a PID for non zero prime ideals. Is this the case you are considering? Commented Jul 27, 2020 at 9:14
• The spectrum, as a topological space, doesn’t see whether a prime is principal or not, so I don’t think you can prove this without talking about rings. Commented Jul 27, 2020 at 9:16
• @SabinoDiTrani The statement holds more generally, just as the questioner stated: Every non-zero principal prime in a domain is a maximal principal ideal, that is maximal amongst all principal ideals. Commented Jul 27, 2020 at 9:36

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.