# Incomparability of Prime ideals in integral Ring Extensions

I'm having an issue with the theorem on incomparability of prime ideals in integral ring extensions, which I have stated as follows:

Let $R \subseteq S$ be an integral extension of commutative rings (with identity), and suppose that $P_{1} \subseteq P_{2}$ are two prime ideals of $S$ such that $R \cap P_{1} = R \cap P_{2}$, then $P_{1} = P_{2}$

The proof of which requires the following lemma:

Let $R \subseteq S$ be an integral extension of commutative rings (with identity), with $S$ an integral domain. If $I$ is a non-zero ideal of $S$, then $R \cap I \neq \{ 0 \}$.

The proof proceeds as follows:

Let $Q = R \cap P_{1} = R \cap P_{2}$. Then $R / Q \subseteq S / P_{1}$ is an integral extension with $P_{2} / P_{1}$ an ideal of the domain $S / P_{1}$ such that $P_{2} / P_{1} \cap R / Q = Q / Q = \{ \bar{0} \}$, and so by the lemma $P_{2} / P_{1}$ is the zero ideal of $S / P_{1}$. Hence $P_{2} \subseteq P_{1} \subseteq P_{2}$ and so $P_{1} = P_{2}$

My issue with this is this proof only seems to require the primality of $P_{1}$, since the only place that primality is required is to ensure that $S / P_{1}$ is a domain so that we can apply the lemma. Am I correct that the primality of $P_{2}$ is not needed in the statement of incomparability? And if so is there a reason it is stated as needing $P_{2}$ to be prime too?

$\textbf{Edit: }$ I posted this as a comment below, but I thought it might be worthwhile to replicate this here, since it serves as my motivation for asking this question, and provides a concrete example for where this stronger / generalised form of the statement of incomparability is useful. Though, I do acknowledge that, as mentioned below, it's not hard to alter this proof to allow us to apply the typical form of incomparability.

$\textbf{Claim: }$ An integral extension of a Jacobson Ring is a Jacobson Ring.

$\textbf{Proof :}$ Let $R \subseteq S$ be an integral extension of rings and let $I$ be a prime ideal of $S$. Then since $R$ is Jacobson, and $(I \cap R)$ is a prime ideal of $R$, there are maximal ideals of $R$, $\{ M_{i} \}_{i=1}^{l}$ say, such that $(I \cap R) = \cap_{i=1}^{l} (M_{i})$. Then for each $i$ the chain $(I \cap R) \subseteq M_{i}$ is a chain of prime ideals of $R$ with $I$ a prime ideal of $S$ laying over $(I \cap R)$, and so by the going up theorem there are primes ideals of $S$, $\{ N_{i} \}_{i=1}^{l}$ say, each containing $I$ and each lying over its respective $M_{i}$. Then $N_{i}$ is a prime ideal lying over a maximal ideal, so is itself maximal. Then let $I' = \cap_{i=1}^{l} (N_{i})$. Then $I \subseteq I'$ are ideals of $S$ with $I$ prime and $$(I' \cap R) = ( \cap_{i=1}^{l} (N_{i})) \cap R = \cap_{i=1}^{l} (N_{i} \cap R) = \cap_{i=1}^{l} (M_{i}) = (I \cap R)$$ and so by the more general form of incomparability, noticing crucially I have not proved primality of $I'$, we see that $I = I'$. Hence $I$ is an finite intersection of maximal ideals of $S$, and so since $I$ was arbitrary in $S$ amongst the prime ideals of $S$, we see that $S$ is Jacobson.

$\textbf{ Remark: }$ My proof here also uses a more specific form of the going up theorem than the one I was provided with. The statement of the going up theorem I have states that if $\{Q_{i}\}$ is a strictly increasing chain of prime ideals in $R$, then there is a strictly increasing chain of prime ideal in $S$, $\{P_{i}\}$ say, such that $P_{i} \cap R = Q_{i}$. However since the proof of this is inductive, and you start by choosing a prime ideal $P_{1}$ (whose existence is guaranteed by a simpler lemma) in $S$ laying over $Q_{1}$, we see that in fact if we already have a partially completed chain in $S$ we can use going up to complete it. Which is exactly what I have done in the proof above.

You are right that the proof doesn't need that $P_2$ is prime for the statement to be correct, as the proof shows.
As for why the statement is not stated in more generality, I can only speculate, but it's probably because it's usually not needed when you apply the theorem. (I have seen quite a few applications of incomparability, but there $P_2$ was always prime.)
Another possible reason why one might state it that way is that when we only state it for prime ideals, there are geometric reformulations. If you're familiar with the Zariski topology, then the statement can be reformulated as saying that the fibers of the induced map $\operatorname{Spec}(S) \to \operatorname{Spec}(R)$ are discrete. This reformulation doesn't really work for the more general statement.
• Ah okay! Thanks! I was trying to prove that an integral extension of a Jacobson Ring was Jacobson. So I let $R \subseteq S$ be integral with $R$ Jacobson, and let $I$ be prime in $S$, then $(I \cap R) = \cap_{i=1}^{l} M_{i}$ for some $M_{i}$ maximal ideas of $R$. Then by going up there are prime (and hence maximal because $M_{i}$ is maximal) ideals $N_{i}$ of $S$ each containing $I$ such that $N_{i} \cap R = M_{i}$. Then $I' = \cap_{i=1}^{l} N_{i}$ is an ideal of $S$ containing prime $I$ such that $I \cap R = I' \cap R$, so would then conclude $I' = I$. – Adam Higgins Jun 6 '18 at 22:25
• @AdamHiggins oh, you're right, that's a good example where the more general version of incomparability is useful. Of course you also reduce to $I=0$ as in the proof of incomparability or do some other proof, but your proof is nice. – MatheinBoulomenos Jun 6 '18 at 22:36