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Let $R \subset S$ be an integral extension of commutative rings. Then if $P \subset R$ is prime, there exists a prime ideal $Q \subset S$ such that $Q \cap R = P$.

My D&F book says look at Corollary 50, but I cannot find it in the book!

Is the proof easy or does it require much build-up lemmas?

Hints, or location of Corollary 50, please.

user 1
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Daniel Donnelly
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2 Answers2

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Here is the proof: Theorem 5.9 of the book "Algebra" by "Hungerford".
you can also find a proof in many books about Commutative Algebra; for example:

  • 5.10 of the book Introduction To Commutative Algebra by Atiyah Macdonald

  • 13.34 of the book Steps in Commutative Algebra by Sharp

user 1
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I can do some of the proof on my own maybe :)

Let $P \subset R$ be an ideal. Then in particular $P$ is a subring of $R$ as well as $S$.

Let $Q = \{ s \in S : s \text{ is integral over } P\}$. Then it too is a subring of $S$ and in particular an additive subgroup and on its way to being an ideal.

Buy an earlier theorem, $Q = Pt_1 + \dots Pt_n$ must be a finitely generated $P$-module. Then so to is $sQ$ for any $s \in S$, namely $Q = P(t_1 s) + \cdots + P(t_n s)$. Thus $Q$ is an ideal of $S$. Since $R \subset S$, the contraction of $Q$ to $R$, $Q \cap R$ is an ideal, being the inverse image of a hom that includes $R$ in $S$.

Notice that $Q \cap R \supset P$, thus there is a surjective hom $R/P \to R/(Q\cap R)$. I don't know how to prove that $Q \cap R$ is also prime, maybe it's not!

Daniel Donnelly
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