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Let $R$ commutative ring with unity, $S\subseteq R$ subring, $p$ minimal prime ideal of $S$. Show there exists a minimal prime ideal $q$ in $R$ with the property that the contraction $q^c=q\cap S=p$.

First of all, I am not sure whether the minimality of $q$ refers to the prime ideals in $R$ or to the prime ideals with such contraction property, if the latter is the case, then we only need to show the existence of such prime ideal, minimal one would be given by Zorn's lemma. Second, if we drop the minimality condition, then the proposition clearly doesn't hold ($\mathbb{Z}\subseteq\mathbb{Q}$ for instance), so this condition must be crucial here.

Kaa1el
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  • The distinction doesn't matter, since once we have one prime ideal $q$ with the desired property, we can choose any minimal prime contained in $q$. By minimality of $p$, this still has the property we want. – Andrew Dudzik Sep 18 '14 at 03:57
  • Geometrically your property means that that the the morphism of affine schemes $Spec(S)\to Spec(R)$ is dominant i.e. that its image is dense (This is not a proof, just a visualization). – Georges Elencwajg Sep 18 '14 at 08:39
  • @GeorgesElencwajg My approach to this problem very much came from the discussion on this question, which was concerned with dominant morphisms (and the original solution used Chevalley). – Andrew Dudzik Sep 18 '14 at 18:58

1 Answers1

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It is enough to find a single prime ideal $q$ satisfying the condition; since any smaller prime ideal will also satisfy the condition (by minimality of $p$), we can find the desired minimal prime using Zorn.

Let $q$ be an ideal of $R$, maximal with respect to the property that $q\cap S \subset p$ (such a $q$ exists because $(0)$ satisfies the condition). It is not so hard to show that such a $q$ is prime, so we are done by minimality of $p$.

(I don't want to write out the last proof in full, since it's equivalent to an easier one with localizations: we have $R_p\neq 0$, so it has a prime ideal; let $q$ be its preimage in $R$, which has the desired property because everything in $S\setminus p$ gets mapped to a unit)

user26857
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Andrew Dudzik
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    Since $p$ may not be a prime in $R$, can we do the localization in this way? – Kaa1el Sep 18 '14 at 13:40
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    @Kaa1el $R$ is an $S$-module, and we can localize any $S$-module at a prime of $S$. Put another way, $S\setminus p$ is a multiplicative subset of $R$, and we have $R_p = (S\setminus p)^{-1} R$ – Andrew Dudzik Sep 18 '14 at 18:49