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Let $S$ be a commutative ring with $1$ and let $R$ be any subring of $S.$ Let $p$ be a minimal prime of $R.$ Then how can I show that there is a prime ideal $q$ of $S$ whose contraction is $p,$ i.e., $q \cap R = p.$

What I am trying to show is the necessary and sufficient condition that $p=p^{ec}.$ Clearly $p \subset p^{ec}.$ But the other way I couldn't complete. I need some help.

user371231
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  • Regarding the notation $p=p^{ec}$, I suspect you have in mind taking the extension of ideal $p\subset R$ to $S$ (finite sums of products of elements from $p$ times elements of $S$), followed by the contraction of that ideal (not necessarily a prime in $S$ even when $p$ is prime in $R$) back to $R$ by intersection, hence $(p^e)^c$. – hardmath Oct 23 '18 at 20:28
  • Yes, you are right. – user371231 Oct 23 '18 at 20:37
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    Hint. Pick a prime ideal in the ring of fractions $(R\setminus p)^{-1}S$. – user26857 Oct 24 '18 at 18:41
  • Yes that helps, follows from the immediate commutative diagram. – user371231 Oct 24 '18 at 19:29

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