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This is a follow up to my previous question here

if $A \subset B$ is a finite ring extension and $P$ is a prime ideal of $A$ show there is a prime ideal $Q$ of $B$ with $Q \cap A = P$. (M. Reid, Undergraduate Commutative Algebra, Exercise 4.12(i))

We have seen that $PB \not= B$ and I found an example where $PB \cap A \not = A$ (the example is $P=(2)$, $B=\mathbb Z[\tfrac{1}{2}]$). This makes me think $Q$ must be a subset of $PB$ instead of my original idea: the maximal ideal containing $PB$.

So I would like to ask advice on:

How should we find a prime ideal $Q$ of $B$? and show that $Q \cap A = P$?

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    $\begingroup$ Your original ideal is the natural one. $\endgroup$ – user26857 Jan 23 '16 at 21:43
  • $\begingroup$ One can prove that $PB\cap A=P$. $\endgroup$ – user26857 Jan 23 '16 at 21:54
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Suppose that $P$ is a maximal ideal, and let $Q\subset B$ be a maximal ideal containing $PB$. Then $Q\cap A\supseteq P$, so $Q\cap A=P$.
If $P$ is not maximal, then localize at $A\setminus P$ and find a finite ring extension $A_P\subseteq B_P$. Now use the previous result for the maximal ideal $PA_P$.

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  • $\begingroup$ I guess that Reid had something else in his mind, as the rings of fractions are introduced later, in Chapter 6. $\endgroup$ – user26857 Jan 23 '16 at 21:46
  • $\begingroup$ Nice idea to assume $P$ maximal! I don't see how to use the localization but I'm happy with a proof that works for dim 1 rings for now. Thank you kindly! $\endgroup$ – Brennan.Tobias Jan 23 '16 at 22:06
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    $\begingroup$ @Brennan.Tobias You are welcome. (Unfortunately I've never seen a proof of that result without using localization.) $\endgroup$ – user26857 Jan 23 '16 at 22:08

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