Showing there is a prime in a ring extension using Nakayama's lemma

Question

Here's the problem that I'm working on:

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))

There was a hint to use Nakayama's lemma to show $PB \not= B$, for that part what I've done is said $B$ is a finitely generated $A$ module, $P$ is an ideal of $A$ therefore I can apply NAK here and it tells me that: $PB = B$ implies that there exists some $r \in A$ satisfying $r \equiv 1 \pmod P$. I would need to get a contradiction from this but I wasn't able to see how.

Any guidance on whether I am approaching this correctly and how to do the part I am stuck on would be appreciated! Thank you

Answer 1

For the first part, the contradiction comes from the fact that there exists $r\in A$ such $r=1+p, p\in P$ and $rB=0$ this implies $(1+p)A=0$ so for every $a\in A, (1+p)a=a+pa=0$ which implies $a\in P$ contradiction since $P$ is different of $A$.

Answer 2

For the second part:
Since $A \subset B$ is a finite ring extension, $B$ is integral over $A$. Now you can use "LYING-OVER THEOREM", (Atiyah-Macdonald's Theorem 5.10).

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If you are interested in a direct proof you can derive it from propositions in section 5 of the book (Atiyah-Macdonald's book: Introduction To Commutative Algebra) that leads to Theorem 5.10. (see 5.1, 5.6, 5.8 and then 5.10. The proofs suggest using localization.)