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I am reading Introduction to Commutative Algebra / Atiyah & Macdonald, Theorem 5.11 ("Going-up theorem").

The statement is:

Let $A \subset B$ be rings, $B$ integral over $A$; let $p_1 \subset \dotsm \subset p_n$ be a chain of prime ideals of $A$ and $q_1 \subset \dotsm \subset q_m$ (m < n) a chain of prime ideals of $B$ such that $q_i \cap A = p_i$ ($1 \leq i \leq m$). Then the chain $q_1 \subset \dotsm \subset q_m$ can be extended to a chain $q_1 \subset \dotsm \subset q_n$ such that $q_i \cap A = p_i$ for $1 \leq i \leq n$.

I read the proof and I don't think the fact that $q_1$ (and hence $p_1$) is prime is used.

My question:

Does the conclusion still hold if we omit the assumption that $q_1$ (and $p_1$) is prime?

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    A related question: http://math.stackexchange.com/questions/150295/a-counterexample-to-the-going-down-theorem2012-05-27

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