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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: math.stackexchange.com/questions/150295/… –  KCd May 27 '12 at 9:10

1 Answer 1

You are right, we donot need that $\mathfrak{q}_1,\ldots,\mathfrak{q}_m$ are prime. In the proof, we need $\mathfrak{p}_{i+1}$ where $i+1\geq m+1$ is prime.

For example, $m=1$, we need $\mathfrak{p}_2$ is prime, and then it follows $B/\mathfrak{q}_1$ is integral over $A/\mathfrak{p}_1$. By lying over theorem, we can find a prime ideal $\mathfrak{q}_2\subset B$ such that $\mathfrak{q}_2\supset \mathfrak{q}_1$ and lie over $\mathfrak{p}_2$.

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