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Theorem: If A is a Noetherian local ring and A[x] catenary, then A is formally catenary.

In the proof, it is assumed that A is integral domain and A* (the completion of A) is not equidimensional and then Lemma 3 is used.

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(Lemma 3. Let (R,m) be a catenary Noetherian local integral domain, and let R* be its completion. $\dim R = n$ and Q is a minimal prime of R* such that $1 < \dim R^*/Q=d < n$. Then for $i=1,2,...,d-1$, the set $Φ_i=\{p\in spec\ R \mid ht\ p=i, ∃ P\in Min Ass(R^*/pR^*); Q\subset P, \dim R^*/P=d-i\}$ is non-empty.)

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Why we have the conditions to use Lemma 3? (Why is there such Q?)

user26857
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azna
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1 Answers1

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I think you missed that $A[X]$ catenary implies $A$ catenary.

A prime ideal $Q$ such that $\dim A^*/Q<n$ there exists since $A^*$ is not equidimensional, and $\dim A^*=\dim A=n$. Instead, the case $\dim A^*/Q=1$ should be treated separately. (For this I suggest you to take a look at http://stacks.math.columbia.edu/tag/0AW1.)

user26857
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