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Let $R$ be a commutative Noetherian ring and $a_1,\ldots,a_n,b_1,\ldots,b_n\in R$ be elements such that the heights of the ideals $A_m:=(a_1,\ldots,a_m)$ and $B_m=(b_1,\ldots,b_m)$ are equal to $m$ for each $m\in\{1,\ldots,n\}$. I want to find elements $c_1,\ldots,c_n\in R$ so that $c_m\in A_m\cap B_m$ and the height of the ideal $(c_1,\ldots,c_m)$ is equal to $m$ for each $m\in\{1,\ldots,n\}$.

Note that we can choose $c_1=a_1b_1$ since if $P$ is any minimal prime ideal over $(c_1)$ then $a_1\in P$ or $b_1\in P$ which means $A_1\subseteq P$ or $B_1\subseteq P$ so that the height of $P$ is at least $1$.

How do we choose $c_2,\ldots,c_n$?

user26857
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jgeilberg
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1 Answers1

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Induction on $n\ge1$.

For the case $n=1$ set $c_1=a_1b_1$.

Inductively there exist $c_1, \dots, c_{n-1}$ such that $c_i \in (a_1, \dots, a_i) \cap (b_1, \dots ,b_i)$ and $\operatorname{ht}(c_1,\dots,c_i)=i$ for all $i$, $1\le i\le n-1$. The height of ideals $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ is $n$. Thus $(a_1,\dots,a_n)\nsubseteq\mathfrak p$ and $(b_1,\dots,b_n)\nsubseteq\mathfrak p$ for every prime $\mathfrak p$ minimal over $(c_1, \dots, c_{n-1})$. So there exist $d\in(a_1,\dots,a_n)$ and $d' \in (b_1,\dots,b_n)$ such that neither $d$ or $d'$ are contained in any minimal prime over $(c_1,\dots,c_{n-1})$. Thus $c_n=dd'$ is the desired element.

user26857
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    Now it's clear to me, thank you. Just one thing to add in case anyone else wonders about it: existence of $d$ and $d'$ is guaranteed by the prime avoidance lemma and the fact that in a Noetherian ring there are only finitely many minimal primes over an ideal. – jgeilberg Sep 07 '15 at 16:51
  • @jgeilberg Right. – user26857 Sep 07 '15 at 16:51