Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension $d$. Show for any chain of prime ideals $\mathfrak p_0\subsetneq…\subsetneq\mathfrak p_n$ with $a∈\mathfrak p_n$ that there is a chain of prime ideals $\mathfrak p_1'\subsetneq…\subsetneq \mathfrak p_n'$ with $a\in\mathfrak p_1'$ and $\mathfrak p_n=\mathfrak p_n'$.
It hints to proceed by induction on the length of chain. The base case $n=1$
is true by assumption.
Best Answer
I found this technique in another text in the proof of Krull's generalised principal ideal theorem and immediately recalled this question:
Assume by induction that for chains of fewer than $n$ prime ideals the result holds. Consider a chain $\mathfrak p_0\subsetneq ...\subsetneq \mathfrak p_n\ni a$. If $a\in \mathfrak p_j$ for $j<n$ already, then the induction hypothesis implies it. So suppose $a\in \mathfrak p_n$ and $a\not\in \mathfrak p_{n-1}$.
Look at the quotient $R/\mathfrak p_{n-2}$. Then we have $$0=\mathfrak p_{n-2} / \mathfrak p_{n-2} \subsetneq (\mathfrak p_{n-2}+aR) / \mathfrak p_{n-2}\subseteq\mathfrak p_{n}/\mathfrak p_{n-2}.$$
By Krull's principal ideal theorem, if a prime $P$ is minimal over $a$, then its height is $\leq 1$. This means that $\mathfrak p_n /\mathfrak p_{n-2}$ cannot be minimal over $a+\mathfrak p_{n-2}$. Hence there exists some prime ideal $\mathfrak p_{n-1}'$ such that $\mathfrak p_{n-2}\subsetneq \mathfrak p_{n-2}+aR\subseteq \mathfrak p_{n-1}'\subsetneq \mathfrak p_n$. In particular, $a\in \mathfrak p_{n-1}'\subsetneq \mathfrak p_n$ and the induction hypothesis gives the result.