# Unique minimal prime over an ideal

Krull's principal ideal theorem:

Let $A$ be a Noetherian ring. Let $\mathfrak p$ be a prime ideal in $A$ of height $r$. Then there exist elements $a_1,...,a_r \in \mathfrak p$ such that $\mathfrak p$ is minimal over $(a_1, . . . ,a_r)$ and further $\mathrm{ht}(a_1,...,a_i) = i$ for every $i, 0 ≤ i ≤ r$.

I wanted to prove that if $A$ is a Noetherian ring and $\mathfrak p$ is a prime ideal in $A$ of height $r$, then there exist $a_1,...,a_r,a_{r+1} \in \mathfrak p$ such that $\mathfrak p$ is the unique minimal prime of height $r$ over $(a_1, . . . ,a_{r+1})$.

I am unable to prove it for the case $r=0$, i.e. if $\mathfrak p$ is a prime of height $0$ then there exist a element $a∈\mathfrak p$ such that $\mathfrak p$ is the unique minimal prime of height $0$ over $(a)$. Any hint or ideas?

• Your question is an immediate consequence of prime avoidance: two prime ideals of the same height can't be contained one into another, so $\mathfrak p$ is not contained into the union of the other height $r$ primes containing $(a_1,\dots,a_r)$. That's all. – user26857 Sep 27 '16 at 7:43

Let $\mathfrak q_1,\dots,\mathfrak q_s$ be the set minimal primes of height $r$ over $(a_1,...,a_r)$. WLOG, let $\mathfrak q_1=\mathfrak p$. Then $\mathfrak p\not\subset\mathfrak q_i$ for each $i, 2 ≤ i ≤ s$. Hence by prime avoidance lemma, $\mathfrak p\not\subset\bigcup\limits_{i=2}^{s}\mathfrak q_{i}$, i.e. there exists $a_{r+1}$ $\in \mathfrak p$ such that $a_{r+1}$ $\not\in\bigcup\limits_{i=2}^{s}\mathfrak q_{i}$.
Now, $\mathfrak p$ is a minimal prime over $(a_1,...,a_{r+1})$ and $\mathrm{ht}(\mathfrak p) =r$. Let $\mathfrak p'$ be a minimal prime of height $r$ over $(a_1,...,a_{r+1})$. Then there exists $i, 1 ≤ i ≤ s$ such that $\mathfrak q_i\subset\mathfrak p'$ and since $\mathrm{ht}(\mathfrak p')=\mathrm{ht}(\mathfrak q_i)$ = $r$ $\implies$ $\mathfrak q_i=\mathfrak p'$. But $a_{r+1}\not\in\mathfrak q_i$ for each $i, 2 ≤ i ≤ s$ $\implies$ $\mathfrak p'=\mathfrak q_1=\mathfrak p$.