Let $R$ be a noetherian ring, $x\in R $ be a non zero divisor, and $P$ a prime ideal of $R$ which is minimal over $(x)$. I'm trying to show that $\operatorname{ht}P=1$. Also if $Q$ is a prime ideal of height $1$, is there a non zero divisor in $Q$?
I know that by Krull's principal ideal theorem that for any non-unit $x$ we have $\mathrm{ht}\;P\le 1$ but I can't show the equality if this $x$ is not a zero divisor.
I'm stuck. Any ideas? Need your help. Thank you.
Suppose $P$ is of height zero, i.e., a minimal prime of $R$. Then $R_P$, the localization of $R$ at $P$, is a Noetherian local ring of dimension zero, that is, an Artin local ring. The maximal ideal $PR_P$ is therefore nilpotent, so the image of $x$ in $PR_P$ is nilpotent, which means there exists $n\geq 1$, an integer, with $x^n/1=0$ in $R_P$. This means there is some $s\notin P$ with $sx^n=1$. Then $s\neq 0$, and if we choose $n$ minimal, i.e., so that $x^{n-1}/1\neq 0$ in $R_P$, then $sx^{n-1}\neq 0$ but $(sx^{n-1})x=sx^n=0$, so $x$ is a zero divisor, contrary to assumption.
To answer your second question: yes, at least when $R$ is reduced (I'm not sure without this hypothesis). When $R$ is reduced, the set of zero divisors is equal to the union of the minimal primes. Since $R$ is Noetherian, there are finitely many minimal primes, say $P_1,\ldots,P_r$. Now if $Q$ has height one but consists entirely of zero divisors, then $Q\subseteq P_1\cup\cdots\cup P_r$, so by prime avoidance, $Q\subseteq P_i$ for some $i$, a contradiction. So $Q$ must contain a non-zero-divisor.
