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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.

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up vote 3 down vote accepted

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.

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thank you for your help, I really appreciate it, do you know whether the converse is true or not? – i.a.m Mar 19 '13 at 1:30
The classical counterexample when $R$ is not reduced is the following: let $k$ be a field, let $R=k[x,y]/(x^2, xy)$, and let $Q=xR$. Then $Q$ is maximal ideal of height $1$, but all its elements are zero-divisors. The keyword for this is associated prime ideals. – user18119 Mar 19 '13 at 16:58
@QiL'8 The height of $xR$ is zero. If $Q=(x,y)$ then it's ok. A primary decomposition of $(x^2,xy)$ is $(x)\cap(x^2,y)$, so $Q$ is associated of height $1$. – user26857 Mar 19 '13 at 17:22
@YACP: you are right, $Q=(x,y)$. Thanks ! – user18119 Mar 19 '13 at 17:27

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