# $x \in R$ nonzero divisor $\Rightarrow$ $\operatorname{height}((x))=1$

Let $R$ be a Noetherian ring, and $x \in R$ be a nonzero divisor, I want to show $$\mathrm{height}((x)):=\inf\{\mathrm{height}(\mathfrak{p})|(x) \subset \mathfrak{p}\}=1.$$ By Krull's theorem, one only need to show: any prime $\mathfrak{p}$ contains $x$ is not a minimum prime of $R$.

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Is there a condition on the $\mathfrak{p}$ on your definition? (E.g., are they supposed to be prime ideals?) –  Arturo Magidin Jan 24 '12 at 21:14
You can use the following fact: the union of the minimal primes of $R$ is contained in the set of of zero divisors of $R$. This is part of exercise 9 of Chapter 3 of Atiyah-Macdonald. –  Bruno Joyal Jan 24 '12 at 22:11
Yes, thank you! –  Li Zhan Jan 26 '12 at 17:53

## 1 Answer

We agree that by Krull's Hauptidealsatz, we already have $height(x)\leq1$, whether $x$ is a zero-divisor or not.

On the other hand, in a noetherian ring $R$ the set of zero-divisors is exactly the union of the prime ideals belonging to the zero ideal $(0)$.
The minimal prime ideals of $R$ are associated to $(0)$ and so if $x$ is not a zero divisor it will be contained in no minimal prime ideal of $R$ and thus $height (x)\geq 1$, which is what you wanted.

The assertion on zero-divisors is true under the sole hypothesis that $(0)$ has a primary decomposition ( automatic in a noetherian ring) and can be found in Atiyah-Macdonald's Commutative Algebra, page 53.

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Thank you so much! I just found a general statement, and I'll put here for the convenience of me and others: Let $A$ noetherian ring and $M$ finite generated $A$ module. Then $\cup_{p \in Ass(M)}{p}=\{zero\ divisor \ of\ M\}\cup\{0\}$. This is Theorem 3.1 in Eisenbud's commutative algebra. –  Li Zhan Jan 26 '12 at 17:54