4
$\begingroup$

Let $R$ be a Noetherian ring, and $x \in R$ be a nonzero divisor. I want to show $$\operatorname{height}\, (x):=\inf\{\operatorname{height}\mathfrak{p}\mid (x) \subseteq \mathfrak{p}\}=1.$$

By Krull's theorem, one only need to show: any prime $\mathfrak{p}$ contains $x$ is not a minimum prime of $R$.

  • 0
    Is there a condition on the $\mathfrak{p}$ on your definition? (E.g., are they supposed to be prime ideals?)2012-01-24
  • 2
    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.2012-01-24
  • 0
    Yes, thank you!2012-01-26

1 Answers 1

3

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.

  • 0
    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.2012-01-26