Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anyone know an "elementary" proof of the fact that a Noetherian local ring has finite Krull dimension?

The one I know is from Atiyah&Macdonald's book Introduction to Commutative Algebra, where they use Hilbert functions (which is not an elementary proof). On the other hand, I am studying the local rings section of Weibel's book Introduction to Homological Algebra, where it says that if $R$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k$ then the Krull dimension of $R$ is bounded by $\dim_k\mathfrak{m}/\mathfrak{m}^2$, and there is not any reference about that. This made me think that this could be easy to prove but I haven't succeeded.

share|cite|improve this question
The principal ideal theorem is also not horrible to prove. I feel like the treatment in Qing Liu's algebraic geometry book is nice and streamlined. – Keenan Kidwell Jan 24 '13 at 2:04
up vote 5 down vote accepted

I suggest you R. Y. Sharp's book: Steps in Commutative Algebra.

By Theorem 15.4 (Krull's generalized principal ideal theorem), $\dim R$ is less than or equal to "the number of elements in each minimal generating set for $m$", which is finite since $R$ is Noetherian. And by Proposition 9.3 this is equal to $\dim_k m/m^2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.