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.

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    $\begingroup$ 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. $\endgroup$ Jan 24, 2013 at 2:04
  • $\begingroup$ Related: math.stackexchange.com/questions/1106866 $\endgroup$
    – Watson
    Jan 24, 2017 at 12:50

1 Answer 1


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 $\dim_k m/m^2$.


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