If we have a commutative ring with 1, what can we say about $\operatorname{dim} S^{-1}A$ for some multiplicative subset $S$, or more specifically, what happens if $S = A \setminus \mathfrak{p}$ for a prime ideal $\mathfrak{p}$?

Do the dimensions of the localization generally coincide for distinct prime or maximal ideals? Thanks!

  • 1
  • 13
  • 62
  • 125
  • 193
  • 1
  • 5
  • 9
    Maybe this is helpful: the primes of $A_\mathfrak{p}$ correspond to the primes of $A$ contained in $\mathfrak{p}$. So the dimension of the localization is the [height](http://en.wikipedia.org/wiki/Height_(ring_theory)) of $\mathfrak{p}$. – Dylan Moreland Jan 15 '12 at 17:14
  • You were both very helpful, thanks. – argon Jan 15 '12 at 18:53

1 Answers1


What is true is that $\dim S^{-1}A\le \dim A$ because $\mathrm{Spec}(S^{-1}A)$ is homeomorphic to a subset of $\mathrm{Spec}(A)$.

For $A_{\mathfrak p}$, the dimension varies: if $\mathfrak p$ is a minimal prime, then the localization has dimension $0$. If $\mathfrak p$ is not maximal and $\dim A$ is finite, then $\dim A_{\mathfrak p}<\dim A$. On the other hand, $\dim A$ is the supremum of the $\dim A_{\mathfrak m}$ when $\mathfrak m$ runs the maximal ideals of $A$. If $A$ is an integral finitely generated algebra over a field, then $\dim A=\dim A_{\mathfrak m}$ for any maximal ideal.