Krull dimension and localization

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!

Solution 1:

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.