Local property of dimension Let $R$ be a commutative ring with unity. If the Krull dimensions of all the localizations $S^{-1}R$ are zero, where $S$ runs among multiplicative subsets of $R$, is it true that the Krull dimension of $R$ is zero? 
If the dimension of a given localization is zero is that of $R$ zero?
For $S$ a multiplicative subset of $R$, it is known that $\dim S^{-1}R \le \dim R$, therefore if the right hand side is zero so is the other side. 
Thanks for any suggestion!
 A: A chain of prime ideals $p_0\subset p_1\subset\dots \subset p_d$ induces a chain of the same length in $S^{-1}R$ if and only if $S\subset R-p_d$. Therefore, if $R$ is of finite dimension, then $\mathrm{dim}(S^{-1}R)=\mathrm{dim}(R)$ if and only if there is a maximal ideal $m\subset R$ of maximal height such that $S\subset R- m$. In particular, we can just take $S=R-m$.
Thus, there is a non-trivial localisation $S^{-1}R\not= R$ of the same (finite) dimension if and only if $R$ is not local. Namely, if $m$ is any prime ideal of maximal height, then $S:=R-m$ will do the job, whilst it can't consist of units only since otherwise $m$ would be local.
To address the two questions directly:


*

*If $S^{-1}R$ has dimension zero for all multiplicative $S\subset R$, then $R$ has dimension zero too, for the trivial reason that we can take $S=\{1\}$. If we want a non-trivial localisation with this property, we have to assume that $R$ is not local.

*Does $\mathrm{dim}(S^{-1}R)=0$ for some $S$ imply that $\mathrm{dim}(R)=0$? No! Take $S:=R-p$ for any minimal prime. Then $\mathrm{dim}(S^{-1}R)=0$, no matter what the dimension of $R$ is.

A: Yes, it is true since $\{1\}$ is a multiplicative subset and the localization of $R$ by $\{1\}$ is $R$.
