A few days ago I asked a similar [question](https://mathoverflow.net/questions/68386/can-we-say-anything-about-the-krull-dimension-of-a-localization) about Krull dimension and got fantastic answers. Unfortunately, for the application I have in mind (a question on ring spectra), Krull dimension doesn't generalize correctly. One must instead use [right global dimension](http://en.wikipedia.org/wiki/Global_dimension), or just global dimension since my ring object is commutative. Note that this has a similar title, but totally different content from [another question](https://mathoverflow.net/questions/59981/global-dimension-and-localization) (different because my $v$ does not generate a maximal ideal). I'm looking for a theorem of the form



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> If R is a nice ring and v is a reasonable element in R then Gl.Dim($R[1/v]$) must be either Gl.Dim($R$) or Gl.Dim($R$)−1.
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By "nice ring" I mean $R$ is commutative and is finitely generated over some base ring (e.g. $Z\_{(2)}$), but we should not assume it's an integral domain. If necessary we can assume it's Noetherian and local, but I'd rather avoid this. As for $v$, it's a non-zero divisor which is not in the base ring and it has only a few relations with other elements in $R$, none of which are in the base ring. If we can't get the theorem above, perhaps we can figure out something to help me get closer:



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> Are there any conditions on $v$ such that the dimension would drop by more than 1 after inverting $v$?
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Here's what I know (from Weibel and Rotman): 


If $f:R \rightarrow S$ is a ring homomorphism and $A$ is a left $S$ module then $pd\_R(A)\leq pd\_S(A)+pd\_R(S)$


If $x$ is a central nonzerodivisor in $R$ and $A$ is a nonzero $R/v$-module with $pd\_{R/v}(A)<\infty$ then $pd\_R(A)=1+pd\_{R/v}(A)$. If $B$ is a nonzero $R$-module then $pd\_R(B)\geq pd\_{R/v}(B/vB)$ 


This result gives me some hope because of the short exact sequence $0\rightarrow R\stackrel{\cdot v}{\rightarrow} R \rightarrow R/v \rightarrow 0$, but I don't see an obvious answer coming from this idea alone. One would also need to relate the dimensions of $R/v$ and $v^{-1}R$.


If $A$ is a projective $R$-module then $v^{-1}A$ is a projective $v^{-1}R$ module. According to [this](http://www.math.washington.edu/~smith/Research/asz7.pdf) paper, injective dimension is not well behaved under localization (but this seems to be because the localization of an injective module need not be injective).



Is there a relationship between the right global dimensions of R and R[1/v]?
In general, if $R$ is any ring and $S$ is any right denominator set, the right globaldimension of the localization $R\_S$ does not exceed that of $R$.


If $R$ is of finite global dimension and Noetherian, and $S$ is left and right denominator set, then $R\_S$ and $R$ have the same dimension iff there is a simple $R$-module $M$ with $\operatorname{gldin}R=\operatorname{pdim}M$ and $M\_S\neq0$. Therefore the dimension *drops* if this does not happen.


In the local commutative case, there is one simple module, so what you want from $v$ to decrease the dimension a lot is for it to turn one of the maps in the minimal resolution of the residue field $k$ of $R$ into a split one. For example, if in that resolution we have a piece looking like $$\cdots \to P\_j\xrightarrow{\phi} P\_{j-1}\to\cdots$$ there is a finite number of elements which, when inverted, turn $\phi$ into an isomorphism (this is called *universal localization* in some contexts) If $v$ is the product of these elements, you get a drop in the dimension.


