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 Kr.Dim$(R[\frac{1}{v}])$ must be either Kr.Dim$(R)$ or Kr.Dim$(R)-1$.
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My attempts to do this purely algebraically are not working, so I started looking into methods from algebraic geometry. I thought that Grothendieck's Vanishing Theorem might help (i.e. if dim$(X)=n$ then $H^i(X,\mathcal{F})=0$ for any sheaf of abelian groups $\mathcal{F}$ and any $i>n$) but the problem is that the converse for this theorem fails, so I can't conclude anything about dimension. Perhaps this theorem could give some sort of test for when dimension drops, but I'm hoping for a better answer.


We'll definitely need some hypotheses. For the application I have in mind we can assume $R$ is commutative and is finitely generated over some base ring (e.g. $\mathbb{Z}\_{(2)}$), 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 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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One thing I know: to have any hope of dimension dropping by $1$ I need to be inverting a maximal irreducible component. I'm curious as to the algebraic condition this puts on $v$. 



Can we say anything about the Krull dimension of a localization?
The dimension of $R[1/v]$ is the biggest height of some prime ideal $P$ such that $v\notin P$. So, let $I\_{d-1}$ be the intersection of all primes of height at least $d-1$ ($d= \dim R$), then



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> $\dim R[1/v] \geq d-1$ if and only if $v\notin I\_{d-1}$. 
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Under a mild condition (all maximal ideals has height at least $d-1$), then $I\_{d-1}$ is inside the Jacobson radical of $R$. So in this case $v$ is not inside the Jacobson radical would suffice. 


EDIT (in response to the comment by the OP): Now let $I\_d$ be the intersection of all primes with height $d$. Then $\dim R[1/v] \leq d-1$ iff $v\in I\_d$. So 



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> $\dim R[1/v] =d-1$ iff $v \in I\_d- I\_{d-1}$.
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Here is a short calculation to show that in Fernando's example above, $I\_1=(x)$. Any prime that does not contain $x$ would have to contain $(x-1, y, z)$, but this is a maximal ideal of height $0$. So $x \in I\_1$. On the other hand $R/(x)$ is $k[y,z]$ so the intersection of all height one there is $0$. Thus $I\_1=(x)$. This gives you precisely what elements satisfies your condition. 


