Can we say anything about the Krull dimension of a localization?

Question

I'm looking for a theorem of the form

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$(R)-1$.

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:

Are there any conditions on $v$ such that the dimension would drop by more than 1 after inverting $v$?

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 $v$.

Answer 1

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

$\dim R[1/v] \geq d-1$ if and only if $v\notin I_{d-1}$.

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

$\dim R[1/v] =d-1$ iff $v \in I_d- I_{d-1}$.

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.

Answer 2

Thinking geometrically, take the disjoint union of a plane and a point in the 3-dimensional affine space. This has dimension 2. If you remove the plane by inverting its equation, you obtain the the point, which is 0-dimensional.

Algebraically, let $k$ be a field, $R = k[x,y,z]/(x^2-x,xy,xz)$, $v=x$. Then $R$ has Krull dimension $2$ and $R[1/v]=k$ is 0-dimensional.

Answer 3

$R[\frac 1v]$ corresponds to the open subset of $\mathrm{Spec}R$ where $v\neq 0$. So, all kind of things can happen. Here is an example: Let $R_1$ be a "nice" ring with a unity $v\in R_1$ of dimension $n$ and $R_2$ a "nice" ring with a unity $w\in R_2$ of dimension $m$. Let $R=R_1\oplus R_2$. Then $\dim R=\max \{n,m\}$ and $\dim R[\frac 1v]=\dim R_1=n$, so the dimension drops by $\max \{n,m\}-n$.