Is there a relationship between the right global dimensions of R and R[1/v]? - MathOverflow

Is there a relationship between the right global dimensions of R and R[1/v]?

2011-06-23T22:59:25Z

A few days ago I asked a similar question 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, or just global dimension since my ring object is commutative. Note that this has a similar title, but totally different content from another question (different because my $v$ does not generate a maximal ideal). I'm looking for a theorem of the form

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.

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:

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

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 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).

Answer by Fernando Muro for Is there a relationship between the right global dimensions of R and R[1/v]?

2011-06-23T23:04:02Z

Under your hypotheses on $R$, the global dimension is a local notion, and so it the gobal dimension. Moreover, since you're assuming finite global dimension, your ring is regular, therefore the Krull and global dimensions coincide locally and globally. This shows that any answer to this questions is an answer to your other question, and vice versa.

Answer by Mariano Suárez-Alvarez for Is there a relationship between the right global dimensions of R and R[1/v]?

2011-06-23T23:07:33Z

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.