All rings are Noetherian and commutative, modules are finitely generated.


It is a theorem of Serre that over a regular ring $R$, every module has a finite projective resolution.


More generally, if $R$ is regular in codimension n, what can we say about projective resolution of modules over $R$? For example, is it true that every ideal with height less than n has a finite projective resolution?


Similarly, over a Noetherian seperated regular scheme $X$, every coherent sheaf has a finite resolution by vector bundles. The same questions can be asked for schemes as for rings.


Examples are extremely appreciated. Thanks!


**Edit:It is not true that every ideal with height less than n has a finite projective resolution.** As inkspot pointed out, if $R$ is normal, excellent, local and all height 1 ideals have finite projective dimension, then $R$ is factorial. So the local ring of a cone at origin gives a counterexample. 


Since factorial is equivalent to $Cl(R)=0$ for $R$ normal, this makes me wonder **for a local ring $R$** whether every ideal with height less than n has a finite projective resolution is equivalent to:



> 
> 1. $R$ is regular in codimension n plus some other condition on the ring such as normal and excellent.
> 2. Some kind of "generalized divisor class group"(may be Chow group) vanishes.
> 
> 
> 


If $R$ is not local, I think condition 2 should be replaced by something like:



> 
> 2'. Some part of $K\_0(R)$ and $G\_0(R)$ are isomorphic.
> 
> 
> 


where $K\_0$ is the Grothendieck group of the category of projective modules over $R$, $G\_0$ is the Grothendieck group of the category of finite generated modules over $R$.


Could above be true?



Projective resolution of modules over rings which are regular in codimension n
If $R$ is normal (so regular in codimension $1$), excellent and local and all height $1$ ideals $I$ have finite projective dimension, then $R$ is factorial. So there are many counter-examples. (I don't have a reference to hand, but the argument is Serre's proof that regular implies factorial. Say $X= Spec\ R$ and $j:U\to X$ is the regular locus. A finite projective (= free) resolution of $I$ restricts to a free resolution of the restriction $\mathcal{I}$ of $I$ to $U$. Now $\mathcal{I}$ is locally free of rank $1$; taking the determinant of its resolution shows that it is free, and then $I=j\_\*\mathcal{I}$ is free, which means that $R$ is factorial.)


