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:
- $R$ is regular in codimension n plus some other condition on the ring such as normal and excellent.
- 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?
Best Answer
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.)