# Projective resolution of modules over rings which are regular in codimension n

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?

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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.)

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Since factorial is equivalent to $Cl(R)=0$ for normal ring. Suppose $R$ is local, is there some kind of "generalized divisor class group" such that if it vanishes then every ideal with height less than n has a finite projective resolution? – Liu Hang Jan 29 '11 at 5:08

Dear Liu,

I like your updated question a lot. To make things easier to discuss, let me define the following properties for a Noetherian local ring $R$ and $n>0$:

($A_n$) every ideal with height less than $n$ has a finite projective resolution.

($B_n$) $R_P$ is regular for each prime $P$ of height at most $n$.

($C_n$) The Chow groups $CH^i(R)=0$ for codimensions $1\leq i \leq n$.

Your question asked whether $A_n$ implies $B_n$ and $C_n$.

It is not hard to see that $A_n$ implies $B_n$: localizing a resolution of $R/P$ over $R$ shows that the residue field of $R_P$ has finite projective dimension, which forces $R_P$ to be regular.

I do not know whether or not $A_n$ implies $C_n$, and I am probably not alone! As far as I known, it is a conjecture that Chow groups of codimensions at least one in any regular local ring are $0$. It is known if $R$ is essentially of finite type over a field (and perhaps a bit more generally, see this paper).

However, it is relatively easy to show that if $A_n$ holds, then $CH^i(R)$ is torsion for codimensions $1\leq i \leq n$. The brief reason is if an ideal $I$ has positive height and finite projective dimension, the class of $R/I$ is $0$ in the Grothendieck group $G_0(R)$, and this group is equal to the total Chow group after tensoring with $\mathbb Q$.

I think $A_n$ for even $n=2$ is a pretty strong condition. I would not be too surprised if it implies regularity of $R$. For example, it is not hard to show that $A_n$ implies $B_{n+1}$ if $R$ is Cohen-Macaulay and $n\geq 2$ (so if $\dim R=3$, $A_2$ forces $R$ to be regular).

UPDATE per the comments: If $R$ is not local, $A_n \Rightarrow B_n$ still holds. However, $C_n$ is hopeless. Just take any non-singular affine curve with non-trivial Picard group. Then any module still has finite projective dimension since the ring is regular. The trouble is that $A_n$ is a local condition, but $C_n$ is not.

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Dear Hailong, thanks for the informative answer. – Liu Hang Feb 10 '11 at 12:08
What about the case when $R$ is not local? – Liu Hang Feb 12 '11 at 2:30
Liu: see my update. – Hailong Dao Feb 23 '11 at 22:47