Finitistic projective dimension of commutative rings of finite weak global dimension

Question

For a commutative (not necessarily Noetherian) ring $R$, define the following two quantities

$$fpd(R):=\sup\{\operatorname{pd}_R(M) | M \text{ is a finitely presented }R\text{-module of finite projective dimension} \}$$

$$FPD(R):=\sup\{\operatorname{pd}_R(M) | M \text{ is an }R\text{-module of finite projective dimension} \}$$

Clearly, $fpd(R)$ is bounded above by $FPD(R)$ which in turn is bounded above by the global dimension of $R$.

I want to ask: If $R$ has finite weak global dimension , then is $FPD(R)$ finite, or at least, is $fpd(R)$ finite?

This is true if $R$ is Noetherian because in that case the weak global dimension and global dimension of $R$ are the same. However, if $R$ is not Noetherian, I don't have any clue. Please help.

Answer 1

I don't know about $\operatorname{fpd}(R)$, but there are examples where $\operatorname{FPD}(R)$ is infinite.

Let $$R=\prod_{\aleph_\omega}k,$$ the product of $\aleph_\omega$ copies of a field $k$.

It follows from the results in Barbara Osofsky's paper Homological dimension and cardinality (Trans. AMS 151, pp. 641-649 (1970)) that for every natural number $n$, $R$ has an ideal with projective dimension $n$, and so $\operatorname{FPD}(R)=\infty$.

Also, $R$ is von Neumann regular, so its weak global dimension is finite (in fact, zero).

This doesn't help for the question about $\operatorname{fpd}(R)$, as all finitely presented modules for von Neumann regular rings are projective.

More generally, if $R$ is coherent, then the projective dimension of a finitely presented module is the same as its weak dimension, so if $R$ has finite weak global dimension then $\operatorname{fpd}(R)=\operatorname{wgldim}(R)<\infty$. So a ring $R$ with finite weak global dimension, but with $\operatorname{fpd}(R)=\infty$, will have to be noncoherent.