Let $R$ be a commutative Noetherian ring, and let $\text{mod } R$ denote the abelian category of finitely generated $R$-module. Consider the bounded derived category $D^b(\text{mod } R) $ which is a triangulated category. Every object $X \in \text{mod } R$ can be naturally identified in $D^b(\text{mod } R) $. For an object $X \in D^b(\text{mod } R)$, let $\text{Thick}_{D^b(\text{mod } R)} X$ denote the intersection of all thick subcategories (https://ncatlab.org/nlab/show/thick+subcategory) of $D^b(\text{mod } R)$ containing $X$. For example, note that $\text{Thick}_{D^b(\text{mod } R)} R$ is the collection of all perfect complexes, hence $M \in \text{mod } R$ belongs to $\text{Thick}_{D^b(\text{mod } R)} R$ if and only if $M$ has finite projective dimension.

Following 4.1 of https://doi.org/10.4171/cmh/56, we say $X\in D^b(\text{mod } R)$ is Virtually small if there exists a non-exact complex $Y \in \text{Thick}_{D^b(\text{mod } R)} X \cap \text{Thick}_{D^b(\text{mod } R)} R$.

My question is: If $M \in \text{mod } R$ is Virtually small in $D^b(\text{mod } R)$, then does $M$ embed into a finitely generated module of finite projective dimension?


1 Answer 1


For every $M$, $M\oplus R$ is virtually small, so your question is equivalent to the question: Does every finitely generated $R$-module embed in a finitely generated module of finite projective dimension?

The answer is no. For example, let $R$ be a local commutative finite dimensional algebra over a field $k$, that is not self-injective, such as $R=k[x,y]/(x^2,xy,y^2)$. Then the only finitely generated modules of finite projective dimension are the finitely generated projective modules, and the injective generator $\operatorname{Hom}_k(R,k)$ does not embed in a projective module.


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