Let $R$ be an associative ring with identity.
Recall that a pair of classes of modules $(\mathcal{A}, \mathcal{B})$ in $R\text{-Mod}$ is called a cotorsion pair if
$$ \mathcal{A} = {}^{\perp}\mathcal{B} \quad\text{and}\quad \mathcal{B} = \mathcal{A}^{\perp}, $$ where
$$ {}^{\perp}\mathcal{B} = \{ M \mid Ext^1_R(M, B) = 0 \text{ for all } B \in \mathcal{B} \}, \qquad \mathcal{A}^{\perp} = \{ N \mid Ext^1_R(A, N) = 0 \text{ for all } A \in \mathcal{A} \}. $$ The cotorsion pair is said to be complete if every module $M$ admits both a special $\mathcal{A}$-precover and a special $\mathcal{B}$-preenvelope.
It is well known that for a commutative domain $R$, the cotorsion pair $(\mathcal{P}_1, \mathcal{D})$ — where
$\mathcal{P}_1$ denotes the class of all modules of projective dimension at most $1$, and
$\mathcal{D}$ denotes the class of all divisible modules — is complete.
In this situation we have the following equivalences: $$ R \text{ is a field } \ \Longleftrightarrow\ \mathcal{D} = R\text{-Mod} \ \Longleftrightarrow\ \text{every module in } \mathcal{P}_1 \text{ is projective.} $$
In the general (not necessarily commutative) case, it is known that
$(\mathcal{P}_1, \mathcal{P}_1^{\perp})$ is a complete cotorsion pair.
Hence the following statements are equivalent:
- $\mathcal{P}_1^{\perp} = R\text{-Mod}$;
- Every module in $\mathcal{P}_1$ is projective;
- Every quotient of a projective module by a projective submodule is projective;
- $R$ has no (non-projective) module of projective dimension $1$.
Question.
Are rings satisfying these equivalent conditions already known or characterized in the literature?
Equivalently,
$$ \text{Which rings have the property that every module of projective dimension at most $1$ is projective?} $$
I could not find a reference or a specific name for such rings.
Are they related to hereditary, semihereditary, or perfect rings, or do they form a strictly smaller or larger class?
