# Heller operator without left adjoint?

Suppose given a noetherian ring $R$. On the stable category $R\text{-}\underline{\text{mod}} := R\text{-mod}/R\text{-proj}$, we have the Heller operator $$\Omega : R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}} \; ,$$ defined by a choice of short exact sequences $\Omega X \to P \to X$ with $P$ projective, accordingly on morphisms.

Does there exist a ring $R$ for which $\Omega$ does not have a left adjoint?

Of course, $R$ should neither be self-injective ($\Omega$ an equivalence) nor hereditary ($\Omega = 0$).

If $\Omega$ has a left adjoint $S$, then $(X,\Omega f) = (SX,f)$ is injective for $X$ an $R$-module and $f$ a monomorphism. So I've been searching (without success) for a monomorphism $f$ in the stable category that is not mapped to a monomorphism under $\Omega$. First of all, $f$ should be a non-split mono, but this is possible, since we are not in the triangulated case.

It was pointed out to me that in "On a theorem of E. Green on the dual of the transpose" by M. Auslander and I. Reiten, it is shown that that Omega lifts to an exact functor isomorphic to tensoring with a bimodule if $R$ is an artin algebra (MR1143845).

The question is related to this question ("Brown, but not Quillen?").

If $R$ is a two-sided noetherian ring and $d \geq 1$ an integer, then the functor $$\operatorname{Tr}\Omega^d\operatorname{Tr} \colon R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}}$$ is a left adjoint of $\Omega^d \colon R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}}$.
In particular $\operatorname{Tr}\Omega\operatorname{Tr}$ is a left adjoint of $\Omega$.
Here $\operatorname{Tr} \colon R\text{-}\underline{\text{mod}} \to R^{\operatorname{op}}\text{-}\underline{\text{mod}}$ denotes the transpose, if $M$ is an $R$-module with projective presentation $$P_1 \to P_0 \to M \to 0,$$ then $\operatorname{Tr} M=\operatorname{coker} (\operatorname{Hom}_R(P_0,R) \to \operatorname{Hom}_R(P_1,R))$.