Two modules are isomorphic in the stable module category iff they are projectively equivalent

Question

Let $R$ be a (not necessarily commutative) ring. Let ${\text{mod-}R}$ be the category of finitely generated right $R$-modules. Let $\underline{\text{mod-}R}$ be the stable module category, with the same objects as ${\text{mod-}R}$ and morphisms given by $\text{Hom}_R(M,N)/ \hspace{-3pt} \sim$, where $f \sim g$ iff $f-g$ factors through a projective module.

We say that two $R$-modules $M,N$ are projectively equivalent iff there exist finitely generated projective modules $P,Q$ with $M \oplus P \cong N \oplus Q$ in ${\text{mod-}R}$.

I'm trying to prove the following:

Let $M,N$ be finitely generated $R$-modules. Then $M$ and $N$ are isomorphic in the stable module category iff they are projectively equivalent.

I know this to be true, but I want to prove it by constructing specific maps. So far I've only attempted the forward implication. So suppose that $M$ and $N$ are isomorphic in the stable module category, so there exist $f : M \to N$ and $g: N \to M$ with \begin{gather*} gf - \text{id}_M = \phi : M \xrightarrow{\phi_1} P \xrightarrow{\phi_2} M \\ fg - \text{id}_N = \psi : N \xrightarrow{\psi_1} Q \xrightarrow{\psi_2} N. \end{gather*} I then want to construct maps $\theta:M \oplus P \to N \oplus Q$ and $\lambda:N \oplus Q \to M \oplus P$ (or perhaps the maps should go between $M \oplus Q$ and $N \oplus P$...?) which are mutual inverses. However, playing around with these maps hasn't led me anywhere. In fact one major issue is that I can't see how I would be obtain the identity in the second component of $\lambda \theta$, and I suspect that I actually need to construct some new maps using the projectivity of $P$ and $Q$.

Any ideas on how to proceed?

Answer 1

Your method is not going to work unless you're careful about the choice of the projective modules $P$ and $Q$ through which $gf-\text{id}_M$ and $fg-\text{id}_N$ factor. Indeed, you could replace $P$ by $P\oplus P'$ for any projective module $P'$, changing the isomorphism type of $M\oplus P$.

Instead, use the fact that $gf-\text{id}_M$ factor through a projective $P$ to construct maps $M\stackrel{f'}{\to}N\oplus P\stackrel{g'}{\to}M$ such that $g'f'=\text{id}_M$ and $f'g'-\text{id}_{N\oplus P}$ factors through a projective, so that $M\oplus\operatorname{ker}(g')\cong N\oplus P$, and then (using the fact that $f'g'-\text{id}_{N\oplus P}$ factors through a projective) that $\operatorname{ker}(g')$ is projective.