Let $A$ be a finite-dimensional algebra over a field. Define the stable category of right $A$-modules $\underline{\text{mod}}-A$ as a category with objects the finitely generated modules over $A$ and $\text{Hom}_{\underline{\text{mod}}-A}(M,N)=\text{Hom}_{{\text{mod}}-A}(M,N)/\mathcal{P}(M,N)$, where $\mathcal{P}(M,N)$ is the subgroup of morphisms which factor through a projective module, so $f\in\mathcal{P}(M,N)$ if and only if there is a projective finitely generated module $P$ and morphisms $i: M\to P, q: P\to M$ such that $f=q\circ i.$
I am trying to show that if $M$ is an indecomposable non-projective module, then it is also indecomposable as an object of the stable category $\underline{\text{mod}}-A$. It seems this should be straightforward, but I am completely at a loss.
I tried proving this by contradiction. Suppose $M$ is not indecomposable in the stable category, then then there is an idempotent endomorphism $f$ of $M$. This translates to $f^2-f$ filtering through some projective module in the category of $A$-modules. But I can't see how one can use the projectivity of the module through which this map filters.
Best Answer
This follows immediately from the following more general fact.
Theorem: Let $R$ be a finite-dimensional algebra over a field $k$, let $S$ be a quotient of $R$, and let $f\in S$ be a nontrivial idempotent. Then $f$ can be lifted to a (necessarily nontrivial) idempotent of $R$.
(To get your statement, apply this with $R$ the endomorphism ring of $M$ and $S$ the endomorphism ring of $M$ in the stable module category.)
Proof of Theorem: Fix a lift $g\in R$ of $f$ and let $p(t)\in k[t]$ be the minimal polynomial of $g$. Since $f$ is a nontrivial idempotent, its minimal polynomial is $t^2-t$, and so $p(t)$ is divisible by $t^2-t$; say $p(t)=t^n(t-1)^mq(t)$ where $q(t)$ is relatively prime to $t$ and $t-1$. By the Chinese remainder theorem, we then have $k[t]/(p(t))\cong k[t]/(t^n)\times k[t]/(t-1)^m\times k[t]/(q(t))$. Let $r(t)\in k[t]$ be a polynomial whose image in $k[t]/(t^n)\times k[t]/(t-1)^m\times k[t]/(q(t))$ via this isomorphism is $(0,1,0)$. Then the image of $r(t)$ in $k[t]/(t^2-t)$ is $t$ (since it is $0$ mod $t$ and $1$ mod $t-1$), which means exactly that $r(g)$ is a lift of $f$. Also, since $(0,1,0)$ is idempotent in $k[t]/(t^n)\times k[t]/(t-1)^m\times k[t]/(q(t))$, so is $r(g)$.