Let R be an Artinian ring with 1.
I know that as a (right) R-module it decomposes into a finite direct sum of projective indecomposable R-submodules.
Let P be a projective R-module. Is it true that P can be analogously decomposed into a direct sum of indecomposable projective R-modules?
If it's true than how can I show this?
Best Answer
This is true.
Let $P$ be a projective right $R$-module, where $R$ is an artinian ring. Then the Jacobson radical satisfies $J(P)=PJ(R)$, and $P/J(P)$ is a semisimple $R/J(R)$-module. Write this as $\bigoplus_iS_i$ with each $S_i$ simple, and let $P_i$ be the projective cover of $S_i$, so an indecomposable projective. This gives another epimorphism $\bigoplus_iP_i\to P/J(P)$, and we can lift these two epimorphisms to maps $\theta\colon P\to\bigoplus_iP_i$ and $\phi\colon\bigoplus_iP_i\to P$. Now $1-\phi\theta$ is an endomorphism of $P$ with image in $J(P)$, so must be nilpotent, and hence $\phi\theta$ is an automorphism. Similarly $\theta\phi$ is an automorphism of $\bigoplus_iP_i$, proving that $P\cong\bigoplus_iP_i$ decomposes into a direct sum of indecomposable projectives.
More generally, one result that I like is the following
P. Prihoda, Projective modules are determined by their radical factors, J. Pure Applied Algebra 210 (2007) 827–835.
This shows that for any ring $R$, if $P$ and $Q$ are projective $R$-modules and $\bar\theta\colon P/J(P)\xrightarrow\sim Q/J(Q)$ is an isomorphism, then we can lift $\bar\theta$ to an isomorphism $\theta\colon P\to Q$. This can be seen as a generalization of Kaplansky’s result that every projective module is a direct sum of countable generated projective modules.
I have some notes explaining the connection here
https://www.math.uni-bielefeld.de/birep/activities/topics/files/ws19-hubery-projectives-over-hereditary-tensor-algebras.pdf