Let $R$ be a ring and assume that for every left $R$-module ${}_RM$, if $M$ has a projective cover, then $M$ is projective.
Can someone help me prove that in that case, $J(R)=0$?
Let $R$ be a ring and assume that for every left $R$-module ${}_RM$, if $M$ has a projective cover, then $M$ is projective.
Can someone help me prove that in that case, $J(R)=0$?
Well, $J(R)$ is a superfluous submodule of $R$, so letting $M=R/J(R)$, we have found that the projective cover of $M$ is $R$. Moreover $M$ is Jacobson semisimple (meaning $J(M)=J(R/J(R))=\{0\}$.)
But by hypothesis we've said that any module with a projective cover is projective, so that it's already it's own projective cover (up to isomorphism.)
Thus $M\cong R$ as left $R$ modules. Since $M$ is Jacobson semisimple, $R$ is too, so $J(R)=\{0\}$ necessarily.