Let $A$ be a commutative ring with unit, $P$ a projective $A$-module and denote by $R(P)$ the intersection of all maximal (with respect to inclusion) submodules of $P$. If $R(A)$ denotes the Jacobson ideal of $A$, how to prove that $R(P)=R(A)P$?
It is easy to show that $R(A)P$ is contained in $R(P)$: if $M$ is a maximal submodule of $P$ then $P/M$ is simple and the annihilator of $P/M$ is a maximal ideal. Hence $R(A)$ annihilates $P/M$, i.e., $R(A)P$ is contained in $M$.
For the other hand I think that commutativity of the R(_) with direct sums helps: if $(M_i)$ is a family of $A$-modules and $M$ its direct sum, then $R(M)=\oplus R(M_i)$.