Proving that P/PJ is a projective right module over R/J If P is a projective right module over a ring R  and J is a two sided ideal of R.  Prove that P/PJ  is   a  projective right module over R/J  .
My idea was trying to proof that " $M$ is an $R$-module iff  $M$ is an module over $R/J$ " where J is an ideal of R .
 A: For all $R$-modules $M$ we have an isomorphism $M/MJ\cong M\otimes_R(R/J)$ as right $R/J$/modules. It follows that the functor $F:M\in\mathrm{Mod}_R\to M/MJ\in\mathrm{Mod}_{R/J}$ is isomorphic to $(\mathord-)\otimes_R(R/J)$, which has a right adjoint $\hom_{R/J}(R/J,\mathord-):\mathrm{Mod}_{R/J}\to\mathrm{Mod}_R$, which is immediately seen to map surjections to surjction
As $F$ is a left adjoint to a functor  which preserves surjections, it maps projectives to projectives.
A: Suppose $M$ is a right $R/J$-module, and view it also as a right $R$-module in the obvious way. An $R$-linear map $f:P\to M$ has to vanish on $PJ$, so it induces a map $\bar f:P/PJ\to M$. This gives us a map $$\phi:f\in\hom_R(P,M)\longmapsto\bar f\in\hom_{R/J}(P/PJ,M).$$
On the other hand, if $g:P/PJ\to M$ is a morphism of $R/J$-modules, then then it is also a morphism of $R$-modules and its composition $g\circ\pi$ with the projection $\pi:P\to P/PJ$ is also a map of $R$-modules. We have therefore a well defined map $$\psi:g\in\hom_{R/J}(P/PJ,M)\longmapsto g\circ\pi\in\hom_R(P,M).$$
You can easily check that $\phi$ and $\psi$ are inverse bijections, and in fact isomorphisms of abelian groups.
Now $P$ was $R$-projective, so $\hom_R(P,\mathord-)$ is an exact functor on $R$-modules. Show using the above that $\hom_{R/J}(P/PJ,\mathord-)$ is an exact functor on $R/J$-modules, so that $P/PJ$ is $R/J$-projective.
A: If $P$ is $R$-projective, there is a free $R$-module $F$ and an idempotent endomorphism $p:F\to F$ such that $p(F)\cong P$.
Now let $J$ be an ideal in $R$. Since $F$ is $R$-free, $F'=F\otimes_R(R/J)$ is $R/J$-free. On the other hand, the map $p'=p\otimes\mathrm{id}_{R/J}:F'=F\otimes_R(R/J)\to F'=F\otimes_R(R/J)$ is an idempotent endomorphism of $F'$, so the image $p'(F')$ is a projective $R/J$-module.
Check now that $P/PJ$ is isomorphic as an $R/J$-module to $p'(F')$.
A: Since $P$ is $R$-projective, there is a dual basis $(p_i,\phi_i)_{i\in I}$. This means: there is a set $I$, elements $p_i\in P$ for all $i\in I$, and morphisms of $R$-modules $\phi_i:P\to R$ for all $i\in I$, such that for all $p\in P$ the set $I_p=\{i\in I:\phi_i(p)\neq0\}$ is finite and $$p=\sum_{i\in I_p}p_i\phi_i(p).$$
Now let $J$ be an ideal in $R$. For all $i\in I$, let $\bar p_i=p_i+PJ$ be the class of $p_i$ in $P/PJ$. On the other hand, the composition $P\xrightarrow{\phi_i}R\to R/J$ (with the second map the canonical projection) vanishes on the $R$-submodule $PJ$, so it induces a map $\bar\phi_i:P/PJ\to R/J$ which is a map of $R$-modules and of $R/J$-modules.
Check that $(\bar p_i,\bar\phi_i)_{i\in I}$ is a dual basis for the $R/J$-module $P/PJ$, so that it is projective.
