If $P$ and $Q$ are two finitely generated projective $A$ modules such that $P/Jac(A)P \cong Q/Jac(A)Q$ then $P \cong Q$. 
Let $A$ be a commutative ring with $1$ and $P$ and $Q$ are two finitely generated projective $A$ modules such that $P/JP \cong Q/JQ$, where $J=Jac(A)$. Then I have to show that $P \cong Q$.

I have no clue to prove it. Only thing I could think about $P \oplus JQ \cong Q \oplus JP$, by Schanuel. I can't proceed further, need some help. Thanks.
 A: To expand on the comment by user Mohan. Let $J$ be the radical of $A$ and note that $P/JP$ is naturally isomorphic to $\overline{P}=A/J\otimes_A P$. I'll use a bar to write these quotients.

*

*Pick an isomorphism $f :\overline{P} \longrightarrow \overline{Q}$. There is a surjective map $$\pi_Q : Q\to  \overline{Q}.$$


*There is a map $P\to \overline{Q}$ obtained by post-composition of the corresponding  projection $\pi_P:P\to \overline{P}$ with $f$.


*Since $P$ is projective, there is a map $F: P\to Q$ with the property that the induced map after tensoring with $A/J$ is precisely $f$.


*By hypothesis, $f$ is an isomorphism and, since $A/J\otimes_A -$ is right exact, you get that $F$ is surjective by Nakayama's lemma. Indeed, it follows that if $C$ is the cokernel of $F$ then $\overline{C}$ is isomorphic to the cokernel of $f$, which is zero.


*Since $F$ is onto, you have a short exact sequence ending at $Q$:
$$0\longrightarrow K \longrightarrow P \longrightarrow Q \longrightarrow 0$$
which remains exact after tensoring with $A/J$ since $Q$ is projective. Hence $\overline{K}$ and so $K=0$ by another use of Nakayama's lemma.
