Let $P \neq 0$ be a projective right module over a ring $R$ with unity. I need to prove that $P$ has a maximal submodule (This is equivalent to saying that the radical of $P$ is a proper submodule of $P$). I could prove it for free right modules based on the fact that $F J(R)=\text{rad}(F)$, where $J(R)$ is the Jacobson radical of $R$. By the way, the aforementioned identity is valid for projective right modules as well.
I tried to give a proof by contradiction as follows: Suppose $P$ has no maximal submodules. Then $rad(P)=P$. Since $P$ is projective, then $P$ is a summand of some free right module $F$. Hence, $F=P \oplus Q$ for some $Q \subseteq F$. Now, $rad(F)=\text{rad}(P) \oplus \text{rad}(Q)=P \oplus \text{rad}(Q)$. If $\text{rad}(Q)=Q$, then $\text{rad}(F)=F$, contradicting the fact that $F$ has a maximal submodule. So, we must have that $\text{rad}(Q) \subsetneqq Q$. What next?!
I appreciate any help?! Thanks in advance.