# Every nonzero projective module has a maximal submodule

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.

1. If $$M$$ is projective then rad$$(R)M$$ is the intersection of the maximal submodules of $$M$$.
1. If $$P$$ is projective and $$P =$$ rad$$(R)P$$ then $$P = 0.$$
Let $$P\neq\{0\}$$ be projective and for a contradiction suppose $$P$$ has no maximal submodules then by the lemma $$rad(R)P=rad(P)=P$$ then by 2. $$P=0.$$