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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.

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1 Answer 1

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I think the easiest solution is to use this lemma:

  1. If $M$ is projective then rad$(R)M$ is the intersection of the maximal submodules of $M$.

And this :

  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. $P=0.$

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