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Show that given a ring $R$ with identity such that every $R$-module is free, then $R$ has no nonzero maximal ideals.

I only know that every ideal of $R$ is a direct summand of $R$. Is it possible to show the proposition without using the fact that if every $R$-module is free, then $R$ is a division ring?

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Let $R$ be a ring such that every left $R$-module is free, and let $I \subset R$ be a maximal left ideal. Then $R/I$ is a simple nonzero $R$-module, and is free by hypothesis, so $R/I$ has a basis. Take any basis element $x$, and let $\varphi \colon R \to R/I$ be the $R$-module homomorphism given by $\varphi(r) = rx$. Since $x$ is nonzero and $R/I$ is simple, $Rx = R/I$, so $\varphi$ is surjective. On the other hand, $\varphi$ must be injective, as $x$ is a basis element, so $r\cdot x \neq 0$ for any nonzero $r \in R$. Hence, $R \cong R/I$ as $R$-modules, so $R$ must also be simple; in particular, it has no nonzero proper left-ideals.

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