Recall that if $R$ is a (nontrivial) ring and $M$ is a $R$ -module, then (nonempty) subset $S\subseteq M$ is called linearly independent if for any $m_1,\ldots,m_n\in M$ and any $r_1,\cdots, r_n\in R$ the equality $$r_1\cdot m_1+\ldots +r_n\cdot m_n=0$$ implies that $r_1=\ldots =r_n=0$ . If $S\subseteq M$ is a linearly independent subset of generators of $M$ , then $S$ is called a basis of $M$ . Of course not every module has a basis (it even doesn't have to have linearly independent subsets). $R$ -module is called free, if it has basis. In particular if $R$ is a field, then it is well known that every $R$ -module is free. What about the converse?

Proposition. Let $R$ be a unital ring. Then $R$ is a division ring if and only if every left $R$ -module is free.

Proof. ,,$\Rightarrow$ '' First assume that $R$ is a divison ring. Then obviously $R$ has only two (left) ideals, namely $0$ and $R$ (because every nontrivial ideal contains invertible element and thus it contains $1$ , so it contains every element of $R$ ). Let $M$ be a $R$ -module and $m\in M$ such that $m\neq 0$ . Then we have homomorphism of $R$ -modules $f:R\to M$ such that $f(r)=r\cdot m$ . Note that $\mathrm{ker}(f)\neq R$ (because $f(1)\neq 0$ ) and thus $\mathrm{ker}(f)=0$ (because $\mathrm{ker}(f)$ is a left ideal). It is clear that this implies that $\{m\}$ is linearly independent subset of $M$ . Now let $$\Lambda=\{P\subseteq M\ \big| \ P\mbox{ is linearly independent}\}.$$ Therefore we proved that $\Lambda\neq\emptyset$ . Note that $(\Lambda,\subseteq)$ is a poset (where ,,$\subseteq$ '' denotes the inclusion) in which every chain is bounded. Thus we may apply Zorn's lemma. Let $P_0\in\Lambda$ be a maximal element in $\Lambda$ . We will show that $P_0$ is a basis (i.e. $P_0$ generates $M$ ). Assume that $m\in M$ is such that $m\not\in P_0$ . Then $P_0\cup\{m\}$ is linearly dependent (because $P_0$ is maximal) and thus there exist $m_1,\cdots,m_n\in M$ and $\lambda,\lambda_1,\cdots,\lambda_n\in R$ such that $\lambda\neq 0$ and $$ \lambda\cdot m + \lambda_1 \cdot m_1+\cdots \lambda_n \cdot m_n = 0.$$ Since $\lambda\neq 0$ , then $\lambda$ is invertible in $R$ (because $R$ is a divison ring) and therefore $$m=(-\lambda^{-1}\lambda_1)\cdot m_1+\cdots +(-\lambda^{-1}\lambda_n)\cdot m_n.$$ Thus $P_0$ generates $M$ , so every $R$ -module is free. This completes this implication.

,,$\Leftarrow$ '' Assume now that every left $R$ -module is free. In particular every left $R$ -module is projective, thus $R$ is semisimple and therefore $R$ is Noetherian. This implies that $R$ has invariant basis number. Let $I\subseteq R$ be a nontrivial left ideal. Thus $I$ is a $R$ -module, so it is free and since all modules are projective (because they are free), then $I$ is direct summand of $R$ . If $I$ is proper, then we have a decomposition of a $R$ -module $$R\simeq I\oplus I',$$ but rank of $R$ is $1$ and rank of $I\oplus I'$ is at least $2$ . Contradiction, because $R$ has invariant basis number. Thus the only left ideals in $R$ are $0$ and $R$ . Now let $x\in R$ . Then $Rx=R$ , so there exists $\beta\in R$ such that $$\beta x=1.$$ Thus every element is left invertible. But then every element is invertible. Indeed, if $\beta x=1$ then there exist $\alpha\in R$ such that $\alpha\beta =1$ and thus $$1=\alpha\beta=\alpha(\beta x)\beta=(\alpha\beta)x\beta=x\beta,$$ so $x$ is right invertible. Thus $R$ is a divison ring. $\square$

Remark. Note that this proof can be dualized to the case of right modules and thus we obtained that a unital ring $R$ is a divison ring if and only if every right $R$ -module is free.