rings whose every module is free

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?

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

 $R\simeq I\oplus I^{\prime},$

but rank of $R$ is $1$ and rank of $I\oplus I^{\prime}$ 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.

Title rings whose every module is free RingsWhoseEveryModuleIsFree 2013-03-22 18:50:20 2013-03-22 18:50:20 joking (16130) joking (16130) 10 joking (16130) Example msc 16D40