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 $m_1,\mathrm{\dots },m_n\in M$ and any $r_1,\mathrm{\cdots },r_n\in R$ the equality

$$r_1\cdot m_1+\mathrm{\dots }+r_n\cdot m_n=0$$

implies that $r_1=\mathrm{\dots }=r_n=0$. 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 $m\ne 0$. Then we have homomorphism of $R$-modules $f:R\to M$ such that $f(r)=r\cdot m$. Note that $\ker (f)\ne R$ (because $f(1)\ne 0$) and thus $\ker (f)=0$ (because $\ker (f)$ is a left ideal). It is clear that this implies that $\{m\}$ is linearly independent subset of $M$. Now let

$$\mathrm{\Lambda }=\{P\subseteq M|P\text{ is linearly independent}\}.$$

Therefore we proved that $\mathrm{\Lambda }\ne \mathrm{\varnothing }$. Note that $(\mathrm{\Lambda },\subseteq )$ is a poset (where ,,$\subseteq$” denotes the inclusion) in which every chain is bounded. Thus we may apply Zorn’s lemma. Let $P_0\in \mathrm{\Lambda }$ be a maximal element in $\mathrm{\Lambda }$. We will show that $P_0$ is a basis (i.e. $P_0$ generates $M$). Assume that $m\in M$ is such that $m\notin P_0$. Then $P_0\cup \{m\}$ is linearly dependent (because $P_0$ is maximal) and thus there exist $m_1,\mathrm{\cdots },m_n\in M$ and $\lambda ,{\lambda }_1,\mathrm{\cdots },{\lambda }_n\in R$ such that $\lambda \ne 0$ and

$$\lambda \cdot m+{\lambda }_1\cdot m_1+\mathrm{\cdots }{\lambda }_n\cdot m_n=0.$$

Since $\lambda \ne 0$, then $\lambda$ is invertible in $R$ (because $R$ is a divison ring) and therefore

$$m=(-{\lambda }^{-1}{\lambda }_1)\cdot m_1+\mathrm{\cdots }+(-{\lambda }^{-1}{\lambda }_n)\cdot m_n.$$

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. $\mathrm{\square }$

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
Canonical name	RingsWhoseEveryModuleIsFree
Date of creation	2013-03-22 18:50:20
Last modified on	2013-03-22 18:50:20
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	10
Author	joking (16130)
Entry type	Example
Classification	msc 16D40