# 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 $\mathrm{ker}(f)\ne R$ (because $f(1)\ne 0$) 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

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

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 |