# free modules over a ring which is not a PID

Let $R$ be a unital ring. In the following modules will be left modules.

We will say that $R$ has the free submodule property if for any free module^{} $F$ over $R$ and any submodule ${F}^{\prime}\subseteq F$ we have that ${F}^{\prime}$ is also free. It is well known, that if $R$ is a PID, then $R$ has the free submodule property. One can ask whether the converse^{} is also true? We will try to answer this question.

Proposition^{}. If $R$ is a commutative ring, which is not a PID, then $R$ does not have the free submodule property.

Proof. Assume that $R$ is not a PID. Then there are two possibilities: either $R$ is not a domain or there is an ideal $I\subseteq R$ which is not principal. Assume that $R$ is not a domain and let $a,b\in R$ be two zero divisors, i.e. $a\ne 0$, $b\ne 0$ and $a\cdot b=0$. Let $(b)\subseteq R$ be an ideal generated by $b$. Then obviously $(b)$ is a submodule of $R$ (regarded as a $R$-module). Assume that $(b)$ is free. In particular there exists $m\in (b)$, $m\ne 0$ such that $r\cdot m=0$ if and only if $r=0$. But $m$ is of the form $\lambda \cdot b$ and because $R$ is commutative^{} we have

$$a\cdot m=a\cdot (\lambda \cdot b)=\lambda \cdot (a\cdot b)=0.$$ |

Contradiction^{}, because $a\ne 0$. Thus $(b)$ is not free although $(b)$ is a submodule of a free module $R$.

Assume now that there is an ideal $I\subseteq R$ which is not principal and assume that $I$ is free as a $R$-module. Since $I$ is not principal, then there exist $a,b\in I$ such that $\{a,b\}$ is linearly independent^{}. On the other hand $a,b\in R$ and $1$ is a free generator^{} of $R$. Thus $\{1,a\}$ is linearly dependent, so

$$\lambda \cdot 1+\alpha \cdot a=0$$ |

for some nonzero $\lambda ,\alpha \in R$ (note that in this case both $\lambda ,\alpha $ are nonzero, more precisely $\lambda =a$ and $\alpha =-1$). Multiply the equation by $b$. Thus we have

$$\lambda \cdot b+(\alpha \cdot b)\cdot a=0.$$ |

Note that here we used commutativity of $R$. Since $\{a,b\}$ is linearly independend (in $I$), then the last equation implies that $\lambda =0$. Contradiction. $\mathrm{\square}$

Corollary. Commutative ring is a PID if and only if it has the free submodule property.

Title | free modules over a ring which is not a PID |
---|---|

Canonical name | FreeModulesOverARingWhichIsNotAPID |

Date of creation | 2013-03-22 18:50:08 |

Last modified on | 2013-03-22 18:50:08 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 13E15 |