# example of injective module

In the category^{} of unitary $\mathbb{Z}$-modules (which is the category of Abelian groups), every divisible Group is injective^{}, i.e. every Group $G$ such that for any $g\in G$ and $n\in \mathbb{N}$, there is a $h\in G$ such that $nh=g$. For example, $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are divisible, and therefore injective.

###### Proof.

We have to show that, if $G$ is a divisible Group, $\phi :U\to G$ is any homomorphism^{}, and $U$ is a subgroup^{} of a Group $H$, there is a homomorphism $\psi :H\to G$ such that the restriction^{} ${\psi |}_{U}=\phi $. In other words, we want to extend $\phi $ to a homomorphism $H\to G$.

Let $\mathcal{D}$ be the set of pairs $(K,\psi )$ such that $K$ is a subgroup of $G$ containing $U$ and $\psi :K\to G$ is a homomorphism with ${\psi |}_{U}=\phi $. Then $\mathcal{D}$ ist non-empty since it contains $(U,\phi )$, and it is partially ordered by

$$(K,\psi )\le ({K}^{\prime},{\psi}^{\prime}):\u27faK\subseteq {K}^{\prime}\text{and}{\psi}^{\prime}{|}_{K}=\psi .$$ |

For any ascending chain

$$({K}_{1},{\psi}_{1})\le ({K}_{2},{\psi}_{2})\le \mathrm{\dots},$$ |

in $\mathcal{D}$, the pair $({\bigcup}_{i\in \mathbb{N}}{K}_{i},{\bigcup}_{i\in \mathbb{N}}{\psi}_{i})$ is in $\mathcal{D}$, and it is an upper bound for this chain. Therefore, by Zorn’s Lemma, $\mathcal{D}$ contains a maximal element $(M,\chi )$.

It remains to show that $M=H$. Suppose the opposite, and let $h\in H\setminus M$. Let $\u27e8h\u27e9$ denote the subgroup of $H$ generated by $h$. If $\u27e8h\u27e9\cap M=\{0\}$, the sum $M+\u27e8h\u27e9$ is in fact a direct sum^{}, and we can extend $\chi $ to $M+\u27e8h\u27e9$ by choosing an arbitrary image of $h$ in $G$ and extending linearly. This contradicts the maximality of $(M,\chi )$.

Let us therefore suppose $\u27e8h\u27e9\cap M$ contains an element $nh$, with $n\in \mathbb{N}$ minimal^{}. Since $nh\in M$, and $\chi $ is defined on $M$, $\chi (nh)$ exists, and furthermore, since $G$ is divisible, there is a $g\in G$ such that $ng=\chi (nh)$. It is now easy to see that we can extend $\chi $ to $M+\u27e8h\u27e9$ by defining $\chi (h):=g$, in contradiction^{} to the maximality of $(M,\chi )$.

Therefore, $M=H$. This proves the statement. ∎

Title | example of injective module |
---|---|

Canonical name | ExampleOfInjectiveModule |

Date of creation | 2013-03-22 17:43:40 |

Last modified on | 2013-03-22 17:43:40 |

Owner | Glotzfrosch (19314) |

Last modified by | Glotzfrosch (19314) |

Numerical id | 5 |

Author | Glotzfrosch (19314) |

Entry type | Example |

Classification | msc 16D50 |