# 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, $\varphi:U\to G$ is any homomorphism, and $U$ is a subgroup of a Group $H$, there is a homomorphism $\psi:H\to G$ such that the restriction $\psi|_{U}=\varphi$. In other words, we want to extend $\varphi$ 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 $\psi|_{U}=\varphi$. Then $\mathcal{D}$ ist non-empty since it contains $(U,\varphi)$, and it is partially ordered by

 $(K,\psi)\leq(K^{\prime},\psi^{\prime}):\Longleftrightarrow K\subseteq K^{% \prime}\text{ and }\psi^{\prime}|_{K}=\psi.$

For any ascending chain

 $(K_{1},\psi_{1})\leq(K_{2},\psi_{2})\leq\dots,$

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 $\langle h\rangle$ denote the subgroup of $H$ generated by $h$. If $\langle h\rangle\cap M=\{0\}$, the sum $M+\langle h\rangle$ is in fact a direct sum, and we can extend $\chi$ to $M+\langle h\rangle$ by choosing an arbitrary image of $h$ in $G$ and extending linearly. This contradicts the maximality of $(M,\chi)$.

Let us therefore suppose $\langle h\rangle\cap M$ 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 $M+\langle h\rangle$ 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 ExampleOfInjectiveModule 2013-03-22 17:43:40 2013-03-22 17:43:40 Glotzfrosch (19314) Glotzfrosch (19314) 5 Glotzfrosch (19314) Example msc 16D50