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 $𝒟$ 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 $𝒟$ ist non-empty since it contains $(U,\phi )$, and it is partially ordered by

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

For any ascending chain

$$(K_1,{\psi }_1)\le (K_2,{\psi }_2)\le \mathrm{\dots },$$

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

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

Let us therefore suppose $⟨h⟩\cap M$ 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 $M+⟨h⟩$ by defining $\chi (h):=g$, in contradiction to the maximality of $(M,\chi )$.

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