There are no maximal $\mathbb{Z}$-submodules in $\mathbb{Q}$

Question

Is it true that $\mathbb{Q}$ viewed as $\mathbb{Z}$-module (i.e. abelian group) has no maximal $\mathbb{Z}$-submodules? Why ?

Answer 1

If you define action of $\mathbb Z$ on $\mathbb Q$ by $q\to nq$ for $q\in \mathbb Q$ and $n \in \mathbb Z $ it is trivial to check that $\mathbb Q$ is an $\mathbb Z$ module.

Now, Let $M$ be maximal submodule of $\mathbb Q$ then $\mathbb Q/M$ is an simple $\mathbb Z$ module. You can also say that it is cyclic module since by simplicity the module generated by elemet should be whole module.

Now, any cyclic $\mathbb Z$ submodule isomorphic to $\mathbb Z/(\langle a \rangle)$ and by simplicity we must have $a$ is prime so;

$$\mathbb Q/M\cong Z_p$$

Thus, we can have homomorphism from $\mathbb Q$ to $Z_p$ with kernel $M$ which is a contradiction.(why? you can simply think the homomorphism between two field $\mathbb Q$ and $Z_p$, it must be the trivial one.)