Is there a $\mathbb{Z}$-module epimorphism $\mathbb{Z} \to \mathbb{Q}$

Question

Consider the right $\mathbb{Z}$-modules $\mathbb{Z}_\mathbb{Z}$ and $\mathbb{Q}_\mathbb{Z}$. Is there any $\mathbb{Z}$-module epimorphism $\mathbb{Z}\to \mathbb{Q}$ ?!. Indeed, if any, it's determined only by its value at $1$.

Note: By an epimorphism $f:A\to B$ of $R$-modules, I mean an $R$-homomorphism which is surjective.

Thanks in advance.

Answer 1

Here is a very easy proof by me:

Suppose there exists a $\mathbb{Z}$-module epimorphism $f:\mathbb{Z} \to \mathbb{Q}$. By the First Isomorphism Theorem, $\mathbb{Z}/\ker f \cong \mathbb{Q}$. Notice that $\ker f$ is either $0$ if $f$ is one-to-one or $\ker f = n\mathbb{Z}$ for some $n\in \mathbb{Z}^*$ (as $\ker f$ is an ideal of $\mathbb{Z}$ and $\mathbb{Z}$ is a PID). Hence, either $\mathbb{Z}/\ker f \cong \mathbb{Z}$ or $\mathbb{Z}/\ker f = \mathbb{Z}_n$. In either cases the isomorphism $\mathbb{Z}/\ker f \cong \mathbb{Q}$ is a contradiction; since $\mathbb{Q}$ is a divisible group whereas $\mathbb{Z}$ and $n\mathbb{Z}$ are not.

Answer 2

The answer is no. Otherwise, let $f\colon \mathbb{Z}\rightarrow\mathbb{Q}$ be an epimorphism since it is a surjective $\mathbb{Z}$-homomorphism then ; $f(\mathbb{Z})=\mathbb{Q}$. On the other hand $f\colon (\mathbb{Z},+)\rightarrow(\mathbb{Q,+)}$ is a group homomorphism and since $\mathbb{Z}$ is cyclic then so is $f(\mathbb{Z})=\mathbb{Q}$; but $\mathbb{Q}$ is not cyclic, contradiction.
Recall that homomorphisms of $\mathbb{Z}$-modules are homomorphisms of abelian groups.

Answer 3

Assuming $f:\mathbb Z\to\mathbb Q$ is surjective, there exists $n\in\mathbb Z$ such that $f(n)=\tfrac12f(1)$. This implies $f(n+n)=f(n)+f(n)=f(1)$, that is, $f(2n-1)=0$. And of course $2n-1\neq0$. But then $f(x)=f(x\bmod(2n-1))$, so there are at most $|2n-1|$ numbers in the image of $f$, which contradicts that $\mathbb Q$ is infinite.