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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.

Hussein Eid
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    Recall the category of $\mathbb{Z}$-modules is equivalent (isomorphic, even) to the category of abelian groups. So then, is there an epimorphism $\mathbb{Z} \to \mathbb{Q}$ of abelian groups? Do you see why the answer must be "no"? If not, what are some of your ideas? Once we have a better idea of exactly where you're struggling, we can help you better ^_^ – HallaSurvivor Apr 04 '23 at 23:42
  • I think the answer is NO. But, unfortunately, I don't know why!. – Hussein Eid Apr 04 '23 at 23:46
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    In the category of abelian groups, epimorphisms are always surjective. – Arturo Magidin Apr 05 '23 at 00:01
  • Indeed, any epimorphism is surjective!!. By an epimorphism $A \to B$ of $R$-modules, we mean that $A\to B$ is an $R$-homomorphism that is surjective. – Hussein Eid Apr 05 '23 at 00:02
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    Can you be specific what your notation means? – Thomas Andrews Apr 05 '23 at 00:08
  • I edited the question. – Hussein Eid Apr 05 '23 at 00:16

3 Answers3

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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.

mr_e_man
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  • Alternatively, $0=f(2n-1)=(2n-1)f(1)$ implies $f(1)=0$ and thus $f(x)=xf(1)=0$ for all $x\in\mathbb Z$. – mr_e_man Apr 10 '23 at 03:50
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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.

Muhammad rahmani
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    It's not clear to me why $f(1_\mathbb{Z})$ must be $1_\mathbb{Q}$. Isn't the map which sends $1$ to $2$ a $\mathbb{Z}$-module homomorphism (even if it isn't a ring homomorphism)? (This doesn't affect the overall idea of your proof.) – Jason DeVito - on hiatus Apr 05 '23 at 01:37
  • @JasonDevito Thanks for this observation, see above. – Muhammad rahmani Apr 05 '23 at 02:40
  • There is no reason your $F$ needs to be well defined. For example, if $f(1) = 2$ as in my previous comment, then $F(1) = f(1) = 2$ by the first rule, but $F(1) = F(1\cdot 1) = f(1)f(1) = 2\cdot 2 = 4$ by the second rule. – Jason DeVito - on hiatus Apr 05 '23 at 02:42
  • @ $f$ is just a group homomorphism, for this reason we must extend it to a ring homomorphism to obtain $f(1_\mathbb{Z})=1_\mathbb{Q}$. But you assumed $f(1)=2$ and $f$ is abstract . – Muhammad rahmani Apr 05 '23 at 02:50
  • Your ring homomorphism extension gives something which is not even a function, much less a ring homomorphism. Allow me to be more direct: As it is written, this answer is incorrect. It is simply not true that $f(1_{\mathbb{Z}})$ must be equal to $1_\mathbb{Q}$. An explicit counterexample is given in my first comment. My second comment indicates that it is, in general, impossible to extend a function from a group homomorphism to a ring homomorphism. That said, I do think this answer is salvagebable, and would encourage you to work on it, when you have time. Here's one idea: – Jason DeVito - on hiatus Apr 05 '23 at 14:14
  • What your argument does establish is that $f(\mathbb{Z})$ must be cyclic. Thus, if $f$ is surjective, $\mathbb{Q} = f(\mathbb{Z})$ must be cyclic. But can you show directly that $\mathbb{Q}$ isn't cyclic? – Jason DeVito - on hiatus Apr 05 '23 at 14:15
  • @JasonDevito thanks for clarificaion, I edited my Answer. – Muhammad rahmani Apr 05 '23 at 17:22
  • @JasonDeVito Yes I know that $(\mathbb{Q},+)$ is not cyclic – Muhammad rahmani Apr 05 '23 at 17:23
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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.

Hussein Eid
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