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I want to find all rings homomorphisms from:

i) $$\mathbb{Z} \rightarrow \mathbb{Z}_m$$

ii) $$\mathbb{Z}_m\rightarrow \mathbb{Z}$$

iii) $$\mathbb{Z}_n \rightarrow \mathbb{Z}_m$$


I don't know how to work this exercise, can someone explain to me how we think about these types of questions?

For the first one I only found $2$ the trivial and $\varphi(m)=m \pmod{n} $

for ii) I think of the trivial and $\varphi(m)=m $

and for iii) I found the trivial and $\varphi(m)=am, a\in \mathbb{Z}_m$ such that $na=0 \pmod{m}$

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    $\begingroup$ They are all cyclic groups, so if you can focus on where a generator goes, you can find everything else easily. $\endgroup$
    – Anurag A
    Jul 3, 2021 at 21:31
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    $\begingroup$ Homomorphisms of what? Homomorphisms of groups are not the same as homomorphisms of rings. $\endgroup$
    – WhatsUp
    Jul 3, 2021 at 21:35
  • $\begingroup$ @WhatsUp I think OP means group homomorphism (from the tags). $\endgroup$
    – Anurag A
    Jul 3, 2021 at 21:36
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    $\begingroup$ There are many types of homeomorphisms. Here, are you talking about homeomorphisms of rings or groups? $\endgroup$ Jul 3, 2021 at 21:49
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    $\begingroup$ This must be a duplicate thread. $\endgroup$ Jul 3, 2021 at 22:24

1 Answer 1

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Assumption: ring homomorphisms send 1 to 1.

The only possible ring homomorphism of type i) is $\varphi(n) = \overline{n}$ - the canonical projection, for $\mathbb{Z}$ is the initial object in the category of rings.

For ring homomorphisms of type ii), note that if $\varphi(\overline{1}) = k \in \mathbb{Z}$, then $n \varphi(\overline{1}) = 0$, which means $\varphi$ is the $0$ function. If you require that ring homomorphisms send 1 to 1, then there are no such ring homomorphisms.

For iii), we have that $\mathbb{Z} \to \mathbb{Z}_m$ must factor through $\mathbb{Z}_n \to \mathbb{Z}_m$ - which means that $(n) \subseteq (m)$ - i.e., $m$ divides $n$.

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