# Find all ring homomorphisms from $\mathbb{Z} \rightarrow \mathbb{Z}_m$

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 $$na=0 \pmod{m}$$

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

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 $$n$$.