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I am trying to find a method to find ring morphism between two rings. I know the definition and all but I am unsure how to go about finding the ring morphisms between $\mathbb{Z}$ and $\mathbb{Z}_6$

If anyone could help me figure that out it would be great.

Thanks in advance

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1 Answer 1

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The general problem is too abstract to deal with. The specific one, homomorphisms between $\mathbb{Z}$ and $\mathbb{Z}_6$, is much simpler:

Because the additive group $\mathbb{Z}$ is generated by $1$, any ring homomorphism with domain $\mathbb{Z}$ is completely determined by the value of $f(1)$. So we just need to find out the possible values of $f(1)$ that will give a ring homomorphism $f\colon\mathbb{Z}\to\mathbb{Z}_6$.

From the additive group point of view, any value for $f(1)$ will give you an additive homomorphism (because $\mathbb{Z}$ is the "free group" on one generator).

Now, if $f(1) = a\in\mathbb{Z}_6$, then you also need $a(rs) = f(rs) = f(r)f(s) = (ar)(as) = a^2rs$. So that means that for every $r,s\in\mathbb{Z}$, we need $a^2rs \equiv ars\pmod{6}$. Taking $r=s=1$, we conclude that we need $a^2\equiv a\pmod{6}$. How many elements of $\mathbb{Z}_6$ have the property that $a^2=a$? (These are called idempotents). Only those can be the value of $f(1)$. And if $a^2\equiv a\pmod{6}$, then $f(1)=a$ will give you a homomorphism (verify that it is multiplicative).

So the problem of determining all ring homomorphisms $\mathbb{Z}\to\mathbb{Z}_6$ (that do not need to send $1$ to $1$) is equivalent to the problem of determining all idempotents of $\mathbb{Z}_6$. Can you do the latter?

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