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How to find all ring homomorphisms from $\mathbb Z_{12} \to \mathbb Z_{30}$?

I know that it is enough to determine $f([1]_{12})$ ; moreover $f([1]_{12})$ should be an idempotent element of $\mathbb Z_{30}$ and $\text{ord}(f([1]_{12})\mid 12$ and $30$, so we must find the idempotents in $\mathbb Z_{30}$ whose additive orders divide $\gcd(12,30)=6$. But this still leaves a lot to check. Is there any further restrictions on homomorphisms of such type?

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    $\begingroup$ If you really mean ring homomorphisms, I believe one part of the definition is that $f(1)=1$, so in that case there will only be one such homomorphism. If instead what you really mean is a group homomorphism only, then $\mathbb Z_{30}$ has unique subgroups of order $6,3,2$ and $1$ respectively. These have $\varphi(6)=2,\varphi(3)=2,\varphi(2)=1$ and $\varphi(1)=1$ generators respectively, and $f$ should map $[1]_{12}$ to a generator. That provides you with six such homomorphisms. $\endgroup$ – String Mar 6 '15 at 9:00
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    $\begingroup$ @String I think there is no unitary ringhomomorphisms here at all. In $\mathbb Z_{12}$ we have $12=0$ so it should be mapped to $0$. However, in $\mathbb Z_{30}$ we do not have $12=0$. $\endgroup$ – drhab Mar 6 '15 at 9:23
  • $\begingroup$ @drhab: Still homomorphisms between cyclic groups map generators to generators and are determined entirely by the image of the standard generator $1$ in the pre-image. $\endgroup$ – String Mar 6 '15 at 10:02
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    $\begingroup$ @String Yes, I agree with that. But not with "so in that case there will be only one such homomorphism." I should say: at most one. $\endgroup$ – drhab Mar 6 '15 at 10:06
  • $\begingroup$ @drhab: Ah, I see. You are absolutely correct! $\endgroup$ – String Mar 6 '15 at 10:07
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I don't know why the OP has thought that there is "a lot to check". There are only six possibilities: $f(1)\in\{0,5,10,15,20,25\}$ and among these the idempotents are $\{0,10,15,25\}$, so there are four ring homomorphisms.

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