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Are there any "homomorphisms" from $Z$ onto finite rings other than $Z/nZ$ ? I think if instead of mapping $k$ to $k$ (mod $p$), you map it to $p - (k$ (mod $p$)$)$ and you get $f(-ab) = f(a)f(b)$. But are there any interesting ones?

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For $0,..,2$ at least the only finite ring structure with standard morphism is $Z_3$. – Enjoys Math Aug 9 '13 at 3:37
Do you mean ring homomorphisms? Assuming your ring has an identity, and you require a ring map to send identity to identity, there is a unique ring homomorphism from $\mathbf{Z}$ to any other ring. The only rings (up to isomorphism) admitting surjective ring homomorphisms from $\mathbf{Z}$ are the rings $\mathbf{Z}/n\mathbf{Z}$ for integers $n$, as follows from the fact that $\mathbf{Z}$ is a PID and the "first isomorphism theorem" for rings. – Keenan Kidwell Aug 9 '13 at 3:51
To back up what @KeenanKidwell correctly said, the integers have only one map into any ring (they are an initial object in the category of rings) depending the characteristic of the ring. Since the image really is just the ring generated by $1$, you clearly only have the choices $\mathbb{Z}/n\mathbb{Z}$. – Alex Youcis Aug 9 '13 at 4:20
Thanks. I guess I'll say that it has to be some other type of morphism. – Enjoys Math Aug 9 '13 at 4:35
But what other type? If the target is a ring, then it seem odd to ask about homomorphisms as something other than rings. – Tobias Kildetoft Aug 9 '13 at 7:52
up vote 1 down vote accepted

The kernel of a homomorphism needs to be an ideal. All proper ideals of $\mathbb{Z}$ are principal, i.e., in one-to-one correspondence to elements $n=0,1,2, \dots$.

If you ask about group homomorphisms, the answer is the same.

Of course, I assume implicitly for $n=0$ that you mean $\mathbb{Z} / 0\;\mathbb{Z} :=\mathbb{Z}$. I seems that you forgot about the map $\mathbb{Z} \rightarrow \mathbb{Z}, n \mapsto n$.

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There is no need to define(!) Z/0Z to be Z. It is defined as a special quotient, and is isomorphic to Z. – Martin Brandenburg Aug 9 '13 at 8:49
Coming from the $\bmod 0$-side, it seems worth mentioning. – Aug 9 '13 at 10:45

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