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If $R$ is a ring, show that there is exactly one ring homomorphism $\phi: \mathbb{Z} \to R$.

I can't grasp the idea that there can be only one ring homomorphism. Aren't there many (at least more than 1) ring homomorphism between any two rings? Thank you!

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take $R=\mathbb{Z}$ and let $\phi_1(n)=0$ and $\phi_2(n)=n$. –  Jonathan Apr 14 '13 at 17:10

3 Answers 3

Hint:

Look at $\,\phi(1)\,$ . What is its role in Im$(\phi)\,$ ? What if there was other homomorphism $\,\psi:\Bbb Z\to R\,$ ? ...

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Hint $\rm\ \ \phi(1) = r\:\Rightarrow\: \phi(k) = \phi(1+\cdots +1) = \phi(1)+\cdots+\phi(1) = r+\cdots+ r = kr$

Remark $\ $ Generally homomorphisms are completely determined by their action on generators. Thus if you are working with rings with $\,1,\,$ where, by definition, ring hom's must map $1$ to $1$, then, as we saw above, $\,\phi\,$ is completely determined by that constraint. Then $\rm\ Im\: \phi \cong \Bbb Z/ker\:\phi = \Bbb Z/n\:$ is called the characteristic subring of $\rm\:R,\:$ and $\rm\:n\:$ is called the characteristic of $\rm\:R.\:$ Therefore the integer $\rm\:n\:$ characterizes the image of $\,\Bbb Z\,$ in $\rm\:R,\:$ i.e. the subring of $\rm\:R\:$ generated by $\rm\:1\in R.$

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It's normal to require that ring homomorphisms map the one-element of the domain to the one-element of the codomain. In this situation it means that $1 \in \mathbb Z$ is forced to be mapped to $1_R \in R$.

Because the additive group $(\mathbb Z,+)$ is generated by $1$ and the map $\phi$ in particular is a group homomorphism from $(\mathbb Z,+)$ to $(R,+)$, this forces $\phi$ to be unique.

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Thanks. I'll just delete that whole business about rngs then. I don't know what I was thinking there. –  kahen Apr 14 '13 at 18:05
    
Don't worry, it happens to everyone. Deleting my comments. –  Andreas Caranti Apr 14 '13 at 18:50

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